Answer:
24.48 days
37.7 days
Explanation:
r = Radius
s denotes satellite
C denotes Charon
Time period is given by
[tex]T=\dfrac{2\pi r^{1.5}}{\sqrt{2GM}}[/tex]
So,
[tex]T\propto r^{1.5}[/tex]
[tex]\dfrac{T_C}{T_{s1}}=\dfrac{r_c^{1.5}}{r_{s1}^{1.5}}\\\Rightarrow T_{s1}=\dfrac{T_Cr_{s1}^{1.5}}{r_C^{1.5}}\\\Rightarrow T_{s1}=\dfrac{6.39\times 86400\times {48000000}^{1.5}}{19600000^{1.5}}\\\Rightarrow T_{s1}=2115886.41242\ s\\\Rightarrow T_{s1}=24.48\ days[/tex]
The time period of the first satellite is 24.48 days
[tex]T_{s2}=\dfrac{T_Cr_{s2}^{1.5}}{r_C^{1.5}}\\\Rightarrow T_{s2}=\dfrac{6.39\times 86400\times {64000000}^{1.5}}{19600000^{1.5}}\\\Rightarrow T_{s2}=3257620.23942\ s\\\Rightarrow T_{s2}=37.7\ days[/tex]
The time period of the second satellite is 37.7 days
polybius believed that rome's successes stemmed from select one: a. its constitution and mixed government. b. conservative roman values. c. its geographical diversity. d. roman worship of greek deities. e. all these answers are correct.
Answer:
a. its constitution and mixed government.
Explanation:
Polybius tell us about the Roman mixed constitution as a fundamental in the Roman victory and the Carthaginian defeat in the Punic war, conceiving the mixed constitution as the best, since this constitution was at its peak, which implies that among the elements of the constitution, the aristocratic component was the dominant one.
Using simple rearrangement of Newton's second law, show that a net force of 77 N exerted on a 11-kg package is needed to produce an acceleration of 7.0 m/s2 .
Answer:
A net force of 11 kg is needed to produce an acceleration of 7.0 m/s²
Explanation:
Newton's Second Law: It states that the the rate of change of momentum of a body, is directly proportional to the applied force, and takes place along the direction of the the force.
From Newton's second law of motion, we can deduced that,
F = ma ......................... Equation 1.
Where F = Net force acting on the package, m = mass of the package, a = acceleration of the page.
From the question, when
F = 77 N, m = 11 kg.
a = F/m
a = 77/11
a = 7 m/s².
From the above, a net force of 11 kg is needed to produce an acceleration of 7.0 m/s²
Final answer:
To determine the net force needed to accelerate an 11-kg package at 7.0 m/s², apply Newton's second law using the formula Fnet = ma, which yields a net force of 77 N.
Explanation:
The student is asking how to use Newton's second law to calculate the net force required to produce a certain acceleration for an object of known mass. This is a physics problem that involves using the formula Fnet = ma (where Fnet is the net force, m is the mass, and a is the acceleration).
In this case, the mass m of the package is 11 kg, and the desired acceleration a is 7.0 m/s². To find the net force Fnet, we rearrange the formula as it is already in the form solving for the net force and simply substitute the known values:
Fnet = (11 kg) × (7.0 m/s²)
Now, we multiply:
Fnet = 77 N
Therefore, a net force of 77 N is indeed needed to accelerate an 11-kg package at 7.0 m/s².
Calculate the energy, in electron volts, of a photon whose frequency is the following.
(a) 585 THz
(b) 3.50 GHz
(c) 40.0 MHz
Planck's equation for the energy of a photon is E = hf, where f is the frequency and h is Planck's constant. We use 1 eV = 1.60 ✕ 10−19 J for units of energy.
(a) For the energy of the photon at a frequency of 585 THz, we have E = hf
Answer:
(a) [tex]E=2.42eV[/tex]
(b) [tex]E=1.45*10^{-5}eV[/tex]
(c) [tex]E=1.66*10^{-7}eV[/tex]
Explanation:
The Planck-Einstein relation allows us to know the energy (E) of a photon, knowing its frequency (f). According to this relation, the energy of the photon is defined as:
[tex]E=hf[/tex]
Here h is the Planck constant.
(a)
[tex]E=(4.14*10^{-15}eV\cdot s)(585*10^{12}Hz)\\E=2.42eV[/tex]
(b)
[tex]E=(4.14*10^{-15}eV\cdot s)(3.50*10^{9}Hz)\\E=1.45*10^{-5}eV[/tex]
(c)
[tex]E=(4.14*10^{-15}eV\cdot s)(40.0*10^{6}Hz)\\E=1.66*10^{-7}eV[/tex]
The graph shows the force on an object of mass M as a function of time. For the time interval 0 to 4 s, the total change in the momentum of the object is?( in kg.m/s)
40
20
-20
0Graph:Square wave from -10 to 10
Answer:
The answer is zero. The total change in momentum is equal to the sum of the areas within the time interview 0-4s. The area under the graph is Force x time (F being the breadth and time the length)
The sun of the areas is zero.
Explanation:
See the attachment below for the full solution.
Thank you for reading this post. I hope it is helpful to you.
A bullet of mass 0.01 kg moving horizontally strikes a block of wood of mass 1.5 kg which is suspended as a pendulum. The bullet lodges in the wood, and together they swing upwards a distance of 0.40 m. What was the velocity of the bullet just before it struck the wooden block
Answer:
423m/s
Explanation:
Suppose after the impact, the bullet-block system swings upward a vertical distance of 0.4 m. That's means their kinetic energy is converted to potential energy:
[tex]E_p = E_k[/tex]
[tex]mgh = mv^2/2[/tex]
where m is the total mass and h is the vertical distance traveled, v is the velocity right after the impact at, which we can solve by divide both sides my m
Let g = 9.81 m/s2
[tex]gh = v^2/2[/tex]
[tex]v^2 = 2gh = 2 * 9.81* 0.4 = 7.848[/tex]
[tex]v = \sqrt{7.848} = 2.8m/s[/tex]
According the law of momentum conservation, momentum before and after the impact must be the same
[tex]m_uv_u + m_ov_o = (m_u + m_o)v[/tex]
where [tex]m_u = 0.01, v_u[/tex] are the mass and velocity of the bullet before the impact, respectively.[tex]m_ov_o[/tex] are the mass and velocity of the block before the impact, respectively, which is 0 because the block was stationary before the impact
[tex]0.01v_u + 0 = (0.01 + 1.5)*2.8[/tex]
[tex]0.01v_u = 4.23[/tex]
[tex]v_u = 4.23 / 0.01 = 423 m/s[/tex]
The initial velocity of the bullet just before it struck the block was 422 m/s.
To determine the velocity of a bullet just before it strikes a block of wood and causes a ballistic pendulum motion, we follow these steps:
First, identify the masses: the bullet (0.01 kg) and the wooden block (1.5 kg). Together, they have a combined mass of (1.51 kg) after the bullet lodges in the wood.Next, use the height they rise to find the velocity after collision. The potential energy at the highest point (0.40 m) is converted from kinetic energy:[tex]p_{initial}[/tex] = [tex]P_{final}[/tex]
(0.01 kg)u = (1.51 kg)(2.8 m/s)
Simplifying this:
u = 1.51 kg * 2.8 m/s / 0.01 kg
u = 422 m/s
Thus, the initial velocity of the bullet just before it struck the block was 422 m/s.
A fly sits on a potter's wheel 0.30 m from its axle. The wheel's rotational speed decreases from 4.0 rad/s to 2.0 rad/s in 5.0 s. Determine the wheel's average rotational acceleration.
Answer:
0.4rad/s²
Explanation:
Angular acceleration is the time rate of change of angular velocity . In SI units, it is measured in radians per second squared (rad/s²)
w1 = 4rad/s, w2 =2rad/s, t = 5sec, r = 0.30m
a = ∆w/t
a = (w2 - w1)/t
a = (2 - 4)/5 = -2/5 =
a = - 0.4rad/s²
The -ve sign indicates a deceleration in the motion
Good luck
A typical cell phone charger is rated to transfer a maximum of 1.0 Coulomb of charge per second. Calculate the maximum number of electrons that can be transferred by this charger in 1.0 hour
Answer:
The maximum no. of electrons- [tex]2.25\times 10^{22}[/tex]
Solution:
As per the question:
Maximum rate of transfer of charge, I = 1.0 C/s
Time, t = 1.0 h = 3600 s
Rate of transfer of charge is current, I
Also,
[tex]I = \frac{Q}{t}[/tex]
Q = ne
where
n = no. of electrons
Q = charge in coulomb
I = current
Thus
Q = It
Thus the charge flow in 1. 0 h:
[tex]Q = 1.0\times 3600 = 3600\ C[/tex]
Maximum number of electrons, n is given by:
[tex]n = \frac{Q}{e}[/tex]
where
e = charge on an electron = [tex]1.6\times 10^{- 19}\ C[/tex]
Thus
[tex]n = \frac{3600}{1.6\times 10^{- 19}} = 2.25\times 10^{22}[/tex]
The unit weight of a soil is 96 lb/ft3 . The moisture content if this soil is 17% when the degree of saturation is 60%. Determine: a. Void ratio b. Specific gravity of solids c. Saturated unit weight.
The void ratio of the soil is 1.2045, the specific gravity of solids is 2.6329, and the saturated unit weight is 82.7586 lb/ft3.
Explanation:The void ratio of the soil can be determined using the formula:
e = (1 + w) / (1 - w)
where e is the void ratio and w is the moisture content. Plugging in the values, we get:
e = (1 + 0.17) / (1 - 0.17) = 1.2045
To calculate the specific gravity of solids, we can use the formula:
Gs = (1 + e) * (1 - S) / (1 - e * S)
where Gs is the specific gravity of solids and S is the degree of saturation. Substituting the given values:
Gs = (1 + 1.2045) * (1 - 0.6) / (1 - 1.2045 * 0.6) = 2.6329
The saturated unit weight can be found using the equation:
Gamma_sat = Gamma_dry / (1 + w)
where Gamma_sat is the saturated unit weight and Gamma_dry is the dry unit weight. Given that the unit weight of the soil is 96 lb/ft3, we have:
Gamma_sat = 96 / (1 + 0.17) = 82.7586 lb/ft3
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A hot-air balloon is rising upward with a constant speed of 2.51 m/s. When the balloon is 3.16 m above the ground, the balloonist accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground
Answer:
t = 1.099 s
Explanation:
given,
constant speed = 2.51 m/s
height of balloon above ground = 3.16 m
time elapsed before it hit the ground = ?
Applying equation of motion to the compass
[tex]y = u t + \dfrac{1}{2}at^2[/tex]
[tex]-3.16 = 2.51 t + \dfrac{1}{2}\times (-9.8)t^2[/tex]
[tex]4.9 t^2 - 2.51 t - 3.16 = 0[/tex]
using quadratic formula to solve the equation
[tex]t = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]t = \dfrac{-(-2.51)\pm \sqrt{2.51^2-4(4.9)(-3.16)}}{2\times 4.9}[/tex]
t = 1.099 s, -0.586 s
hence, the time elapses before the compass hit the ground is equal to 1.099 s.
When it rises, air has more weight above it, and the higher pressure allows the air to expand.A.TrueB.False
Answer:
False
Explanation:
There is less pressure higher in the atmosphere, which means that air will expand, and thus cool
A certain car has a fuel efficiency of 39.2 miles per gallon (mi/gal). Express this efficiency in kilometers per liter (km/L).
Answer:
e = 16.67 km/L
Explanation:
given,
fuel efficiency = 39.2 mi/gal
we need to convert fuel efficiency in Km/l
1 Km = 0.621 mile
1 gal = 3.786 L
now, efficiency
[tex]e = 39.2 \dfrac{mi}{gal}\times \dfrac{1\ km}{0.621\ mile}\times\dfrac{1\ gal}{3.786\ L}[/tex]
e = 16.67 km/L
hence, the efficiency of the car in km/L is equal to 16.67 km/L
The normal force equals the magnitude of the gravitational force as a roller coaster car crosses the top of a 33-mm-diameter loop-the-loop. Part A What is the car's speed at the top? Express your answer to two significant figures and include the appropriate units.
The car's speed at the top of the loop is approximately 0.41 m/s.
At the top of the loop-the-loop, the normal force equals the magnitude of the gravitational force, which means the net force acting on the roller coaster car is zero.
This condition occurs when the car is just about to lose contact with the track due to insufficient normal force.
Using the centripetal force formula, [tex]\( F_{\text{net}} = \frac{mv^2}{r} \)[/tex], where \( m \) is the mass of the car, v is its speed, and \( r \) is the radius of the loop.
At the top of the loop, the net force is zero, so we equate the gravitational force and the centripetal force:
[tex]\[ mg = \frac{mv^2}{r} \][/tex]
Solving for v, we get:
[tex]\[ v = \sqrt{gr} \][/tex]
Substituting the given radius [tex]\( r = \frac{33 \, \text{mm}}{2} = 0.0165 \, \text{m} \)[/tex] and the acceleration due to gravity [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex], we find:
[tex]\[ v = \sqrt{(9.8 \, \text{m/s}^2)(0.0165 \, \text{m})} \approx 0.41 \, \text{m/s} \][/tex]
Therefore, the car's speed at the top of the loop is approximately 0.41 m/s.
Suppose that we use a heater to boil liquid nitrogen (N2 molecules). 4480 J of heat turns 20 g of liquid nitrogen into gas. Note that the latent heat is equal to the change in enthalpy, and that liquid nitrogen boils at 77 K. The system is kept at a constant pressure of 1 atm. 20) Assuming that you can treat the gas as ideal gas and that the volume of the liquid compute the binding energy of a nitrogen molecule in the liquid. (the binding energy is the difference in internal energy per molecule between the liquid and gas) approximately zero,
a. 9.4 x 10-21 J
b. 3.8 х 1027 J
c. 4.2 x 10-18 J
d. 10-20 J e. 2.1 x 10-19 J
Answer:
The energy is [tex]9.4\times10^{-21}\ J[/tex]
(a) is correct option
Explanation:
Given that,
Energy = 4480 j
Weight of nitrogen = 20 g
Boil temperature = 77 K
Pressure = 1 atm
We need to calculate the internal energy
Using first law of thermodynamics
[tex]Q=\Delta U+W[/tex]
[tex]Q=\Delta U+nRT[/tex]
Put the value into the formula
[tex] 4480=\Delta U+\dfrac{20}{28}\times8.314\times77[/tex]
[tex]\Delta U=4480-\dfrac{20}{28}\times8.314\times77[/tex]
[tex]\Delta U=4022.73\ J[/tex]
We need to calculate the number of molecules in 20 g N₂
Using formula of number of molecules
[tex]N=n\times \text{Avogadro number}[/tex]
Put the value into the formula
[tex]N=\dfrac{20}{28}\times6.02\times10^{23}[/tex]
[tex]N=4.3\times10^{23}[/tex]
We need to calculate the energy
Using formula of energy
[tex]E=\dfrac{\Delta U}{N}[/tex]
Put the value into the formula
[tex]E=\dfrac{4022.73}{4.3\times10^{23}}[/tex]
[tex]E=9.4\times10^{-21}\ J[/tex]
Hence, The energy is [tex]9.4\times10^{-21}\ J[/tex]
The binding energy of a nitrogen molecule in the liquid is calculated using the formula for latent heat and Avogadro's number. Once the latent heat and the number of molecules are determined, we can find the energy per molecule and hence, the binding energy. After performing the calculations, we find the binding energy of a nitrogen molecule to be approximately 5.2 x 10^-21 J.
Explanation:To calculate the binding energy of a nitrogen molecule in the liquid, we can use the formula for latent heat. This is given by:
Q = mL, where,
Q = Heat energy applied (color change or state change)
m = mass
L = Latent heat
After calculating the latent heat, we can determine the number of moles of nitrogen gas formed and subsequently the number of molecules using Avogadro's number. Then, the energy per molecule is calculated by dividing the total heat absorbed by the number of molecules. Therefore, the binding energy will be the difference in energy per molecule (latent heat per molecule) between the liquid and the gaseous states.
The amount of heat given (Q) is 4480 J and the mass (m) of liquid nitrogen boiled is 20 g. The molar mass (M) of nitrogen (N2) is approximately 28 g/mol. Using this, we can rearrange the formula to find L which is the energy/mass.
Substituting the values, we get L = Q/m = 4480J / 20g = 224 J/g
The number of moles of N2(n) = m/M = 20g / 28 g/mol ~ 0.714 mol
The number of N2 molecules(N) = n x Avogadro's number = 0.714 mol x 6.022 x 10^23 mol^-1 ~ 4.3 x 10^23
Therefore, the binding energy of a nitrogen molecule (E) = L/N = 224J/g / 4.3 x 10^23 = ~5.2 x 10^-21 J.
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A charge of 0.51 C is spread uniformly throughout a 33 cm rod of radius 4 mm. What are the volume and linear charge densities
The definition of volumetric charge density is given as the ratio between the load per unit volume, while the linear load is the same ratio of the load but per unit length. Applying these concepts then we have that the volumetric density of charge is,
Here,
q = Charge
V = Volume
Replacing we have,
[tex]\gamma = \frac{0.51}{\pi r^2 l}[/tex]
[tex]\gamma = \frac{0.51}{\pi (4*10^{-3})^2(0.33)}[/tex]
[tex]\gamma =30475.84C/m^3[/tex]
And the linear charge density is
[tex]\rho = \frac{q}{l}[/tex]
[tex]\rho = \frac{0.51}{0.33}[/tex]
[tex]\rho = 1.54C/m[/tex]
To determine the volume and linear charge densities, divide the total charge by the rod's volume and length, respectively, after calculating the volume using the formula for a cylinder.
Explanation:The volume charge density (ρ) and the linear charge density (λ) are physical quantities that represent the distribution of electric charge in a material. To find the volume charge density, we divide the total charge (Q) by the volume (V) of the rod. The formula is ρ = Q/V. The linear charge density is found by dividing the charge by the length (L) of the rod, using the formula λ = Q/L.
Given that the charge Q is 0.51 C, distributed uniformly along a cylindrical rod with a length of 33 cm (or 0.33 m) and a radius of 4 mm (or 0.004 m), we first need to calculate the volume of the cylinder using the formula V = πr²h, where r is the radius and h is the height (length) of the cylinder. The volume V is then π ×(0.004 m)² ×0.33 m.
After computing the volume, we calculate ρ as ρ = 0.51 C / V. To find λ, we simply divide the charge by the length, λ = 0.51 C / 0.33 m. These computations will yield the volume and linear charge densities for the rod.
Stretched 1 cm beyond its natural length, a rubber band exerts a restoring force of magnitude 2 newtons. Assuming that Hooke's Law applies, answer the following questions:(a) How far (in units of meters) will a force of 3 newtons stretch the rubber band?
The extension in the rubber band for the restoring force of 3 newtons is 1.5 cm.
The force according to Hooke's law is given as:
[tex]F=kx[/tex]
Here F is the restoring force, k is the proportionality constant and x is the stretch or compression.
Given:
Restoring force, [tex]F= 2\ N[/tex]
Stretch in the band, [tex]x=1\ cm[/tex]
The value of the k is computed as:
[tex]F=kx\\2=k \times 1cm\\k= 2\ N/cm[/tex]
The stretch in the band for a restoring force of 3 N is computed as:
[tex]F=kx\\x=\frac{F}{k}\\x= \fract{3}{2}\\x = 1.5 cm[/tex]
Therefore, the extension in the rubber band for the restoring force of 3 newtons is 1.5 cm.
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According to Hooke's Law, a force of 3 newtons will stretch the rubber band 1.5 centimeters.
Explanation:To answer this question, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. In this case, we are given that when a rubber band is stretched 1 cm beyond its natural length, it exerts a restoring force of 2 newtons. This means that the force constant of the rubber band is 2 newtons per centimeter (N/cm).
To find out how far a force of 3 newtons will stretch the rubber band, we can use the formula for Hooke's Law: F = kx, where F is the force, k is the force constant, and x is the displacement. Rearranging the formula, we have x = F/k. Plugging in the values, we get:
x = 3 N / (2 N/cm) = 1.5 cm
Therefore, a force of 3 newtons will stretch the rubber band 1.5 centimeters.
Consider a cloudless day on which the sun shines down across the United States. If 2073 kJ of energy reaches a square meter ( m 2 ) of the United States in one hour, how much total solar energy reaches the entire United States per hour
The total amount of energy per hour is [tex]2.039\cdot 10^{16} kJ[/tex]
Explanation:
In this problem we are told that the amount of energy reaching a square meter in the United States per hour is
[tex]E_1 = 2073 kJ[/tex]
The total surface area of the United States is
[tex]A=9.834\cdot 10^6 km^2[/tex]
And converting into squared metres,
[tex]A=9.834\cdot 10^6 \cdot 10^6 = 9.834\cdot 10^{12} m^2[/tex]
Therefore, the total energy reaching the entire United States per hour is given by:
[tex]E=AE_1 = (9.834\cdot 10^{12})(2073)=2.039\cdot 10^{16} kJ[/tex]
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A uniform, solid metal disk of mass 6.10 kgkg and diameter 30.0 cmcm hangs in a horizontal plane, supported at its center by a vertical metal wire. You find that it requires a horizontal force of 4.29 NN tangent to the rim of the disk to turn it by 3.40 ∘∘, thus twisting the wire. You now remove this force and release the disk from rest.
Answer:
10.84406 Nm/rad
0.068625 kgm²
2.00066 rad/s
0.49983 s
Explanation:
F = Force = 4.29 N
R = Radius = [tex]\dfrac{30}{2}=15\ cm[/tex]
[tex]\theta[/tex] = Angle = [tex]3.4\ ^{\circ}[/tex]
m = Mass of disk = 6.1 kg
Torsional constant is given by
[tex]J=\dfrac{\tau}{\theta}\\\Rightarrow J=\dfrac{FR}{\theta}\\\Rightarrow J=\dfrac{4.29\times 0.15}{3.4\times \dfrac{\pi}{180}}\\\Rightarrow J=10.84406\ Nm/rad[/tex]
The torsion constant is 10.84406 Nm/rad
Moment of inertia is given by
[tex]I=\dfrac{1}{2}mr^2\\\Rightarrow I=\dfrac{1}{2}6.1\times 0.15^2\\\Rightarrow I=0.068625\ kgm^2[/tex]
The moment of inertia is 0.068625 kgm²
Frequency is given by
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{J}{I}}\\\Rightarrow f=\dfrac{1}{2\pi}\sqrt{\dfrac{10.84406}{0.068625}}\\\Rightarrow f=2.00066\ rad/s[/tex]
The frequency is 2.00066 rad/s
Time period is given by
[tex]T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{2.00066}\\\Rightarrow T=0.49983\ s[/tex]
The time period is 0.49983 s
If the energy stored in the fully charged battery is used to lift the battery with 100-percent efficiency, what height is attained? Assume that the acceleration due to gravity is 9.8 m/s2m/s2 and is constant with height
Answer:
h= 32059.37 m
Explanation:
Assuming the missing in formation as
A certain lead-acid storage battery has a mass of 33 kg . Starting from a fully charged state, it can supply 6 A for 20 hours with a terminal voltage of 24 V before it is totally discharged.
Now, Applying energy conservation ( Electrical to potential)
Electrical Energy E= I×V×t
I = correct , V= voltage , t= time of flow of current
E = 6×24×20×60×60.
E= 10368 KJ
Now this energy is used to lift the battery with 100% efficiency
Hence,
electrical energy E= potential energy P
P= mgh
m=mass of the battery , g= the acceleration due to gravity is 9.8 m/s^2
h= height
mgh = 10368 kJ
33×9.8×h= 10368×1000
h = 10368×1000/(33×9.8)
h= 32059.37 m
Find the kinetic energy K of a satellite with mass m in a circular orbit with radius R. Express your answer in terms of m, M, G, and R.
Answer:
[tex]K=\frac{GmM}{2R}[/tex]
Explanation:
The kinetic energy is defined as:
[tex]K=\frac{mv^2}{2}(1)[/tex]
Here, m is the object's mass and v its speed. In this case the speed of the satellite is the orbital speed, which is given by:
[tex]v_{orb}=\sqrt\frac{GM}{R}(2)[/tex]
Here, G is the gravitational constant, M is the mass of the object that the satellite is orbiting and R is the radius of its circular orbit. Replacing (2) in (1):
[tex]K=\frac{mv_{orb}^2}{2}\\K=\frac{m(\sqrt\frac{GM}{R})^2}{2}\\K=\frac{GmM}{2R}[/tex]
The kinetic energy of a satellite in a circular orbit can be found using the equation K=GMm/2r, involving the gravitational constant, mass of the satellite, mass of the body being orbited, and the radius of the orbit. This kinetic energy is half the potential energy and equal to the total energy of the satellite.
Explanation:The kinetic energy K of a satellite with mass m in a circular orbit with radius R can be defined using the equation of the kinetic energy K = 1/2 * mv², where m is the mass of the satellite and v is its speed. However, in the context of gravitational forces, we must consider that the gravitational force provides the centripetal force necessary for the satellite to maintain its orbit with speed v. Therefore, we have GMm/r² = mv²/r where G is the gravitational constant and M is the mass of the body the satellite is orbiting.
When we solve for the speed v, we find that v=sqrt(GM/R), which leads us to an equation for the kinetic energy of K = GMm/2r. These equations demonstrate that the kinetic energy of the satellite is dependent not only on its own mass and speed, but also on the mass of the body it is orbiting and the radius of its orbit.
The concept that the kinetic energy of a satellite in circular orbit is half the magnitude of the potential energy, and the same as the magnitude of the total energy, is an important aspect of understanding these calculations. We also note that the gravitational constant G is by far the least well determined of all fundamental constants in physics.
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a. How much work is done when a 185g tomato is lifted 15.0m? b. The tomato is dropped. What is the velocity, v, of the tomato when it hits the ground? Assume 82.1 % of the work done in Part A is transferred to kinetic energy, E, by the time the tomato hits the ground.
Final answer:
The work done when a 185g tomato is lifted 15.0m is 27.261 J. If 82.1% of this work is converted to kinetic energy, the velocity of the tomato when it hits the ground is approximately 15.54 m/s.
Explanation:
To determine the work done when lifting a 185g tomato 15.0m, we use the formula for gravitational potential energy (GPE), which is equivalent to the work done against gravity.
Work (W) = mass (m) × gravitational acceleration (g) × height (h)
m = 185 g = 0.185 kg (since 1g = 0.001 kg)
g = 9.8 m/s²
h = 15.0 m
W = 0.185 kg × 9.8 m/s² × 15.0 m
W = 27.261 J
To determine the velocity (v) when the tomato hits the ground, assuming 82.1% of the work done is converted to kinetic energy (KE), we calculate:
KE = 0.821 × W
KE = 0.821 × 27.261 J
KE = 22.379 J
The formula for KE is:
KE = 1/2 m v²
Solving for v gives:
v = √(2 × KE / m)
v = √(2 × 22.379 J / 0.185 kg)
v ≈ 15.54 m/s
Thus, the velocity of the tomato when it hits the ground, assuming 82.1% conversion of energy, is approximately 15.54 m/s.
the equitorial diameter of the moon is 3476 kilometers. if a kilometer equals 0.6214 miles, what is the moon's diameter in miles?
To convert the equatorial diameter of the Moon from kilometers to miles, you can use the given conversion factor:
Equatorial diameter in miles = Equatorial diameter in kilometers × Conversion factor
Given:
Equatorial diameter of the Moon = 3476 kilometers
Conversion factor = 0.6214 miles/kilometer
Now, plug in the values:
Equatorial diameter in miles = 3476 km × 0.6214 miles/km
[tex]\[ \text{Equatorial diameter in miles} \approx 2160.9264 \, \text{miles} \][/tex]
So, the Moon's equatorial diameter is approximately 2160.93 miles.
Final answer:
To convert the Moon's diameter from 3476 kilometers to miles, multiply by the conversion factor of 0.6214 miles per kilometer, resulting in approximately 2160.6344 miles.
Explanation:
The question asks for the conversion of the Moon's diameter from kilometers to miles. The given diameter of the Moon is 3476 kilometers. To convert kilometers to miles, we use the provided conversion factor where 1 kilometer equals 0.6214 miles.
Here's how you can calculate the Moon's diameter in miles:
1. Multiply the Moon's diameter in kilometers by the conversion factor:
3476 kilometers × 0.6214 miles/kilometer
2. Calculating this gives:
2160.6344 miles
So, the Moon's diameter in miles is approximately 2160.6344.
What is the potential difference between the terminals of an ordinary AA or AAA battery? (If you’re not sure, find one and look at the label.)
The potential difference between the terminals of an ordinary AA or AAA battery is usually 1.5 volts. Voltage is the work required to move a charge, while current is the charge flow rate. Batteries can provide considerable energy for their size, analogous to lifting a significant mass against gravity.
Explanation:The potential difference between the terminals of an ordinary AA or AAA battery is the electromotive force (emf) when the battery is not part of a complete circuit, and is typically about 1.5 volts (V). If the battery is in a complete circuit, the potential difference is known as the terminal potential difference, which is also measured in volts but may differ slightly from the emf due to internal resistance and load on the battery. Voltage is the measure of work required to move a charge between two points, while current is the rate of charge flow, measured in amperes.
When a battery is used in a circuit, the conventional current flows from the positive terminal to the negative terminal, and this flow of charge creates the potential difference that does work in the circuit. An ammeter, which must be connected in series, is used to measure current. The common AA battery not only has a voltage of 1.5 V but also a significant amount of stored energy, which could be likened to a substantial mass being lifted against gravity, highlighting the compact energy storage capability of such batteries.
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When jumping, a flea accelerates at an astounding 1000 m/s2, but over only the very short distance of 0.50 mm. If a flea jumps straight up, and if air resistance is neglected (a rather poor approximation in this situation), how high does the flea go?
Answer:
The flea will move high to a height of 0.05 meters.
Explanation:
Given that,
Acceleration of the flea, [tex]a=1000\ m/s^2[/tex]
Distance, d = 0.5 mm = 0.0005 m
Let u and v are the initial and final velocity of the flea. Using equation of motion as :
[tex]v^2-u^2=2ad[/tex]
[tex]v^2-u^2=2\times 1000\times 0.0005[/tex]
[tex]v^2-u^2=1[/tex]..........(1)
Using conservation of energy, we get :
[tex]\dfrac{1}{2}mu^2=\dfrac{1}{2}mv^2+mgh[/tex]
[tex]\dfrac{1}{2}u^2=\dfrac{1}{2}v^2+(-g)h[/tex]
[tex]\dfrac{1}{2}u^2=\dfrac{1}{2}v^2-gh[/tex]
[tex]\dfrac{1}{2g}(u^2-v^2)=-h[/tex]
[tex]h=\dfrac{1}{2g}[/tex]
[tex]h=\dfrac{1}{2\times 9.8}[/tex]
h = 0.05 meters
So, the flea will move high to a height of 0.05 meters. Hence, this is the required solution.
An attacker at the base of a castle wall 3.95 m high throws a rock straight up with speed 5.00 m/s from a height of 1.60 m above the ground.
(a) Will the rock reach the top of the wall?
(b) If so, what is its speed at the top? If not, what initial speed must it have to reach the top?
Answer:
a) No
b) the rock must have a minimum initial speed of 6.79m/s for it to reach the top of the building.
Explanation:
Given:
Height of the wall = 3.95m
Initial height = 1.60m
Initial speed = 5.00m/s
distance between the initial height and wall top = 3.95 - 1.60 = 2.35m
Using the formula;
v^2 = u^2 + 2as ....1
Where v = final velocity, u = initial velocity, a = acceleration, s = distance travelled
From equation 1
s = (v^2 - u^2)/2a ...2
Since the rock t moving up,
the acceleration = -g = -9.8m/s2
s = maximum height travelled
v = 0 (at maximum height velocity is zero)
Substituting into equation 2
s = (0 - 5^2)/(2×-9.8) = 1.28m
Therefore, the maximum height is 1.28 from his initial height Which is less than the 2.35m of the wall from his initial height. So the rock will not reach the top of the wall
b) Using equation 1:
u^2 = v^2 - 2as
v = 0
a = -9.8m/s
s = 2.35m. (distance between the initial height and wall top)
u^2 = 0 - 2(-9.8 × 2.35)
u^2 = 46.06
u = √46.06
u = 6.79m/s
Therefore, the rock must have a minimum initial speed of 6.79m/s
A human hair has a thickness of about 60 um. What is this in millimeters
Answer:
0.06
Explanation:
The thickness of a human hair is about 60 micrometers. When converted to millimeters, this measures to 0.06 millimeters.
Explanation:The thickness of a human hair is about 60 micrometers (um). To convert this to millimeters (mm), you need to understand that 1 millimeter is equal to 1000 micrometers. Therefore, you can convert 60 um to mm by dividing 60 by 1000.
This gives an answer of 0.06 mm. So, the thickness of a human hair is 0.06 millimeters.
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A typical adult human lung contains about 330 million tiny cavities called alveoli. Estimate the average diameter of a single alveolus. Assume the alveoli are spherical and a typical human lung is about 1.9 liters.
Answer:
The average diameter of a single alveolus is 0.0222 cm.
Explanation:
Volume of the lung ,V= 1.9 L
[tex]1 L = 1000 cm^3[/tex]
[tex]1.9 L=1.9\times 1000 cm^3=1900 cm^3[/tex]
Number of alveoli in a human lung = [tex]330\times 10^6[/tex]
Volume of single alveoli =v
[tex]v\times 330\times 10^6=V[/tex]
[tex]v=\frac{1900 cm^3}{330\times 10^6}[/tex]
[tex]v=5.7575\times 10^{-6} cm^3[/tex]
The alveoli are spherical.
Radius of an alveolus = r
Volume of the sphere = [tex]\frac{4}{3}\pi r^3[/tex]
[tex]v=\frac{4}{3}\pi r^3[/tex]
[tex]5.7575\times 10^{-6} cm^3=\frac{4}{3}\times 3.14\times r^3[/tex]
[tex]r=0.0111 cm[/tex]
Diameter of the alveolus =d
d = 2r = 2 × 0.0111 cm = 0.0222 cm
The average diameter of a single alveolus is 0.0222 cm.
The question seeks to find the average diameter of a single alveolus in the human lung, using information about the total volume of the lung and the number of alveoli. By using the formula for the volume of a sphere, one can calculate the volume of a single alveolus and derive its radius and thereby its diameter.
Explanation:The problem is essentially asking to find the average diameter of a single alveolus, given we know the total volume of the lung and the number of alveoli. Using the formula for the volume of a sphere, V=4/3πr³, where V is the volume and r is the radius, we can find the volume of a single alveolus by dividing the total volume of the lungs (1.9 liters, which equals to 1.9 x 10^9 cubic millimeters) by the number of alveoli (approximately 330 million). We can then calculate the radius by rearranging the sphere volume formula to r = ((3*V)/4π)^1/3. The diameter would be double the radius.
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An 8.0 cm diameter, 400 g sphere is released from rest ta the tip of a 2.1 m long, 25 degree incline. It rolls, without slipping, to the bottom.a) What is the sphere's angular velocity at the bottom of the incline?b) What fraction of its kinetic energy is rotational?
Answer:
a) 88.1 rad/s
b) 0.286
Explanation:
given information:
diameter, d = 8 cm = 0.08 m
sphere's mass, m = 400 g = 0.4 kg
the distance from rest to the tip, h = 2.1 m
incline angle, θ = 25°
a) What is the sphere's angular velocity at the bottom of the incline?
mg(h sinθ) = 1/2 Iω² + 1/2mv²
I of solid sphere = 2/5 mr², thus
mg(h sinθ) = 1/2 (2/5 mr²) ω² + 1/2 mv², now we can remove the mass
g h sin θ = 1/5 r² ω² + 1/2 v²
ω = v/r, v = ωr
so,
g h sin θ = 1/5 r² ω² + 1/2 (ωr)²
g h sin θ = (7/10) r² ω²
ω² = 10 g h sin θ/7 r²
ω = √10 g h sin θ/7 r²
= √10 (9.8) (2.1) sin 25° / 7 (0.04)²
= 88.1 rad/s
b) What fraction of its kinetic energy(KE) is rotational?
fraction of its kinetic energy = rotational KE / total KE
total KE = total potential energy
= m g h sin θ
= 0.4 x 9.8 x 2.1 sin 25°
= 3.48 J
rotational KE = 1/2 Iω²
= 1/5 mr²ω²
= 1/5 0.4 (0.04)²(88.1)²
= 0.99
fraction of its KE = 0.99/3.48
= 0.286
A) The sphere's angular velocity at the bottom of the incline is; ω = 88.1 rad/s
B) Fraction of its kinetic energy that is rotational is; 0.286
What is the angular velocity?We are given;
Diameter; d = 8 cm = 0.08 m
Mass of sphere; m = 400 g = 0.4 kg
Distance from rest to the tip; h = 2.1 m
Angle of inclination; θ = 25°
a) To get the sphere's angular velocity at the bottom of the incline, we will use the expression;
mg(h*sinθ) = ¹/₂Iω² + ¹/₂mv²
where;
I of solid sphere = ²/₅mr²
Thus;
mg(h*sinθ) = ¹/₂(²/₅mr²)ω² + ¹/₂mv²
The mass m will cancel out to give;
gh*sin θ = ¹/₅r²ω² + ¹/₂v²
where v = ωr
Thus;
gh*sin θ = ¹/₅r²ω² + ¹/₂r²ω²
gh*sin θ = ⁷/₁₀r²ω²
ω = √[(¹⁰/₇)*g*h*(sin θ)/r²]
ω = √[(¹⁰/₇)*9.8*2.1*(sin 25)/(0.04)²]
ω = 88.1 rad/s
b) Fraction of its kinetic energy that is rotational = rotational KE/total KE
But, total KE = total potential energy
Thus;
KE_tot = mgh*sin θ
KE_tot = 0.4 * 9.8 * 2.1 sin 25°
KE_tot = 3.48 J
KE_rot = ¹/₂Iω²
I of solid sphere = ²/₅mr². Thus;
KE_rot = ¹/₅mr²ω²
KE_rot = ¹/₅ * 0.4 * 0.04² * 88.1²
KE_rot = 0.99 J
Fraction of its kinetic energy that is rotational = 0.99/3.48 = 0.286
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Oil and water don’t mix, and the mass density of oil is smaller than that of water. Suppose water is poured into a U-shaped tube that is open at both ends until the water surface is halfway up each leg of the tube, and then some oil is poured on top of the water in the right leg. Once the system comes to equilibrium, are the top of the oil column in the right leg and the top of the water column in the left leg at the same height? If not, which is higher?
The right leg (oil on top) is higher
The left leg (no oil) is higher
The two legs are the same height
Answer:
The right leg (oil on top) is higher
Explanation:
Given:
The mass density of oil is lesser than the mass density of water.
When we pour water in a u-tube that is open at both the ends then the water on both the sides of the tube will rise up to the same height because the algebraic sum of the pressure exerted by the water column and the pressure of atmosphere on both the openings is equal.When we pour oil in the right side of the u-tube we observe that the column of liquid on the right side rises more than the column of the liquid on the left side. However we observe that there is rise on both sides of the u-tube.This is justified by the equation:
[tex]P=\rho.g.h[/tex]
where:
[tex]\rho =[/tex] density of the liquid
[tex]g=[/tex] acceleration due to gravity
[tex]h=[/tex] height of the liquid column
Imagine that two charged objects are the system of interest. When the objects are infinitely far from each other, the electric potential energy of the system is zero. When the objects are close to each other, the electric potential energy is positive. Which of the following statements is(are) incorrect
(a) Both objects are positively charged.
(b) Both objects are negatively charged.
(c) One object is negatively charged and the other one is positively charged.
Final answer:
Statement (c) is incorrect because if one object is negatively charged and the other is positively charged, they would attract each other, resulting in a negative potential energy as they come closer, not a positive one.
Explanation:
The question is addressing the concept of electric potential energy between two charged objects. When the electric potential energy of the system is positive as the two objects come close, we can infer that the objects have like charges, either both positive or both negative. This is due to the fact that work needs to be done against the electrical repulsion to bring like charges together, increasing the system's potential energy.
This makes statements (a) Both objects are positively charged and (b) Both objects are negatively charged possibly correct scenarios, as they would lead to a positive potential energy when the objects are brought together. Statement (c) One object is negatively charged and the other one is positively charged would be incorrect in this context, because a positive and a negative charge would attract each other, and the system would do work on the surroundings as they come closer to each other making the potential energy negative. Therefore, the incorrect statement, given that the electric potential energy is positive when they are near each other, is (c).
An average human weighs about 650 N. If each of two average humans could carry 1.0 C of excess charge, one positive and one negative, how far apart would they have to be for the electric attraction between them to equal their 650-N weight?
Answer:
r = 3721.04 m
Explanation:
Given that,
Weight of human, F = 650 N
Charge on two humans, [tex]q_1=q_2=1\ C[/tex]
We need to find the distance between charges if the electric attraction between them to equal their 650 N weight. It is given by :
[tex]F=\dfrac{kq^2}{r^2}[/tex]
[tex]r=\sqrt{\dfrac{kq^2}{F}}[/tex]
[tex]r=\sqrt{\dfrac{9\times 10^9\times 1^2}{650}}[/tex]
r = 3721.04 m
So, the distance between charges is 3721.04 m if the electric attraction between them to equal their 650 N weight. Hence, this is the required solution.