1. 13,500 cal
First of all, we need to find the amount of heat needed to raise the temperature of the ice from -20°C to 0°C. This is given by
[tex]Q_1 = m C_i \Delta T[/tex]
where
m = 150 g is the mass of the ice
C_i = 0.5 cal/g·C° is the specific heat capacity of the ice
[tex]\Delta T=0 C-(-20 C)=20^{\circ}C[/tex] is the change in temperature of the ice
Substituting,
[tex]Q_1 = (150 g)(0.5 cal/gC)(20 C)=1500 cal[/tex]
Now we have to find the amount of heat needed to melt the ice, which is
[tex]Q_2 = m \lambda_f[/tex]
where
m = 150 g is the mass of the ice
[tex]\lambda_f = 80 cal/g[/tex] is the latent heat of fusion
Substituting,
[tex]Q_2 = (150 g)(80 cal/g)=12,000 cal[/tex]
So the total heat required is
[tex]Q_3 = 1500 cal + 12,000 cal = 13,500 cal[/tex]
2. 3750 cal
The additional amount of heat required to heat the water to 25°C is
[tex]Q_4 = m C_w \Delta T[/tex]
where
m = 150 g is the mass of water
C_w = 1 cal/g·C is the speficic heat capacity of water
[tex]\Delta T=25 C-0 C=25^{\circ}C[/tex] is the change in temperature
Substituting,
[tex]Q_4 = (150 g)(1 cal/gC)(25 C)=3,750 cal[/tex]
3. 9200 cal
First of all, we need to find the amount of heat needed to raise the temperature of the ice from -20°C to 0°C. As at point 1., this is given by
[tex]Q_1 = m C_i \Delta T[/tex]
where
m = 80 g is the mass of the ice
C_i = 0.5 cal/g·C° is the specific heat capacity of the ice
[tex]\Delta T=0 C-(-20 C)=20^{\circ}C[/tex] is the change in temperature of the ice
Substituting,
[tex]Q_1 = (80 g)(0.5 cal/gC)(20 C)=800 cal[/tex]
Now we have to find the amount of heat needed to melt the ice:
[tex]Q_2 = m \lambda_f[/tex]
where
m = 80 g is the mass of the ice
[tex]\lambda_f = 80 cal/g[/tex] is the latent heat of fusion
Substituting,
[tex]Q_2 = (80 g)(80 cal/g)=6,400 cal[/tex]
Finally, the amount of heat required to heat the water to 25°C is
[tex]Q_3 = m C_w \Delta T[/tex]
where
m = 80 g is the mass of water
C_w = 1 cal/g·C is the speficic heat capacity of water
[tex]\Delta T=25 C-0 C=25^{\circ}C[/tex] is the change in temperature
Substituting,
[tex]Q_3 = (80 g)(1 cal/gC)(25 C)=2,000 cal[/tex]
So the total heat required is
[tex]Q=Q_1+Q_2+Q_3=800 cal+6,400 cal+2,000 cal=9,200 cal[/tex]
4. No
Explanation:
The total heat required for this process consists of 3 different amounts of heat:
1- The heat required to bring the ice at melting temperature
2- The heat required to melt the ice, while its temperature stays constant
3- The heat required to raise the temperature of the water
However, computing how much heat is required to melt the ice and adding the amount of heat required to raise the temperature of 80 g of water by 45°C is not equivalent: in fact, the calculation of point 1) requires to use the specific heat capacity of ice, not that of water, therefore the two are not equivalent.
The plane of a loop of wire is perpendicular to a magnetic field. Rank, from greatest to least, the magnitudes of the loop's induced emf for each situation. A. The magnetic field strength increases from 0 T to 1 T in 6 s. B. The magnetic field strength increases from 1 T to 4 T in 2 s. C. The magnetic field strength remains at 4 T for 1 min. D. The magnetic field strength decreases from 4 T to 3 T in 4 s. E. The magnetic field strength decreases from 3 T to 0 T in 1 s.
Answer:
E - B - D - A - C
Explanation:
The magnitude of the emf induced in the loop of wire is given by Faraday-Newmann-Lenz
[tex]\epsilon=\frac{\Delta \Phi_B}{\Delta t}[/tex] (1)
where
[tex]\Delta \Phi_B[/tex] is the variation of magnetic flux
[tex]\Delta t[/tex] is the time interval
Rewriting the flux as product between magnetic field strength (B) and area enclosed by the coil (A):
[tex]\Phi_B = BA[/tex]
and since the area of the coil does not change, the variation of flux can be rewritten as
[tex]\Delta \Phi_B = \Delta B A[/tex]
So (1) becomes
[tex]\epsilon=\frac{\Delta B}{\Delta t}A[/tex]
Which means that the induced emf is proportional to the rate of change of the magnetic field, [tex]\frac{\Delta B}{\Delta t}[/tex]. So we just need to calculate this quantity for each scenario, and rank them from greatest to latest.
We have:
A) [tex]\frac{\Delta B}{\Delta t}=\frac{1 T - 0T}{6 s}=0.167 T/s[/tex]
B) [tex]\frac{\Delta B}{\Delta t}=\frac{4 T - 1T}{2 s}=1.500 T/s[/tex]
C) [tex]\frac{\Delta B}{\Delta t}=\frac{4 T - 4T}{60 s}=0 T/s[/tex]
D) [tex]\frac{\Delta B}{\Delta t}=\frac{3 T - 4T}{4 s}=-0.250 T/s[/tex]
E) [tex]\frac{\Delta B}{\Delta t}=\frac{0 T - 3T}{1 s}=-3.000 T/s[/tex]
So, from greatest to least magnitude, we have:
E - B - D - A - C
The plane of a loop of wire is perpendicular to a magnetic field. Rank, from greatest to least, the magnitudes of the loop's induced emf for each situation will be E - B - D - A-C.
What is magnetic field strength?The number of magnetic flux lines on a unit area passing perpendicular to the given line direction is known as induced magnetic field strength .it is denoted by B.
The magnitude of the induced emf in the loop of wire is;
[tex]\rm \epsilon = \frac{\triangle \phi }{ \triangle t} \\\\[/tex]
The megnetic flux is given by;
[tex]\phi_B= \triangle BA[/tex]
[tex]\rm \epsilon = \frac{ \triangle BA }{ \triangle t} \\\\[/tex]
The above expression shows that the induced emf is proportional to the rate of change of the magnetic field,
[tex]\rm \frac{ \triangle B }{ \triangle t} = \frac{ 1T-0T }{ 6} =0.167\ T/sec[/tex]
[tex]\rm \frac{ \triangle B }{ \triangle t} = \frac{ 4T-1T }{ 2} =1.500\ T/sec[/tex]
[tex]\rm \frac{ \triangle B }{ \triangle t} = \frac{ 4T-4T }{ 60} =0\ T/sec[/tex]
[tex]\rm \frac{ \triangle B }{ \triangle t} = \frac{ 3T-4T }{ 4} =-0.25 \ T/sec[/tex]
[tex]\rm \frac{ \triangle B }{ \triangle t} = \frac{ 0T-3T }{ 1} =-3 \ T/sec[/tex]
From the above, it is observed that Rank, from greatest to least, the magnitudes of the loop's induced emf for each situation will be E - B - D - A-C.
Hence the order will be E - B - D - A-C.
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Two identical conducting spheres carry charges of +5.0 μC and –1.0 μC, respectively. The centers of the spheres are initially separated by a distance L. The two spheres are brought together so that they are in contact. The spheres are then returned to their original separation L. What is the ratio of the magnitude of the electric force on either sphere after the spheres are touched to that before they were touched? A) 1/1 B) 4/5 C) 9/5 D) 5/1 E) 4/9
Answer:
B) 4/5
Explanation:
The magnitude of the electric force between the two spheres is given by
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where
k is the Coulombs' constant
q1 and q2 are the charges on the two spheres
r is the distance between the two spheres
Initially, we have
[tex]q_1 = 5.0\mu C=5.0\cdot 10^{-6}C\\q_2 = 1.0 \mu C=1.0 \cdot 10^{-6}C\\r = L[/tex]
So the force is
[tex]F_1=k\frac{(5.0\cdot 10^{-6}C)(1.0\cdot 10^{-6}C)}{L^2}=(5.0 \cdot 10^{-12} C^2)\frac{k}{L^2}[/tex]
Later, the two spheres are brought together so they are in contact: this means that the total charge will redistribute equally on the two spheres (because they are identical).
The total charge is
[tex]Q=q_1 + q_2 = +4.0 \mu C=4.0\cdot 10^{-6}C[/tex]
So each sphere will have a charge of
[tex]q=\frac{Q}{2}=2.0\cdot 10^{-6} C[/tex]
So, the new force will be
[tex]F_2=k\frac{(2.0\cdot 10^{-6}C)(120\cdot 10^{-6}C)}{L^2}=(4.0 \cdot 10^{-12} C^2)\frac{k}{L^2}[/tex]
And so the ratio of the two forces is
[tex]\frac{F_2}{F_1}=\frac{4.0\cdot 10^{-12} C}{5.0\cdot 10^{-12} C}=\frac{4}{5}[/tex]
After two identical conducting spheres carrying different charges come into contact and separate, their total charge is distributed evenly. Using Coulomb's Law, the ratio of the forces after and before contact is 4/5. The answer is B) 4/5.
Explanation:When two identical conducting spheres come into contact, their total charge is redistributed evenly between them. In this case, we have one sphere with a charge of +5.0 μC and another with –1.0 μC, totaling +4.0 μC for both spheres. After contact and redistribution, each sphere will have half of the total charge, which is +2.0 μC per sphere.
To find the ratio of the magnitudes of the electric forces before and after contact, we use Coulomb's Law, which states that the electric force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between their centers:F = k * |q1 * q2| / L2
Where F is the force, k is Coulomb's constant, q1 and q2 are the charges, and L is the separation distance. Before contact, the force is: Fbefore = k * |(+5.0 μC) * (–1.0 μC)| / L2 = k * 5.0 μC / L2
After contact, the force is: Fafter = k * |(+2.0 μC) * (+2.0 μC)| / L2 = k * 4.0 μC / L2
The ratio of Fafter to Fbefore is therefore (4.0/5.0 μC), which simplifies to 4/5. The answer to the question is option B) 4/5.
It is known that birds can detect the earth's magnetic field, but the mechanism of how they do this is not known. It has been suggested that perhaps they detect a motional EMF as they fly north to south, but it turns out that the induced voltages are small compared to the voltages normally encountered in cells, so this is probably not the mechanism involved. To check this out, calculate the induced voltage for a wild goose with a wingspan of 1.2 m flying due south at 13 m/s at a point where the earth's magnetic field is 5 x 10-5 T directed downward from horizontal by 40-degrees. The expected voltage would be about:A) .50 mV B) .25 mV C) .60 mV D) .060 mV E) .78 mV
Answer:
A) 0.50 mV
Explanation:
In this problem, we can think the wings of the bird as a metal rod moving across a magnetic field. So, and emf will be induced into the wings of the bird, according to the formula:
[tex]\epsilon = BvL sin \theta[/tex]
where
[tex]B=5\cdot 10^{-5} T[/tex] is the strength of the magnetic field
v = 13 m/s is the speed of the bird
L = 1.2 m is the wingspan of the bird
[tex]\theta=40^{\circ}[/tex] is the angle between the direction of motion and the direction of the magnetic field
Substituting numbers into the formula, we find
[tex]\epsilon = (5.0\cdot 10^{-5} T)(13 m/s)(1.2 m) sin 40^{\circ}=0.00050 V = 0.50 mV[/tex]
An aquarium 4 m long, 1 m wide, and 1 m deep is full of water. Find the work needed to pump half of the water out of the aquarium. (Use 9.8 m/s2 for g and the fact that the density of water is 1000 kg/m3.)
Explanation:
Let's begin by explaining that the Work [tex]W[/tex] done by a Force [tex]F[/tex] refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a distance [tex]y[/tex]. When the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:
[tex]W=F.y[/tex] (1)
In the case of this rectangular aquarium (see figure attached), we know its initial volume [tex]V_{i}[/tex] (which is the same volume occupied by water, because the tank is full) is:
[tex]V_{i}=4m.1m.1m=4m^{3}[/tex] (2)
In addition, we know the force needed to pump a small amount of water is:
[tex]F=m_{water}.g[/tex] (3)
On the other hand, we know the density of the water [tex]\rho_{water}[/tex] is given by the following equation:
[tex]\rho_{water}=\frac{m_{water}}{V}[/tex] (4)
Where [tex]m_{water}[/tex] is the mass of water and [tex]V=4m.1m.dy[/tex] is the volume of a thin "sheet" of water.
Finding [tex]m_{water}[/tex]:
[tex]m_{water}=\rho_{water}.V[/tex] (5)
Substituting (5) in (3):
[tex]F=\rho_{water}.V.g[/tex] (6)
And substituting (6) in (1):
[tex]W=\rho_{water}.V.g.y[/tex] (7)
Now, we are asked to find the work needed to pump half of the water out of the aquarium. So, if the aquarium is [tex]1m[/tex] deep, the half is [tex]0.5m[/tex]:
[tex]W=(1000\frac{kg}{m^{3}})(4m.1m.dy)(9.8\frac{m}{s^{2}})y[/tex] (8)
[tex]W=39200\frac{kg.m^{6}}{s^{2}}ydy[/tex] (9)
Well, in order to solve this, we have to write the definite integral from y=0 mto y=0.5m:
[tex]W=39200\frac{kg.m^{6}}{s^{2}}\int\limits^{0.5}_0 {y} \, dy[/tex] (10)
Knowing [tex]\int\limits^b_a {f(y)} \, dy= F(b)-F(a)[/tex]:
[tex]W=39200\frac{kg.m^{6}}{s^{2}}(\frac{(0.5m)^{2}}{2}-\frac{(0m)^{2}}{2})[/tex] (11)
[tex]W=4900kg\frac{m^{2}}{s^{2}}[/tex] (12)
[tex]W=4900J[/tex] >>>>This is the work needed to pump half of the water out of the aquarium
Work is the product of force and displacement. The amount of work needed to pump half of the water out of the aquarium is 4900 J.
What is work done?Work done can be defined as the amount of force needed to displace an object from one location to another.
[tex]W = F \cdot ds[/tex]
Given to us
Volume of the aquarium, V = 4 x 1 x s = 4m³
Height of the aquarium, s = 1 m
Acceleration due to gravity, g = 9.8 m/s²
Density of water, ρ = 1000 kg/m³
We know about work done, it is given as,
[tex]W = F \cdot ds[/tex]
We also know that force can be written as,
[tex]F = m \cdot a[/tex]
Also, the mass can be written as,
[tex]m = \rho \times V[/tex]
Therefore, work can be written as,
[tex]W = F\cdot ds\\\\W = m \cdot a \cdot ds\\\\W = (\rho \times V \times a )ds\\\\W = \int (\rho \times V \times a )ds[/tex]
As we need to pump half of the water, therefore, from below the tank to half the distance
[tex]W = \int_0^{\frac{1}{2}} (\rho \times V \times a )ds\\\\W = \int_0^{\frac{1}{2}} (\rho \times (4 \times 1 \times s) \times a )ds\\\\W = (\rho \times 4 \times a )[s^2]_0^{\frac{1}{2}\\\\[/tex]
Substitute all the values,
[tex]W = 1000\times 4 \times g \times[(\dfrac{0.5}{2}^2) -0^2]\\\\W = 1000\times 4 \times 9.8 \times 0.125\\\\W = 4900\ J[/tex]
Hence, the amount of work needed to pump half of the water out of the aquarium is 4900 J.
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A plant box falls from the windowsill 25.0 m above the sidewalk and hits the cement 3.0 s later. What is the box's velocity when it hits the ground?
24 m/s
22 m/s
120 m/s
Answer:
22m/s
Explanation:
To find the velocity we employ the equation of free fall: v²=u²+2gh
where u is initial velocity, g is acceleration due to gravity h is the height, v is the velocity the moment it hits the ground, taking the direction towards gravity as positive.
Substituting for the values in the question we get:
v²=2×9.8m/s²×25m
v²=490m²/s²
v=22.14m/s which can be approximated to 22m/s
A bucket weighing 2 kg and a rope of negligible mass is used todraw water from a well that is 40.8 meters deep. Just before thebucket starts being raised from the bottom of the well, the bucket isfilled with 6 kg of water. Unfortunately, there is a hole in the bucketand water leaks out at the rate of 0.12 kg per second. The bucketis pulled up at a constant rate of 0.8 meters per second.(a)How much water is in the bucket when the bucket reaches thetop of the well?(b)Find the work done pulling the bucket to the top of the well.(Assume that the acceleration of gravity is 9.8 meters per secondper second.)
Answer:
(a) There will be no water in the bucket by the time the bucket reaches the top of the well.
(b) The work done in this process will be approximately 1.98 × 10³ J.
Explanation:
(a)How long will it take for the bucket to reach the top of the well?
[tex]\displaystyle t = \frac{s}{v} = \rm \frac{40.8}{0.8} = 51.0\;s[/tex].
How much water will leak out during that [tex]51.0[/tex] seconds?
[tex]\rm 51.0\;s \times 0.12\;kg\cdot s^{-1} = 6.12\;kg > 6\;kg[/tex].
That's more than all the water in the bucket. In other words, all [tex]\rm 6\;kg[/tex] of water in the bucket will have leaked out by the time the bucket reaches the top of the well. There will be no water in the bucket by the time the bucket reaches the top of the well.
(b)How long will it take for all water to leak out of the bucket?
[tex]\displaystyle \rm \frac{6\; kg}{0.12\;kg \cdot s^{-1}} = 50\;s[/tex].
In other words,
For the first 50 seconds, water will leak out of the bucket. The mass of the bucket will decrease from [tex]\rm 6\;kg[/tex] to [tex]\rm2\;kg[/tex] at a constant rate.For the last 1 second, the mass of the bucket will stay constant at [tex]\rm 2\;kg[/tex].Express the mass [tex]m[/tex] of the bucket about time [tex]t[/tex] as a piecewise function:
[tex]\displaystyle m(t) = \left\{\begin{aligned} &6 - 0.12\;t,&&\;0\le t < 50\\& 2,&&\; 50\le t <51\end{aligned}[/tex].
Gravity on the bucket:
[tex]W(t) = m(t)\cdot g[/tex].
However, the bucket is moving at a constant velocity. There's no acceleration. By Newton's Second Law, the net force will be zero. Forces on the bucket are balanced. As a result, the size of the upward force shall be equal to that of gravity.
[tex]\displaystyle F(t) = W(t) = m(t)\cdot g[/tex].
The speed of the bucket [tex]v[/tex] is constant. Thus the power [tex]P[/tex] that pulls the bucket upward at time [tex]t[/tex] will be:
[tex]P(t) = F(t)\cdot v = m(t)\cdot v\cdot g[/tex].
Express work as a definite integral of power with respect to time:
[tex]\displaystyle \begin{aligned}W &= \int_{t_0}^{t_1}{P(t)\cdot dt} = \int_{t_0}^{t_1}{m(t)\cdot g \cdot v\cdot dt}\end{aligned}[/tex].
Both [tex]g[/tex] and [tex]v[/tex] here are constants. Factor them out:
[tex]\displaystyle \begin{aligned}W &= \int_{t_0}^{t_1}{m(t)\cdot g \cdot v\cdot dt} = v\cdot g\cdot \int_{t_0}^{t_1}{m(t)\cdot dt} \end{aligned}[/tex].
Integrate the piecewise function [tex]m(t)[/tex] piece-by-piece:
[tex]\displaystyle \begin{aligned}W &= v\cdot g\cdot \int_{t_0}^{t_1}{m(t)\cdot dt}\\ &=0.8\times 9.8 \left [\int_{0}^{50}(8-0.12\;t)\cdot dt + \int_{50}^{51}2\cdot dt\right]\\ &=0.8\times 9.8 \left [\left(8\;t - \frac{0.12}{2}\;t^{2}\right)\bigg|^{50}_{0} + (2\;t)\bigg|^{51}_{50}\right] \\&=0.8\times 9.8 \left [\left(8\times 50-\frac{0.12}{2}\times 50^{2}\right)+ (51\times 2 - 50\times 2)\right]\\&=\rm 1975.68\;J \end{aligned}[/tex].
Hence the work done pulling the bucket to the top of the well is approximately [tex]\rm 1.98\times 10^{3}\;J[/tex].
Which statement is true regarding DC current? A. The electrons move back and forth. B. There can only be one voltage supplied. C. A high voltage current can be transformed into a low voltage current. D. A low voltage current can be transformed into a high voltage current. E. This is what is sent from the power plants to your home.
Answer:
B
Explanation:
In a DC current, the current supplied is steady motion (a straight horizontal line in a graph) and this is what makes it had to transform. The alternating current takes a sine wave motion in a graph. This means is voltage varies from zero to peak and reverses polarity. The rate at which it achieves this is its frequency in Hertz.
Final answer:
DC current is a unidirectional flow of electric charge, unlike AC current which periodically reverses direction. The true statement about DC current is that there can only be one voltage supplied in a DC circuit (option B).
Explanation:
The question concerns the characteristics of DC current. Direct current, or DC, is the flow of electric charge in only one direction, which is characteristic of systems with a constant voltage source, such as a battery. Unlike alternating current, or AC, which periodically reverses direction, DC's flow is unidirectional. The given statements regarding DC current must be assessed for their correctness. To clarify:
A. The electrons in DC current do not move back and forth; they move in a single, constant direction.B. In a DC circuit, there can indeed be only one voltage supplied at any point in time.C. DC current cannot be easily transformed from high voltage to low voltage like AC can using transformers.D. Conversely, DC current cannot be transformed from low voltage to high voltage as easily as AC.E. Electrical power transmitted from power plants to homes is typically AC, not DC.Given this information, we can conclude that the true statement regarding DC current is B. There can only be one voltage supplied in a DC circuit.
Nerve cells in your body can be electrically stimulated; a large enough change in a membrane potential triggers a nerve impulse. Certain plants work the same way. A touch to mimosa pudica, the "sensitive plant," causes the leaflets to fold inward and droop. We can trigger this response electrically as well. In one experiment, investigators placed electrodes on the thick tissue at the base of a leaf. The electrodes were 3.5 mm apart. When the electrodes were connected to a 47 μF capacitor charged to 1.5 V, this stimulated a response from the plant.Eventually, all the charge on the capacitor was transferred to the plant. How much charge was transferred?q=What was the approximate electric field between the electrodes?E=
1. [tex]70.5\cdot 10^{-6} F[/tex]
The relationship between capacitance, charge and voltage across a capacitor is:
[tex]Q=CV[/tex]
where
Q is the charge
C is the capacitance
V is the voltage
In this problem,
[tex]C=47\mu F=47\cdot 10^{-6}F[/tex] is the capacitance
V = 1.5 V
Substituting, we find the charge on the capacitor, that is eventually transferred to the plant:
[tex]Q=(47\cdot 10^{-6}F)(1.5 V)=70.5\cdot 10^{-6} F[/tex]
2. 428.6 V/m
The electric field between the electrodes is given by
[tex]E=\frac{V}{d}[/tex]
where
V = 1.5 V is the potential difference across the electrodes
d = 3.5 mm = 0.0035 m is the distance between the electrodes
Substituting into the equation, we find
[tex]E=\frac{1.5 V}{0.0035 m}=428.6 V/m[/tex]
The charge transferred to the plant is 70.5 μC, and the electric field between the electrodes is approximately 428.57 V/m.
To determine how much charge was transferred to the plant, we can use the formula for the charge stored in a capacitor: q = C * V, where C is the capacitance and V is the voltage.
Given a capacitance C of 47 μF (which is 47 x 10⁻⁶ F) and a voltage V of 1.5 V:
q = 47 * 10⁻⁶ * 1.5
Thus, the charge q is:
q = 70.5 * 10⁻⁶ C or 70.5 μC
To find the electric field between the electrodes, use the formula: E = V / d, where V is the voltage and d is the distance between the electrodes.
Given V is 1.5 V and d is 3.5 mm (which is 3.5 x 10⁻³ m):
E = 1.5 / 3.5 * 10⁻³
Thus, the electric field E is: E = 428.57 V/m
A vertical spring (spring constant =160 N/m) is mounted on the floor. A 0.340-kg block is placed on top of the spring and pushed down to start it oscillating in simple harmonic motion. The block is not attached to the spring. (a) Obtain the frequency (in Hz) of the motion. (b) Determine the amplitude at which the block will lose contact with the spring.
(a) 3.5 Hz
The angular frequency in a spring-mass system is given by
[tex]\omega=\sqrt{\frac{k}{m}}[/tex]
where
k is the spring constant
m is the mass
Here in this problem we have
k = 160 N/m
m = 0.340 kg
So the angular frequency is
[tex]\omega=\sqrt{\frac{160 N/m}{0.340 kg}}=21.7 rad/s[/tex]
And the frequency of the motion instead is given by:
[tex]f=\frac{\omega}{2\pi}=\frac{21.7 rad/s}{2\pi}=3.5 Hz[/tex]
(b) 0.021 m
The block is oscillating up and down together with the upper end of the spring. The block will lose contact with the spring when the direction of motion of the spring changes: this occurs when the spring is at maximum displacement, so at
x = A
where A is the amplitude of the motion.
The maximum displacement is given by Hook's law:
[tex]F=kA[/tex]
where
F is the force applied initially to the spring, so it is equal to the weight of the block:
[tex]F=mg=(0.340 kg)(9.81 m/s^2)=3.34 N[/tex]
k = 160 N/m is the spring constant
Solving for A, we find
[tex]A=\frac{F}{k}=\frac{3.34 N}{160 N/m}=0.021 m[/tex]
a. The frequency (in Hz) of the motion is equal to 3.45 Hertz.
b. The amplitude at which the block will lose contact with the spring is 0.021 meters.
Given the following data:
Spring constant =160 N/mMass of block = 0.340 Kga. To find the frequency (in Hz) of the motion:
Since the spring initiates simple harmonic motion, we would determine its angular frequency.
Mathematically, angular frequency of a spring is given by the formula:
[tex]\omega = \sqrt{\frac{k}{m} }[/tex]
Where:
k is the spring constant.m is the mass of the spring.Substituting the given parameters into the formula, we have;
[tex]\omega = \sqrt{\frac{160}{0.340} }\\\\\omega = \sqrt{470.59}\\\\\omega = 21.69 \;rad/s[/tex]
Now, we can find the frequency of the motion by using the formula:
[tex]F = \frac{\omega}{2\pi} \\\\F = \frac{21.69}{2 \times 3.142} \\\\F = \frac{21.69}{6.284}[/tex]
Frequency, F = 3.45 Hz
b. To determine the amplitude at which the block will lose contact with the spring:
[tex]Force = kA\\\\mg = kA\\\\A = \frac{mg}{k}[/tex]
Where:
A is the amplitude.k is the spring constant.m is the mass of the spring.Substituting the parameters into the formula, we have;
[tex]A = \frac{0.340 \times 9.8}{160} \\\\A = \frac{3.332}{160}[/tex]
Amplitude, A = 0.021 meters
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The effective area of each plate of a parallel plate capacitor is 2.1 m2. The capacitor is filled with neoprene rubber (κ = 6.4). When a 6.0-V potential difference exists across the plates of the capacitor, the capacitor stores 4.0 µC of charge. Determine the plate separation of the capacitor.
Answer:
[tex]1.78\cdot 10^{-4} m[/tex]
Explanation:
The capacitance of the capacitor is given by:
[tex]C=\frac{Q}{V}[/tex]
where
[tex]Q=4.0\mu C=4.0\cdot 10^{-6} C[/tex] is the charge stored on the plates
V = 6.0 V is the potential difference across the capacitor
Substituting, we find
[tex]C=\frac{4.0\cdot 10^{-6} C}{6.0 V}=6.7\cdot 10^{-7} F[/tex]
The capacitance of a parallel-plate capacitor is also given by
[tex]C=k\frac{\epsilon_0 A}{d}[/tex]
where
k = 6.4 is the dielectric constant
[tex]\epsilon_0[/tex] is the vacuum permittivity
[tex]A=2.1 m^2[/tex] is the area of each plate
d is the separation between the plates
Solving for d, we find
[tex]d=\frac{k\epsilon_0 A}{C}=\frac{(6.4)(8.85\cdot 10^{-12} F/m)(2.1 m^2)}{6.7\cdot 10^{-7}F}=1.78\cdot 10^{-4} m[/tex]
An attacker at the base of a castle wall 3.62 m high throws a rock straight up with speed 8.60 m/s at a height of 1.59 m above the ground. (a) Will the rock reach the top of the wall? Yes No (b) If so, what is the rock's speed at the top? If not, what initial speed must the rock have to reach the top? m/s (c) Find the change in the speed of a rock thrown straight down from the top of the wall at an initial speed of 8.60 m/s and moving between the same two points. m/s (d) Does the change in speed of the downward-moving rock agree with the magnitude of the speed change of the rock moving upwards between the same elevations? Explain physically why or why not.
Providing a technical understanding of the results from the aspects of physics. Through the use of relevant formulae and principles, it's been deduced under the right conditions and assuming no externalities, the rock could reach the top and the change in speed for the upward and downwards thrown rocks would theoretically be the same.
Explanation:The subject of the question is related to the concepts of kinematics and energy in Physics.
(a) To determine if the rock can reach the top of the castle wall, we need to calculate the maximum possible height. By taking into account the initial speed and height of the rock, we can find out the greatest height it can reach. Using the formula for time-independent vertical motion: Δy = v₀t + 0.5gt² , where Δy is the maximum height, v₀ is the initial velocity (8.60m/s), 'g' is the acceleration due to gravity (≈9.80m/s²), and 't' is time.
(b) In order to calculate the speed of the rock at the top of the wall, we first need to determine if the rock reaches the top using the formula above. If the rock reaches the top, we can use energy considerations (similar to the 15.0 m/s rock thrown from the 20.0 m bridge) to determine the speed.
(c) When calculating the change in speed for a rock thrown straight down from the top of the wall, when the initial speed is 8.60 m/s, we can use similar calculations with respective sign conventions and constraints.
(d) The change in speed for both the upward and the downward-moving rock would theoretically be the same, assuming no other external forces (like air resistance or wind). This is primarily because the law of conservation of energy states that energy cannot be created or destroyed – only transformed.
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Two workers are sliding 470 kg crate across the floor. One worker pushes forward on the crate with a force of 450 N while the other pulls in the same direction with a force of 330 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor?
Answer:
0.169
Explanation:
There are three forces acting on the crate along the horizontal direction:
- The pushing force of the first worker, F1 = 450 N forward
- The pushing force of the second worker, F2 = 330 N forward
- The frictional force [tex]F_f[/tex] acting backward
The crate slides with constant speed, so its acceleration is zero: a = 0. This means that we can write Newton's second law as
[tex]\sum F = ma = 0\\F_1 + F_2 - F_f = 0[/tex]
The frictional force can be rewritten as
[tex]F_f = \mu mg[/tex]
where
[tex]\mu[/tex] is the coefficient of kinetic friction
m = 470 kg is the mass of the crate
g = 9.8 m/s^2 is the acceleration due to gravity
Substituting everything into the previous equation, we find:
[tex]F_1 + F_2 - \mu mg = 0\\\mu = \frac{F_1 + F_2}{mg}=\frac{450 N+330 N}{(470 kg)(9.8 m/s^2)}=0.169[/tex]
The coefficient of kinetic friction for the crate on this particular floor surface, given that the total force propelling the crate by the workers is balanced by the frictional force, is approximately 0.169.
Explanation:In this problem, two workers are sliding a 470 kg crate across the floor. One pushes with a force of 450 N, and the other pulls with a force of 330 N, with both forces applied horizontally. Given that the crate slides at a constant speed, we can infer that the total force propelling the crate (450 N + 330 N) is balanced by the frictional force.
Since the crate moves at a constant pace, the force of friction equals the total force exerted by the workers, 780 N (450 N + 330 N). The frictional force is generally given as Ff = μkN, where μk is the coefficient of kinetic friction, and N is the normal force. For a crate on a horizontal surface, the normal force equals the weight of the crate, which is mass (m) times gravity (g), or 470 kg * 9.8 m/s², or approximately 4606 N.
So we can set up the equation as follows: 780 N = μk * 4606 N. Solving for μk gives us μk = 780 N / 4606 N = 0.169. Therefore, the coefficient of kinetic friction for the crate on this particular floor surface is approximately 0.169.
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A bicyclist notes that the pedal sprocket has a radius of rp = 9.5 cm while the wheel sprocket has a radius of rw = 6.5 cm. The two sprockets are connected by a chain which rotates without slipping. The bicycle wheel has a radius R = 65 cm. When pedaling the cyclist notes that the pedal rotates at one revolution every t = 1.1 s. When pedaling, the wheel sprocket and the wheel move at the same angular speed.
The angular speed of the wheel when pedaling is approximately [tex]\(3.73 \, \text{s}^{-1}\)[/tex].
Given the information about the radii and the rotations, we can analyze the relationships between the angular speeds of the sprockets and the wheel.
The angular speed of the pedal sprocket can be found using the information that it rotates at one revolution every t = 1.1 seconds.
The formula for angular speed omega is:
[tex]\[ \omega = \frac{{2\pi \text{ (revolutions)}}}{{\text{time}}} \][/tex]
For the pedal sprocket:
[tex]\[ \omega_{\text{pedal}} = \frac{{2\pi}}{{1.1 \, \text{s}}} \][/tex]
Now, knowing the relationship between the angular speeds of the wheel sprocket and the pedal sprocket:
[tex]\[\omega_{\text{wheel sprocket}} = \omega_{\text{pedal}}\][/tex]
The linear speed of a point on the wheel (where the chain makes contact) is given by the product of the angular speed and the radius:
[tex]\[ v = \omega \times r \][/tex]
Since the chain rotates without slipping, the linear speed of the chain where it contacts the wheel is equal to the linear speed of the wheel.
For the wheel sprocket:
[tex]\[ \omega_{\text{wheel sprocket}} \times r_{\text{wheel sprocket}} = \omega_{\text{wheel}} \times r_{\text{wheel}} \][/tex]
Given:
- [tex]\( r_{\text{pedal}} = 9.5 \, \text{cm} \)[/tex]
- [tex]\( r_{\text{wheel sprocket}} = 6.5 \, \text{cm} \)[/tex]
- [tex]\( R_{\text{wheel}} = 65 \, \text{cm} \)[/tex]
We have:
[tex]\[ \omega_{\text{pedal}} = \frac{{2\pi}}{{1.1 \, \text{s}}} \][/tex]
[tex]\[ \omega_{\text{wheel sprocket}} = \omega_{\text{pedal}} = \frac{{2\pi}}{{1.1 \, \text{s}}} \][/tex]
[tex]\[ \omega_{\text{wheel sprocket}} \times r_{\text{wheel sprocket}} = \omega_{\text{wheel}} \times R_{\text{wheel}} \][/tex]
Let's solve for the angular speed of the wheel.
[tex]\[ \omega_{\text{wheel sprocket}} \times r_{\text{wheel sprocket}} = \omega_{\text{wheel}} \times R_{\text{wheel}} \][/tex]
Given:
- [tex]\( r_{\text{wheel sprocket}} = 6.5 \, \text{cm} \)[/tex]
- [tex]\( R_{\text{wheel}} = 65 \, \text{cm} \)[/tex]
We found earlier that [tex]\( \omega_{\text{wheel sprocket}} = \frac{2\pi}{1.1 \, \text{s}} \)[/tex]. Substituting the known values into the equation:
[tex]\[ \frac{2\pi}{1.1 \, \text{s}} \times 6.5 \, \text{cm} = \omega_{\text{wheel}} \times 65 \, \text{cm} \][/tex]
Now, solve for [tex]\( \omega_{\text{wheel}} \)[/tex], the angular speed of the wheel:
[tex]\[ \omega_{\text{wheel}} = \frac{\frac{2\pi}{1.1 \, \text{s}} \times 6.5 \, \text{cm}}{65 \, \text{cm}} \][/tex]
[tex]\[ \omega_{\text{wheel}} = \frac{2\pi \times 6.5}{1.1 \times 65} \, \text{s}^{-1} \][/tex]
[tex]\[ \omega_{\text{wheel}} = \frac{13 \pi}{11} \, \text{s}^{-1} \][/tex]
[tex]\[ \omega_{\text{wheel}} \approx 3.73 \, \text{s}^{-1} \][/tex]
Therefore, the angular speed of the wheel when pedaling is approximately [tex]\(3.73 \, \text{s}^{-1}\)[/tex].
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Approximately 8.35 rad/s is the angular speed of the wheel sprocket in radians per second.
We'll first find the angular speed of the pedal sprocket, then use that to find the angular speed of the wheel sprocket and the bicycle wheel.
Given:
- Radius of the pedal sprocket, [tex]\( r_p = 9.5 \, \text{cm} \)[/tex]
- Radius of the wheel sprocket, [tex]\( r_w = 6.5 \, \text{cm} \)[/tex]
- Radius of the bicycle wheel, R = 65 cm
- Time taken for one revolution of the pedal, t = 1.1 s
1. Calculate the angular speed of the pedal sprocket [tex]\( \omega_p \)[/tex]:
[tex]\[ \omega_p = \frac{2\pi}{1.1 \, \text{s}} \][/tex]
[tex]\[ \omega_p \approx 5.71 \, \text{rad/s} \][/tex]
2. Calculate the linear speed of a point on the pedal sprocket:
[tex]\[ v_p = r_p \cdot \omega_p \][/tex]
[tex]\[ v_p = 9.5 \, \text{cm} \cdot 5.71 \, \text{rad/s} \][/tex]
[tex]\[ v_p \approx 54.295 \, \text{cm/s} \][/tex]
3. Find the angular speed of the wheel sprocket and the bicycle wheel:
[tex]\[ \omega_w = \frac{v_w}{r_w} = \frac{v_p}{r_w} \][/tex]
[tex]\[ \omega_w = \frac{54.295 \, \text{cm/s}}{6.5 \, \text{cm}} \][/tex]
[tex]\[ \omega_w \approx 8.35 \, \text{rad/s} \][/tex]
The angular speed of the wheel sprocket and the bicycle wheel is approximately 8.35 rad/s.
Complete Question:
A bicyclist notes that the pedal sprocket has a radius of rp = 9.5 cm while the wheel sprocket has a radius of rw = 6.5 cm. The two sprockets are connected by a chain which rotates without slipping. The bicycle wheel has a radius R = 65 cm. When pedaling the cyclist notes that the pedal rotates at one revolution every t = 1.1 s. When pedaling, the wheel sprocket and the wheel move at the same angular speed.
Calculate the angular speed of the wheel sprocket in radians per second.
During a test, a NATO surveillance radar system, operating at 37 GHz at 182 kW of power, attempts to detect an incoming stealth aircraft at 104 km. Assume that the radar beam is emitted uniformly over a hemisphere. (a) What is the intensity of the beam when the beam reaches the aircraft's location? The aircraft reflects radar waves as though it has a cross-sectional area of only 0.22 m2. (b) What is the power of the aircraft's reflection? Assume that the beam is reflected uniformly over a hemisphere. Back at the radar site, what are (c) the intensity, (d) the maximum value of the electric field vector, and (e) the rms value of the magnetic field of the reflected radar beam?
(a) [tex]2.68\cdot 10^{-6} W/m^2[/tex]
The intensity of an electromagnetic wave is given by
[tex]I=\frac{P}{A}[/tex]
where
P is the power
A is the area of the surface considered
For the waves in the problem,
[tex]P=182 kW = 1.82\cdot 10^5 W[/tex] is the power
The area is a hemisphere of radius
[tex]r=104 km=1.04\cdot 10^5 m[/tex]
so
[tex]A=2\pi r^2=2\pi (1.04\cdot 10^5 m)^2=6.8\cdot 10^{10} m^2[/tex]
So, the intensity is
[tex]I=\frac{1.82\cdot 10^5 W}{6.8\cdot 10^{10}m^2}=2.68\cdot 10^{-6} W/m^2[/tex]
(b) [tex]5.9\cdot 10^{-7} W[/tex]
In this case, the area of the reflection is
[tex]A=0.22 m^2[/tex]
So, if we use the intensity of the wave that we found previously, we can calculate the power of the aircraft's reflection using the same formula:
[tex]P=IA=(2.68\cdot 10^{-6} W/m^2)(0.22 m^2)=5.9\cdot 10^{-7} W[/tex]
(c) [tex]8.7\cdot 10^{-18} W/m^2[/tex]
We said that the power of the waves reflected by the aircraft is
[tex]P=5.9\cdot 10^{-7} W[/tex]
If we assume that the reflected waves also propagate over a hemisphere of radius
[tex]r=104 km=1.04\cdot 10^5 m[/tex]
which has an area of
[tex]A=2\pi r^2=2\pi (1.04\cdot 10^5 m)^2=6.8\cdot 10^{10} m^2[/tex]
Then the intensity of the reflected waves at the radar site will be
[tex]I=\frac{P}{A}=\frac{5.9\cdot 10^{-7} W}{6.8\cdot 10^{10} m^2}=8.7\cdot 10^{-18} W/m^2[/tex]
(d) [tex]8.1\cdot 10^{-8} V/m[/tex]
The intensity of a wave is related to the maximum value of the electric field by
[tex]I=\frac{1}{2}c\epsilon_0 E_0^2[/tex]
where
c is the speed of light
[tex]\epsilon_0[/tex] is the vacuum permittivity
[tex]E_0[/tex] is the maximum value of the electric field vector
Solving the equation for [tex]E_0[/tex],
[tex]E_0=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(8.7\cdot 10^{-18} W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12} F/m)}}=8.1\cdot 10^{-8} V/m[/tex]
(e) [tex]1.9\cdot 10^{-16} T[/tex]
The maximum value of the magnetic field vector is given by
[tex]B_0 = \frac{E_0}{c}[/tex]
Substituting the values,
[tex]B_0 = \frac{(8.1\cdot 10^{-8} V/m)}{3\cdot 10^8 m/s}=2.7\cdot 10^{-16} T[/tex]
And the rms value of the magnetic field is given by
[tex]B_{rms} = \frac{B_0}{\sqrt{2}}=\frac{2.7\cdot 10^{-16} T}{\sqrt{2}}=1.9\cdot 10^{-16} T[/tex]
If you use an extension cord, current travels from the 120 V outlet, along one wire inside the cord, through the appliance you've plugged into the cord, and back to the outlet through a second wire in the cord. The total resistance of the two wires in a light-duty extension cord is 0.44Ω; the current through such a cord should be limited to 13 A. A 13 A shop vacuum is powered using this cord. What is the voltage drop across the vacuum?
The vacuum experiences a voltage drop of 114.28 V after accounting for the 5.72 V drop due to the resistance of the extension cord.
To calculate the voltage drop across the vacuum, we can use Ohm's law, which states that Voltage (V) = Current (I) times Resistance (R). Given that the total resistance of the wires in the extension cord is 0.44Ω and the current is limited to 13 A, we first find the voltage drop across the extension cord itself:
V = I times R = 13 A times 0.44 Ω = 5.72 V.
Since the outlet provides 120 V and the voltage drop across the cord is 5.72 V, the voltage drop across the vacuum is:
120 V - 5.72 V = 114.28 V.
This voltage drop across the vacuum is how much voltage is available for the vacuum to operate with.
It is important to note that the higher the resistance of the cord or the greater the current, the larger the voltage drop will be, leading to a decrease in the appliance's performance.
6. A large balloon containing 2.00×103 m3 of helium gas at 1.00 atm and a temperature 15.0◦ C rises rapidly from ground level to an altitude where the atmospheric pressure is only 0.900 atm. Assume that the helium behaves as an ideal gas and that the balloon rises so rapidly that no heat is exchanged with the surrounding atmosphere. a. Calculate the volume of the helium gas at the higher altitude. b. Calculate the temperature of the gas at the higher altitude. c. What is the change in the internal energy of the helium as the balloon rises to the higher altitude?
Answer:
2130 m³, 3.11 °C, 1.25×10⁷ J
Explanation:
No heat is exchanged, so this is an adiabatic process. For an ideal gas, this means:
PV^((f+2)/f) = constant,
where P is pressure, V is volume, and f is degrees of freedom.
For a monatomic gas, f=3. For a diatomic gas, f=5. Since helium is monatomic, f=3.
Therefore:
PV^(5/3) = PV^(5/3)
(1.00 atm) (2.00×10³ m³)^(5/3) = (0.900 atm) V^(5/3)
V = 2130 m³
For an ideal gas in an adiabatic process, we can also say:
VT^(f/2) = constant
Therefore:
(2.00×10³ m³) (15.0 + 273.15 K)^(3/2) = (2130 m³) T^(3/2)
T = 276.26 K
T = 3.11 °C
Finally, the change in internal energy is:
ΔU = (f/2) nRΔT
We need to find the number of moles, n, using ideal gas law:
PV = nRT
n = PV/(RT)
n = (1.00 atm) (2.00×10³ m³) / ((8.21×10⁻⁵ atm m³ / mol / K) (15.0 + 273.15 K))
n = 84,500 mol
So the change in internal energy is:
ΔU = (3/2) (84,500 mol) (8.314 J/mol/K) (15.0 - 3.11) K
ΔU = 1.25×10⁷ J
Does a swimmer underwater observe the same color or a different color for this light?
1. A swimmer underwater observes a different color because the wavelength changes.
2. It is not possible to know what a swimmer underwater will observe.
3. A swimmer underwater observes a the same color because the wavelength does not change.
4. A swimmer underwater observes the same color because the frequency does not change.
5. A swimmer underwater observes a different color because the frequency changes.
Explanation:
When a swimmer is underwater and a light pasess from air to water, there is a change in the medium and its index of refraction, hence the light is refracted. This means it changes its direction.
Nevertheless, in this process the refracted ray of light does not change its frequency [tex]f[/tex], because frequency is:
[tex]f=\frac{1}{T}[/tex]
Where [tex]T[/tex] is the period of the wave and this remains unchanged, hence the frequency, as well.
So, as the frequency does not change and the color of light depends on frequency, the color of the light remains the same.
in an eslastic ,the momentum is_______and the mechanical energy is
Answer:
In an elastic collision, the momentum is conserved and the mechanical energy is conserved too.
Explanation:
There are two types of collisions:
- Elastic collision: in an elastic collision, the total momentum before and after the collision is conserved; also, the total mechanical energy before and after the collision is conserved.
- Inelastic collision: in an inelastic collision, the total momentum before and after the colllision is conserved, while the total mechanical energy is not conserved (in fact, part of the energy is converted into other forms of energy such that thermal energy, due to the presence of frictional forces)
You charge a parallel-plate capacitor, remove it from the battery, and prevent the wires connected to the plates from touching each other. When you pull the plates apart to a larger separation, do the following quantities increase, decrease, or stay the same? C decrease Correct: Your answer is correct. Q decrease Incorrect: Your answer is incorrect. E between the plates stay the same Correct: Your answer is correct. ΔV increase Correct: Your answer is correct.
Answer:
Q: remains the same
C: decreases
ΔV: increases
Explanation:
When the capacitor is disconnected from the battery, and the wire connected to the plates are not touching anything else, it means that the charge cannot flow out from the capacitor: so, the charge stored on the plates of the capacitor, Q, will not change, regardless of the distance between the plates.
The capacitance of a parallel plate is given by
[tex]C=\frac{\epsilon_0 A}{d}[/tex]
where A is the area of the plates and d the separation between the plates. As we see from the formula, C is inversely proportional to d: so, if the plates are pulled apart to a larger separation, it means that d increases, and so C decreases.
Finally, the voltage across the capacitor is given by
[tex]\Delta V=\frac{Q}{C}[/tex]
and since we said that Q does not change while C decreases, it means that [tex]\Delta V[/tex] increases, since [tex]\Delta V[/tex] is inversely proportional to C.
The motion of a car on a position-time graph is represented with a horizontal line. What does this indicate about the car’s motion?
If the line on the graph is horizontal, the car is at the same position no matter what time you look at it. That means that the car is sitting motionless in one place, and not moving. The car's speed, velocity, and acceleration are all zero, although its resale value may be decreasing.
A team of astronauts is on a mission to land on and explore a large asteroid. In addition to collecting samples and performing experiments, one of their tasks is to demonstrate the concept of the escape speed by throwing rocks straight up at various initial speeds. With what minimum initial speed ????esc will the rocks need to be thrown in order for them never to "fall" back to the asteroid? Assume that the asteroid is approximately spherical, with an average density ????=2.02×106 g/m3 and volume ????=3.09×1012 m3 . Recall that the universal gravitational constant is ????=6.67×10−11 N·m2/kg2 .
Answer:
9.60 m/s
Explanation:
The escape speed of an object from the surface of a planet/asteroid is given by:
[tex]v=\sqrt{\frac{2GM}{R}}[/tex]
where
G is the gravitational constant
M is the mass of the planet/asteroid
R is the radius of the planet/asteroid
In this problem we have
[tex]\rho = 2.02\cdot 10^6 g/m^3[/tex] is the density of the asteroid
[tex]V=3.09\cdot 10^{12}m^3[/tex] is the volume
So the mass of the asteroid is
[tex]M=\rho V=(2.02\cdot 10^6 g/m^3)(3.09\cdot 10^{12} m^3)=6.24\cdot 10^{18} g=6.24\cdot 10^{15} kg[/tex]
The asteroid is approximately spherical, so its volume can be written as
[tex]V=\frac{4}{3}\pi R^3[/tex]
where R is the radius. Solving for R,
[tex]R=\sqrt[3]{\frac{3V}{4\pi}}=\sqrt[3]{\frac{3(3.09\cdot 10^{12} m^3)}{4\pi}}=9036 m[/tex]
Substituting M and R inside the formula of the escape speed, we find:
[tex]v=\sqrt{\frac{2(6.67\cdot 10^{-11})(6.24\cdot 10^{15})}{(9036)}}=9.60 m/s[/tex]
Monochromatic light is incident on a metal surface, and electrons are ejected. If the intensity of the light increases, what will happen to the ejection rate of the electrons? Monochromatic light is incident on a metal surface, and electrons are ejected. If the intensity of the light increases, what will happen to the ejection rate of the electrons? The ejection rate will remain constant. Changes in the ejection rate cannot be determined without additional information. The ejection rate will decrease. The ejection rate will increase.
The photoelectric effect is a phenomenon that consists of the emission of electrons by certain metals when a beam of light impacts on its surface.
For this phenomenon to occur, certain conditions must be met, such as when the photon collides with the electron, in order to "pull it" from the metal, the photon must have a minimum energy equal to the ionization energy of the atom, so that the electron can leave the influence of the nucleus.
This is achieved with the adequate intensity of the incident radiation, which is related to the number of photons that impact the metal.
This means:
The greater the intensity, the greater the number of photons, hence the greater number of electrons emitted.
Increasing the intensity of monochromatic light incident on a metal surface causes an increase in the rate of electron ejection due to the larger number of photons arriving on the surface per unit time. However, the kinetic energy of the ejected electrons doesn't change with changing light intensity.
Explanation:The phenomenon described here is called the photoelectric effect, which occurs when monochromatic light is incident on a metal surface causing the ejection of electrons, also known as photoelectrons.
When the intensity of the light increases, the rate of electron ejection also increases. This is because the increased intensity of light means a larger number of photons are arriving on the surface per unit time, thus more electrons can gain energy from the photons and be ejected.
The kinetic energy of the ejected electrons, however, does not change with changing light intensity. The energy of an ejected electron equals to the energy of a single photon (which depends only on the light's frequency, not its intensity) minus the binding energy of the electron. This means, even though there are more electrons ejected per unit time due to increased intensity, they have the same kinetic energy as before.
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An element has a half-life of 2.5 days. If it begins with 100 atoms, how many will be left after 5 days?
Answer:
25
Explanation:
Half life equation is:
A = A₀ (½)^(t / T)
where A is the final amount, A₀ is the initial amount, t is time, and T is the half life.
Given that A₀ = 100, t = 5, and T = 2.5:
A = 100 (½)^(5 / 2.5)
A = 100 (½)^2
A = 100 (¼)
A = 25
There will be 25 atoms left.
There’s 25 atoms left
Atom A consists of 10 protons, 12 neutrons, and 10 electrons. Atom B consists of 10 protons, 10 neutrons, and 12 electrons. (5 points) The atoms are isotopes of each other. The atoms are not isotopes of each other.
Answer:
The atoms are isotopes of each other
Explanation:
- Two atoms are said to be isotopes of each other if they have same atomic number Z (which corresponds to the number of protons) but different mass number A (which corresponds to the sum of protons and neutrons in the nucleus).
For atom A, we have:
Z = 10 (10 protons)
A = 10+12 = 22 (10 protons + 12 neutrons)
For atom B, we have:
Z = 10 (10 protons)
A = 10+10 = 20 (10 protons + 10 neutrons)
The two atoms have same atomic number Z but different mass number A, so they are isotopes of each other.
Atom A and Atom B are isotopes of the same element because they have an equal number of protons (10 each), but differ in the number of neutrons (12 in Atom A and 10 in Atom B).
Explanation:The atoms identified as A and B are isotopes of each other. Isotopes are variations of a chemical element, which means they have the same number of protons (and thus belong to the same element) but differ in the number of neutrons. In this case, both atoms A and B share the same number of protons (10) which identifies the element, but differ in the number of neutrons (12 in atom A and 10 in atom B).
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Suppose your vehicle is moving 12m/s as it crosses the 7m line. It took 6 seconds to get to the 7m line. What was its acceleration?
Answer:
3.61 m/s²
Explanation:
Given:
v = 12 m/s
x = 7 m
x₀ = 0 m
t = 6 s
Find:
a
x = x₀ + vt - ½ at²
7 m = 0 m + (12 m/s) (6 s) - ½ a (6 s)²
a = 3.61 m/s²
A vector points into the third quadrant, and its x and y components are both negative. If its magnitude is equal to two times the magnitude of its x component, what is the angle between the vector and the positive x axis?
The angle between the vector and the positive x axis is approximately 60°.
Explanation:The vector points into the third quadrant, which means its x and y components are both negative. Let's assume that the magnitude of the vector is represented by |A| and the magnitude of the x component is represented by |Ax|. The given condition states that |A| = 2 * |Ax|. Using this information, we can find the angle between the vector and the positive x axis.
First, we need to find the values of |A| and |Ax|. Since |A| = 2 * |Ax|, we can substitute this into the Pythagorean theorem:
|A|² = |Ax|² + |Ay|²
Now let's substitute the given information into the equation:
(2 * |Ax|)² = |Ax|² + (|Ay|)²
Expanding and simplifying the equation:
4 * |Ax|² = |Ax|² + (|Ay|)²
3 * |Ax|² = (|Ay|)²
Now, we can take the square root of both sides:
√(3 * |Ax|²) = √((|Ay|)²)
√3 * |Ax| = |Ay|
Since both |Ax| and |Ay| are negative, we can ignore the negative sign. Therefore, |Ax| = -1 and |Ay| = -√3.
Now, we have all the information we need to find the angle between the vector and the positive x axis. We can use the formula: tan(A) = |Ay|/|Ax| = (-√3)/(-1) = √3
The angle A is the inverse tangent of √3: A = atan(√3) ≈ 60°
So, the angle between the vector and the positive x axis is approximately 60°.
Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth to study the earliest stages of star formation, before a star begins to emit visible light. Suppose an infrared telescope is tuned to detect infrared radiation with a frequency of 25.9 THz . Calculate the wavelength of the infrared radiation. Round your answer to 3 significant digits.
Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth to study the earliest stages of star formation, before a star begins to emit visible light, the wavelength is 2.325 x 10¹ um.
The wavelength can be calculated by the information given:
- Frequency f = 25.9 THz = [tex]\(25.9 \times 10^{12}\)[/tex] Hz
- Speed of light c = [tex]\(3.00 \times 10^8\)[/tex] m/s
We'll use the formula [tex]\(\lambda = \frac{c}{f}\)[/tex] to calculate the wavelength ([tex]\(\lambda\)[/tex]).
Substitute the values:
[tex]\[\lambda = \dfrac{3.00 \times 10^8 \, \text{m/s}}{25.9 \times 10^{12} \, \text{Hz}}.\][/tex]
Calculate the value of [tex]\(\lambda\)[/tex]:
[tex]\[\lambda = 1.157 \times 10^{-5} \, \text{m}.\][/tex]
Now, convert the wavelength to micrometers:
[tex]\[\lambda = 1.157 \times 10^{-5} \, \text{m} \times \dfrac{10^6 \, \mu m}{1 \, \text{m}}.\][/tex]
Calculate the value in micrometers:
[tex]\[\lambda \approx 2.325 \, \mu m.\][/tex]
Thus, the wavelength is 2.325 x 10¹ um or 2.325μm.
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The wavelength of the infrared radiation emitted from young stars with a frequency of 25.9 THz can be calculated using the formula relating the speed of light, wavelength, and frequency (c = λv). By rearranging the formula to λ = c/v and substituting the values, we find that the wavelength is approximately 11.6 micrometers.
Explanation:To calculate the wavelength of the infrared radiation, we must use the formula for the relation between the speed of light, wavelength, and frequency. This formula is c = λv, where c is the speed of light, λ (lambda) is the wavelength, and v is the frequency.
The speed of light is a constant and is approximately 3.00 x 108 m/s. The frequency provided is 25.9 THz or 25.9 x 1012 Hz.
To find the wavelength, we rearrange the formula to λ = c/v. We substitute the given values to get λ = (3.00 x 108 m/s) / (25.9 x 1012 Hz). After performing the calculation, the wavelength comes out to approximately 1.16 x 10-5 meters, or 11.6 micrometers when rounded to three significant figures.
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The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.00 X 10^3 N with an effective perpendicular lever arm of 3.10 cm, producing an angular acceleration of the forearm of 125 rad/s2. What is the moment of inertia of the boxer's forearm? kg · m2
Answer:
[tex]0.496 kg m^2[/tex]
Explanation:
The torque exerted is given by
[tex]\tau = Fd[/tex]
where
[tex]F=2.00 \cdot 10^3 N[/tex] is the force applied
d = 3.10 cm = 0.031 m is the length of the lever arm
Substituting,
[tex]\tau=(2.00\cdot 10^3 N)(0.031 m)=62 Nm[/tex]
The equivalent of Newton's second law for rotational motion is:
[tex]\tau = I \alpha[/tex]
where
[tex]\tau = 62 Nm[/tex] is the net torque
I is the moment of inertia
[tex]\alpha = 125 rad/s^2[/tex] is the angular acceleration
Solving the equation for I, we find
[tex]I=\frac{\tau}{\alpha}=\frac{62 Nm}{125 rad/s^2}=0.496 kg m^2[/tex]
A stone is thrown vertically downwards from the top of a skyscraper with an initial speed of 5.10m/s. The stone hits the street below 7.60 seconds after being thrown.How tall is the skyscraper?
Answer:
322 m
Explanation:
y = y₀ + v₀ t + ½ gt²
If we take down to be positive, then:
y = 0 + (5.10) (7.60) + ½ (9.8) (7.60)²
y = 322 m
The skyscraper is 322 m tall.
A electric field propagating in air is incident onto an perfectly conducting interface has the following phasor field In terms of the amplitude Eo and the wavevector components: 1. find a phasor expression for the reflected magnetic field 2. find a phasor expression for the total magnetic field in air item find a phasor expression for the total electric field in air. 3. find a phasor expression for the current induced on the surface of the conductor. 4. find a space-time expression for the current induced on the surface of the the conductor. ind a phasor expression for the induced electric charge on the conductor 6. find a space-time expression for the induced electric charge in the conductor.
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