Initial current = 0
Final current = (15 V) / (60 ohms) = 0.25 Ampere
Current along the way = 0.25 · (1 - e^- time / time-constant)
"time-constant" = L/R = (0.045 / 60) = 7.5 x 10⁻⁴ second
Current = 0.25 · (1 - e^-10,000t/7.5)
When t = 7 ms,
Current = 0.25 · ( 1 - e^-70/7.5)
Current = 0.25 · (1 - e^-9.33)
Current = 0.25 · (1 - 8.84 x 10⁻⁵)
Current = 0.25 · (0.9999)
Current = so close to 250 mA that you can't tell the difference.
The reason is that 7.0 mS is 9.3 time-constants, and during EVERY time-constant, the current grows by 37% of the distance it still has left to go. So after 9.3 of these, it's practically AT the target.
I have a feeling that the time in the question is SUPPOSED TO BE 7 microseconds. If that's true, then
Current = 0.25 · (1 - e^-[ 7 x 10⁻⁶ / 7.5 x 10⁻⁴ ]
Current = 0.25 · (1 - e^-0.00933)
Current = 0.25 · (1 - 0.9907)
Current = 0.25 · (0.0093)
Current = 2.32 mA ?
No, that can't be it either.
Well ! Now, I'm going to determine the true and correct final answer in the only cheap and sleazy way I have left ... by looking at all the choices offered, and eliminating the absurd ones.
The effect of an inductor in the circuit is to resist any change in current. The final current in this circuit is when it's not trying to change any more. So the final current is just the battery with a resistor across it ... (12 V) / (60 ohms). That's 0.25 Ampere, or 250 mA. The current starts at zero when the switch closes, and it builds up and builds up to 250 mA. It's never more than 250 mA.
So look at the choices ! The only one that not more than 250 mA is choice-A .
THAT has to be it. 7.0 mS is a no-brainer. It's 9.3 time-constants after the switch closes, the current has built up to 99.99% of its final value by then, it's not really trying to change much any more, the inductor is just about finished having any effect on the current, and the current is essentially at its final value of 250 mA. The action is all over.
Now, I fully realize that Mister "Rishwait" is a bot and all, and nobody really needs the answer to this question. But every cloud has a silver lining. It's a numskull question, but it earned me 10 points, and it's been a truly fascinating trip down Memory Lane.
the current approximately 7.0 ms after closing the switch is about 250 mA, which is option (a).
To find the current through the circuit 7.0 ms after the switch is closed, we can use the concept of an RL circuit. The current in an RL circuit follows an exponential growth equation, given by:
I(t) = (V/R)(1 - e^(-t/τ))
Where:
I(t) is the current at time t.
V is the voltage from the power supply (15 V in this case).
R is the resistance (60 Ω).
τ (tau) is the time constant of the circuit, given by L/R, where L is the inductance (45 mH = 0.045 H).
First, calculate the time constant τ:
τ = L/R = 0.045 H / 60 Ω = 0.00075 s.
Now, plug in the values into the equation to find I(7.0 ms):
t = 7.0 ms = 0.007 s.
I(0.007 s) = (15 V / 60 Ω) * (1 - e^(-0.007 s / 0.00075 s))
I(0.007 s) = (0.25 A) * (1 - e^(-9.333...))
Now, calculate the current:
I(0.007 s) ≈ (0.25 A) * (1 - e^(-9.333...))
I(0.007 s) ≈ (0.25 A) * (1 - 0.0000962) [Using e^(-9.333...) ≈ 0.0000962]
I(0.007 s) ≈ (0.25 A) * (0.9999038)
I(0.007 s) ≈ 0.24998 A
I(0.007 s) ≈ 250 mA
So, the current approximately 7.0 ms after closing the switch is about 250 mA, which is option (a).
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Which of these sentences features an endorsement? Select all that apply.
Pete's Pizza Parlor delivers hot and delicious pizza in a flash!
The crust on the pizza from Pete's Pizza Parlor is as thin and bland as cardboard.
Lou's Dress Boutique features all organic and high-quality clothing materials.
Lunch Bunch is the perfect portable lunch for home, work, or school.
Answer:
a
Explanation:
A lightbulb is rated by the power that it dissipates when connected to a given voltage. For a lightbulb connected to 120 V household electricity, decreasing the resistance of the filament will _____ the current through the bulb and _____ the power dissipated by the bulb.
Decrease, decrease
Decrease, increase
Increase, increase
Increase, decrease
Answer:
Increase, increase
Explanation:
For a light bulb, voltage, resistance and current follows the Ohm's law:
[tex]V=RI[/tex]
where
V is the voltage
R is the resistance
I is the current
In this problem, the voltage is fixed: V=120 V. We can rewrite the equation as
[tex]I=\frac{V}{R}[/tex] (1)
we see that the current is inversely proportional to the resistance: so, if we decrease the resistance of the filament, the current will increase.
Moreover, the power dissipated by the bulb is
[tex]P=I^2 R[/tex]
which can be rewritten using (1) as
[tex]P=\frac{V^2}{R}[/tex]
We see that the power dissipated is inversely proportional to the resistance: therefore, if we decrease the resistance of the filament, the power dissipated will increase.
Infrared rays have a shorter wavelength than A. radar waves. B. ultraviolet rays. C. X-rays. D. gamma rays.
Answer:
radar waves
Explanation:
Wavelength chart.
Infrared is barely outside of the visible spectrum, but radar is (comparatively) way outside the visible spectrum on the side of longer wavelength.
Longest . . . radar waves
then . . . infrared
then . . . ultraviolet
then . . . X-rays
Shortest . . . gamma rays
An object is placed 8.5 cm in front of a convex spherical mirror of focal length −14.0 cm. What is the image distance? Show all work and include units of measure.
Given:
u = 8.5 cm
f = -14.0 cm
v = image distance
Using the mirror formula 1/u + 1/v = 1/f
1/8.5 + 1/v = 1/-14.0
Rewrite to solve for v:
v = (8.5 * -14.0) / (8.5 - (-14.0))
v = -119 / 22.5
v = -5.29 cm ( round answer as needed.)
A thermally isolated system is made up of a hot piece of aluminum and a cold piece of copper; the aluminum and the copper are in thermal contact. The specific heat of aluminum is more than double that of copper. Which object experiences the greater temperature change during the time the system takes to reach thermal equilibrium?
a) It is impossible to tell without knowing the volumes.
b) It is impossible to tell without knowing the masses.
c) The copper experiences a greater temperature change.
d) Neither; both objects experience the same magnitude temperature change.
e) The aluminum experiences a greater temperature change.
Answer:
b) It is impossible to tell without knowing the masses.
Explanation:
The temperature change of a substance when it receives/gives off a certain amount of heat Q is given by
[tex]\Delta T= \frac{Q}{m C_s}[/tex]
where
Q is the amount of heat
m is the mass of the substance
Cs is the specific heat capacity of the substance
In this case, we have a hot piece of aluminum in contact with a cold piece of copper: the amount of heat given off by the aluminum is equal to the amount of heat absorbed by the copper, so Q is the same for the two substances. However, we see that the temperature change of the two substances depends on two other factors: the mass, m, and the specific heat, Cs. So, since we know only the specific heat of the two substances, but not their mass, we can't tell which object will experience the greater temperature change.
A 3 kg ball is thrown into the air. It has an initial speed of 10 m/s. What is the ball’s gravitational
potential energy at the top of its motion? Bonus: what hieght does the ball reach
Conservation of energy means that initial kinetic energy =maximum gravitational potential energy. 1/2 * 3 * 10^2 = 150J. The gravitational potential energy at the top is 150 joules. The gravitational potential energy is calculated through mgh, so 150 = 3 * 10 * h. Height is 5 meters
How do you do this problem?
Answer:
C
Explanation:
When A and B come in contact with each other, +12 - 12 = 0 so their changes cancel.
Now C has a charge of +12
When A and C come together they each have an equal share of that 12, so each of them has 6
So the answer is
A B C
6 0 6
which is C
What is meant by the ground state of an atom?
Answer:
Let's start by explaining that each atom in its natural state has a specific structure of its energy levels. Where the lowest energy level is called the ground state.
At this level, as long as no energy is communicated to the atom, the electron will remain in the ground state.
What does this mean?
When an atom is in its ground state, its electrons fill the lower energy orbitals completely before they begin to occupy higher energy orbitals.
Walking across a carpet is an example of charge being transferred by
static electricity and friction
How is the atomic number of a nucleus changed by alpha decay?
Answer:
The atomic number of a nucleus will go down by two in alpha decay
20.What factor is more important in determining electrical force
21.Why do we need constants like k and G?
Answer:
20.The first factor is the amount of charge on each object. The greater the charge, the greater the electric force. The second factor is the distance between the charges. The closer together the charges are, the greater the electric force is
Explanation:
What happens when a particle of matter meets its corresponding antiparticle of antimatter?
A) They can form a complete atom. B) The combined mass of the two particles is completely transformed into energy (photons). C) They fuse to make a heavier particle. D) The question makes no sense, since antimatter does not really exist.
The correct answer is B) the combined mass of two particles is completely transformed into energy (photons).
Which physicist first identified the electron as an independent part of the atom?
Thomson conducted a series of experiments on cathode ray tubes, which led him to the discovery of electrons.
This is how on April 30th, 1897 Thomson announced the discovery of the electron (although at that time he called it a corpuscle)
This discovery was considered one of the milestones in science of the late nineteenth century, since it allowed to have a new conception of the structure of matter and its interaction with energy.
It should be noted that from this discovery, Thomson proposed a new model of the atom that although erroneous, remained valid for years and served as the basis for new postulates and research on the subject.
1) The electric field that is .25 m from a small sphere is 350 N/C towards the sphere. What is the charge on the sphere?
2) How much force would a 2.1x10-6 C charge feel if it was .23 m to the right of a 3.6x10-6 C charge?
3) How much power is generated in a circuit if a voltage of 9 V causes a current of .37 A?
4) A negative charge of -6.0 x 10-6 C exerts an attractive force of 15 N on a second charge that is .02 m away. What is the magnitude of the second charge?
5) Sphere A is located at the origin and has a charge of 2.0x10-6 C. Sphere B is located at .6 m on the x-axis and has a charge of -3.6x10-6 C. Sphere C is located at .8 m on the x-axis and has a charge of 4.0x10-6 C.
a) Find the net force on Sphere A.
b) Find the net force on Sphere B.
c) Find the net force on Sphere C.
1) [tex]2.43\cdot 10^{-9} C[/tex]
The electric field produced by a small charged sphere outside sphere is identical to that of a single point charge:
[tex]E=k\frac{Q}{r^2}[/tex]
where
k is the Coulomb constant
Q is the charge on the sphere
r is the distance from the sphere
Here we have
E = 350 N/C is the electric field
r = 0.25 m is the distance from the sphere
Solving for Q, we find the charge:
[tex]Q=\frac{Er^2}{k}=\frac{(350 N/C)(0.25 m)^2}{9\cdot 10^9 Nm^2 C^{-2}}=2.43\cdot 10^{-9} C[/tex]
2) 1.29 N
First of all, we need to find the magnitude of the electric field produced by the charge:
[tex]E=k\frac{Q}{r^2}[/tex]
where here we have
[tex]Q=3.6\cdot 10^{-6}C[/tex] is the charge source of the field
r = 0.23 m is the distance at which we have to calculate the field
Substituting,
[tex]E=(9\cdot 10^9 Nm^2 C^{-2}) \frac{3.6\cdot 10^{-6}C}{(0.23 m)^2}=6.12\cdot 10^5 N/C[/tex]
Now we can calculate the electric force exerted on the second charge, which is given by
[tex]F=qE[/tex]
where
[tex]q=2.1\cdot 10^{-6} C[/tex] is the magnitude of the second charge
Substituting,
[tex]F=(2.1\cdot 10^{-6} C)(6.12\cdot 10^5 N/C)=1.29 N[/tex]
3) 3.33 W
The power generated in a circuit is given by:
P = V I
where
V is the voltage
I is the current
In this circuit, we have
V = 9 V
I = 0.37 A
So the power generated is
[tex]P=(9 V)(0.37 A)=3.33 W[/tex]
4) [tex]+1.1\cdot 10^{-7}C[/tex]
The magnitude of the force between two electric charge is given by Coulomb's law:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where q1 and q2 are the two charges, and r is their separation.
In this problem we have
[tex]q_1 = 6.0\cdot 10^{-6}C[/tex]
[tex]F=15 N[/tex]
[tex]r=0.02 m[/tex]
Solving for q2, we find the magnitude of the second charge
[tex]q_2 = \frac{Fr^2}{k q_1}=\frac{(15 N)(0.02 m)^2}{(9\cdot 10^9 Nm^2 C^{-2})(6.0\cdot 10^{-6}C)}=1.1\cdot 10^{-7}C[/tex]
Moreover, the force is attractive: this means that the two charges must have opposite signs (in fact, like charges repel each other, while unline charges attract each other). In this problem, q1 is negative, so q2 must be positive.
5a) -0.07 N (towards negative x-direction)
First of all, we need to calculate the force exerted on sphere A by sphere B; this is given by
[tex]F_{AB} = k \frac{q_A q_B}{r_{AB}^2}[/tex]
where
[tex]q_A = 2.0\cdot 10^{-6}C\\q_B = -3.6\cdot 10^{-6}C\\r_{AB}=0.6 m[/tex]
Substituting,
[tex]F_{AB} = k \frac{(2.0\cdot 10^{-6} C)(-3.6\cdot 10^{-6} C)}{(0.6 m)^2}=-0.18 N[/tex]
(the negative sign means the force is towards the left)
The force exerted on sphere A by sphere C is
[tex]F_{AC} = k \frac{q_A q_C}{r_{AB}^2}[/tex]
where
[tex]q_A = 2.0\cdot 10^{-6}C\\q_C = 4.0\cdot 10^{-6}C\\r_{AC}=0.8 m[/tex]
Substituting,
[tex]F_{AC} = k \frac{(2.0\cdot 10^{-6} C)(4.0\cdot 10^{-6} C)}{(0.8 m)^2}=+0.11 N[/tex]
So the net force on sphere A is
[tex]F=F_{AB}+F_{AC}=-0.18 N+0.11 N=-0.07 N[/tex]
5b) +3.06 N (towards positive x-direction)
The force exerted on sphere B by sphere A has been calculate at step 5a), and it is
[tex]F_{AB} = k \frac{(2.0\cdot 10^{-6} C)(-3.6\cdot 10^{-6} C)}{(0.6 m)^2}=-0.18 N[/tex]
and the force is attractive, so towards the left
The force exerted on sphere B by sphere C is
[tex]F_{BC} = k \frac{q_B q_C}{r_{BC}^2}[/tex]
where
[tex]q_B = -3.6\cdot 10^{-6}C\\q_C = 4.0\cdot 10^{-6}C\\r_{BC}=0.8 m-0.6 m=0.2 m[/tex]
Substituting,
[tex]F_{BC} = k \frac{(-3.6\cdot 10^{-6} C)(4.0\cdot 10^{-6} C)}{(0.2 m)^2}=-3.24 N[/tex]
and the force is still attractive, so towards the right
So the net force on sphere B is
[tex]F=F_{AB}+F_{BC}=-0.18 N+3.24 N=3.06 N[/tex]
5c) -3.13 N (towards negative x-direction)
The force exerted on sphere C by sphere A has been calculate at step 5a), and it is
[tex]F_{AC} = k \frac{(2.0\cdot 10^{-6} C)(4.0\cdot 10^{-6} C)}{(0.8 m)^2}=+0.11 N[/tex]
and the force is repulsive, so towards the right
Similarly, we found the force exerted on sphere C by sphere B at step 5b)
[tex]F_{BC} = k \frac{(-3.6\cdot 10^{-6} C)(4.0\cdot 10^{-6} C)}{(0.2 m)^2}=-3.24 N[/tex]
and the force is attractive, so towards the left
So the net force on sphere C is
[tex]F=F_{AC}+F_{BC}=+0.11+(-3.24 N)=-3.13 N[/tex]
Match the number with the correct statement
1. line of best fit
2. slope
3. variable
(a line that best represents the trend of the data)
(the change in the y variable compared to the change in the x variable between two points on a graph)
(a quantity that changes during an experiment)
Answer:
1. line of best fit---(a line that best represents the trend of the data)
2. slope---(the change in the y variable compared to the change in the x variable between two points on a graph)
3. variable---(a quantity that changes during an experiment)
Explanation:
Technically the variable one could be considered wrong, as some variables stay the same, but otherwise these are the answers.
If i measure the recession velocities of 2 galaxies and galaxy a has a velocity twice that of galaxy b, how far away is b if a is 200 mpc away? (remember that v = h0 x d
If i measure the recession velocities of 2 galaxies and galaxy a has a velocity twice that of galaxy b, the far of b if a is 200 mpc away is 100 Mpc
Further explanationThe parsec ([tex]pc[/tex]) is a unit of length used to measure the large distances to astronomical objects outside the Solar System.
Although parsecs are used for the shorter distances within the Milky Way, multiples of parsecs are also required for the larger scales in the universe such as Kiloparsecs (kpc) is for the more distant objects within and around the Milky Way, megaparsecs (Mpc) for mid-distance galaxies, and gigaparsecs (Gpc) for many quasars and the most distant galaxies.
If i measure the recession velocities of 2 galaxies and galaxy a has a velocity twice that of galaxy b, how far away is b if a is 200 mpc away? Remember that [tex]v = h_0 * d[/tex]
galaxy A: [tex]2(galaxy B's V) = H_0 * 200[/tex]
[tex]galaxy B's V = H_0 * 100[/tex]
galaxy B: 100 Mpc
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Answer details
Grade: 9
Subject: physics
Chapter: the recession velocities
Keywords: galaxy, the recession velocities, velocity, mpc, measure, far away
An electromagnetic wave with an electric field that oscillates with a period of 8.43 ps is travelling in a vacuum. What is the wavelength of this electromagnetic wave?
"ps" stands for "pico-second". It means 10⁻¹² second.
The period of this wave is 8.43 x 10⁻¹² second.
In a vacuum, the speed of any electromagnetic wave is 3 x 10⁸ m/s.
Wavelength = (speed) / (frequency)
But Frequency = 1 / period
So Wavelength = (speed) x (period)
Wavelength = (3 x 10⁸ m/s) x (8.43 x 10⁻¹² sec)
Wavelength = 2.5 x 10⁻³ meter
Wavelength = 2.5 millimeters
You can find the wavelength of an electromagnetic wave with a given period by first calculating frequency using the formula f = 1/T. Then, use the universal wave equation c = λv, to find the wavelength, λ = c/f. Plugging all the values, we find the wavelength is about 2.53 meters.
Explanation:The subject of this question revolves around an
electromagnetic wave
with a given oscillation period and we need to find its wavelength. To solve this, we first have to convert the period into frequency, as the two are inverse to each other.
The frequency (f) can be calculated by taking the reciprocal of the period (T), so f = 1/T. Given that T = 8.43 ps (ps is picoseconds and 1 ps equals 10^-12 seconds), the frequency would therefore be f = 1/(8.43x10^-12 s) which equals approximately 1.19x10^11 Hz (Hz is the unit of frequency).
Now we use the universal wave equation, c = λv, where c is the speed of light (approximately 3 x 10^8 ms^-1 in a vacuum), v is the frequency, and λ is the wavelength we're trying to find. Solving for λ, we get λ = c/f. Substituting the values, we get λ = (3x10^8 m/s) / (1.19x10^11 Hz), which computes to approximately λ = 2.53m. Therefore, the wavelength of this electromagnetic wave is approximately 2.53 meters.
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The lowest frequency in the audible range is 20 Hz.a)What is the length of the shortest open-open tube needed to produce this frequency?b)What is the length of the shortest open-closed tube needed to produce this frequency?
a) 8.58 m
For an open-open tube, the fundamental frequency is given by
[tex]f=\frac{v}{2L}[/tex]
where
f is the fundamental frequency (the lowest frequency)
v is the speed of sound
L is the length of the tube
In this problem, we have
f = 20 Hz
v = 343 m/s (speed of sound in air)
Solving the equation for L, we find the shortest length of the tube:
[tex]L=\frac{v}{2f}=\frac{343 m/s}{2(20 Hz)}=8.58 m[/tex]
(b) 4.29 m
For an open-closed tube, the fundamental frequency is instead given by
[tex]f=\frac{v}{4L}[/tex]
Where in this problem, we have
f = 20 Hz
v = 343 m/s (speed of sound in air)
Solving the equation for L, we find the shortest length of the tube:
[tex]L=\frac{v}{4f}=\frac{343 m/s}{4(20 Hz)}=4.29 m[/tex]
two charged objects are a certain distance apart. what happens to the force between them if one charge is tripled?
Answer:
The force is tripled too.
Explanation:
The magnitude of the electric force (repulsion or attraction) between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance that separates them (Coulomb's Law).
[tex]F=k_{e} \frac{[q_{1}][q_{2}]}{r^{2}}[/tex]
Where
[tex]k_{e}[/tex] is the Coulomb's constante.
[tex]q_{1}[/tex] and [tex]q_{2}[/tex] are the signed magnitude of the charges.
[tex]r[/tex] the distance between charges.
So, if one of the charge is tripled, due the directly proportional product of the charges, the force will be tripled.
[tex]F=k_{e} \frac{[3q_{1}][q_{2}]}{r^{2}}[/tex]
or
[tex]F=k_{e} \frac{[q_{1}][3q_{2}]}{r^{2}}[/tex]
The result will be still the force tripled.
Which quantities appear in the equation for the magnetic field component of an electromagnetic wave?
In the equation for the magnetic field component of an electromagnetic wave, the quantities that appear include the electric field amplitude (E) and the speed of light (c), using the relationship B = E/c to relate the electric and magnetic field strengths.
Explanation:The question pertains to the quantities that appear in the equation for the magnetic field component of an electromagnetic wave. In electrodynamics, the magnetic field component B of an electromagnetic wave can be calculated using the relationship “B = E/c”, where E is the electric field amplitude of the wave, and c is the speed of light in a vacuum, approximately 3 x 108 m/s. When a plane electromagnetic wave moves in a certain direction, its electric and magnetic fields oscillate perpendicular to the direction of wave propagation and to each other, a characteristic of a transverse wave. The strength of the magnetic field is directly proportional to the electric field and inversely proportional to the speed of light.
For example, if an electromagnetic wave has an electric field magnitude of 6.0 V/m, the magnitude of the magnetic field B can be found using the equation B = E/c. Similarly, for a more complex expression such as E = (6.0 × 10⁻³ V/m) sin(27 × 1¹/m - ωt), the associated magnetic field can be derived using the same principle, adjusted for the wave's specific amplitude, frequency, and phase.
Final answer:
The equation for the magnetic field component of an electromagnetic wave includes the electric field strength, the speed of light, and the permeability of free space.
Explanation:
The quantities that appear in the equation for the magnetic field component of an electromagnetic wave are the electric field strength (E), the speed of light (c), and the permeability of free space (μ₀). In the context of an electromagnetic wave traveling in a vacuum, the magnetic field component B can be expressed as B = E/c, assuming that both E and B are at right angles to the direction of propagation of the wave and to each other as per the transverse nature of electromagnetic waves. This relationship indicates that the magnetic field strength is directly proportional to the electric field strength and inversely proportional to the speed of light, highlighting the inherent interdependence between electric and magnetic fields in electromagnetic waves.
1) Which is a characteristic of a metal? A) It looks dull. B) It feels brittle. C) It is a good insulator. D) It is a good conductor of electricity. 2)
D) It is a good conductor of electricity
A single circular loop with a radius of 35 cm is placed in a uniform external magnetic field with a strength of 0.50 T so that the plane of the coil is perpendicular to the field. The coil is pulled steadily out of the field in 15 s. Find the magnitude of the average induced emf during this interval. Show all work and include units of measure.
Answer:
0.0129 V
Explanation:
The average emf induced in the coil is given by Faraday-Newmann-Lenz:
[tex]\epsilon=-\frac{\Delta \Phi_B}{\Delta t}[/tex]
where
[tex]\Delta \Phi_B[/tex] is the variation of magnetic flux through the coil
[tex]\Delta t = 15 s[/tex] is the time interval
Here we have
B = 0.50 T is the strength of the magnetic field
The radius of the coil is
r = 35 cm = 0.35 m
So the area is
[tex]A=\pi r^2 = \pi (0.35 m)^2=0.385 m^2[/tex]
The initial flux through the coil is
[tex]\Phi_i = BA = (0.50 T)(0.385 m^2)=0.193 Wb[/tex]
while the final flux is zero, since the coil has been completely removed from the magnetic field region; so, the variation of magnetic flux is
[tex]\Delta \Phi = \Phi_f - \Phi_i = -0.193 Wb[/tex]
and so, the average emf induced is
[tex]\epsilon=-\frac{-0.193 Wb}{15 s}=0.0129 V[/tex]
A 0.50-kg ball that is traveling at 6.0 m/s collides head-on with a 1.00-kg ball moving in the opposite direction at a speed of 12.0 m/s. The 0.50-kg ball bounces backward at 14 m/s after the collision. Find the speed of the second ball after the collision.
Answer:
-2 m/s
Explanation:
Assuming the collision is elastic, the total momentum must be conserved:
[tex]p_i = p_f[/tex]
[tex]m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2[/tex]
where
m1 = 0.50 kg is the mass of the first ball
m2 = 1.00 kg is the mass of the second ball
u1 = +6.0 m/s is the initial velocity of the first ball
u2 = -12.0 m/s is the initial velocity of the second ball
v1 = -14 m/s is the final velocity of the first ball
v2 = ? is the final velocity of the second ball
By re-arranging the equation, we can find the final velocity of the 1.00 kg ball:
[tex]v_2 = \frac{m_1 u_1 + m_2 u_2 - m_1 v_1}{m_2}=\frac{(0.50 kg)(6.0 m/s)+(1.00 kg)(-12 m/s)-(0.50 kg)(-14 m/s)}{1.00 kg}=-2 m/s[/tex]
which means, 2 m/s in the same direction the second ball was travelling before the collision.
The speed of the second ball is 2 m/s
To calculate the speed of the second ball after the collision, we use the formula below.
Formula:
mu+m'u' = mv+m'v'................. Equation 1Where:
m = mass of the first ballu = initial speed of the first ballm' = mass of the second ballu' = initial speed of the second ballv = final speed of the first ballv' = final speed of the second ball.From the question,
Given:
m = 0.50 kgm' = 1.00 kgu = 6 m/su' = -12 m/s (opposite)v = -14 m/s. (backward)Substitute these values into equation 1
(0.5×6)+[1(-12)] = [0.5×(-14)]+(1×v')Solve for v'
(3-12) = -7+v'v'-7 = -9v' = -9+7v' = -2 m/sNote: The negative sign can be ignored in the final answer
Hence, the speed of the second ball is 2 m/s.
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please help on this one?
Speed because all waves in the electromagnetic spectrum travel at the speed of light
Atoms of the same element with different numbers of neutrons are called
Answer: isotopes
An isotope is an atom of the same chemical element that has the same number of protons in its nucleus, but not the same number of neutrons.
Therefore, this isotope atom will have the same atomic number and will be located in the same place of the periodic table of the main chemical element, but different mass number.
In other words, the isotopes of a chemical element have a different mass although they have the same atomic number, which is directly related to the number of protons in the nucleus of the atom.
Final answer:
Atoms of the same element that possess different numbers of neutrons are known as isotopes, and these isotopes have the same atomic number but a different mass number.
Explanation:
Atoms of the same element with different numbers of neutrons are called isotopes. For instance, within hydrogen, there are notable examples like deuterium and tritium, which respectively have one and two neutrons, while the most common hydrogen isotope has none. Isotopes of an element share the same atomic number due to having an identical number of protons but possess a different mass number, which is the sum of protons and neutrons in the atom's nucleus.
How do you find the approximate frequency for this graph?
Answer:
Explanation:
Good thing it says approximately. Your best guess is just a guess. It takes 1 hz (or 1 sine wave) about 0.62 seconds before it begins to repeat itself.
The frequency is # of cycles / time = 1 hz / 0.62 = 1.61 hz/second.
The decay constant of a radioactive nuclide is 4.6 x 10^-3 s^-1. What is the half-life of the nuclide?
2.5 min
2.0 min
1.4 min
3.6 min
3.1 min
Answer:
2.5s
Explanation:
Given parameters:
Decay constant of the nuclide = 4.6 x 10⁻³s⁻¹
Unkown:
Half life =?
The half life is the time take for half of the radioactive nuclei to disintegrate. It is mathematically expressed as:
Half life = 0.693/λ
Where λ is the decay constant
Since we have been given λ we plug it into the equation:
Half life = 0.693/4.6 x 10⁻³
Half life = 0.693/0.0046
Half life =150. 65s
Now we convert to minutes because the options are in minutes:
60s = 1 minute
150.65s = 150.65/60 = 2.5s
The half life of the nuclide is 2.5s
hotspots in a microwave oven are caused by the formation of standing waves within the cooking chamber. What is the separation between hotspots in a 2.45 GHz microwave oven?
Answer:
0.122 m
Explanation:
We can imagine the hotspots in the microwave oven to correspond to the crests of the waves, so the distance between two hotspot is just the wavelength of the waves.
The wavelenght of an electromagnetic wave is given by
[tex]\lambda=\frac{c}{f}[/tex]
where
[tex]c=3.0\cdot 10^8 m/s[/tex] is the speed of light
f is the frequency of the wave. In this case,
[tex]f=2.45 GHz = 2.45 \cdot 10^9 Hz[/tex]
so the wavelength is
[tex]\lambda=\frac{3\cdot 10^8 m/s}{2.45\cdot 10^9 Hz}=0.122 m[/tex]
The separation between hotspots in a 2.45 GHz microwave oven is approximately half the wavelength of the microwaves, which is around 6.1 cm. This distance is derived from the speed of light and the frequency of the microwaves used.
Explanation:The separation between hotspots in a microwave oven can be determined by understanding the wavelength of the microwaves used, which for most household microwave ovens is set at a frequency of 2.45 GHz. When these microwaves are generated within the oven, they reflect off the interior walls, creating standing waves. At points where the crests or troughs of these microwaves align, hotspots are formed due to the concentration of energy.
This heating effect occurs because microwaves at a frequency of 2.45 GHz have the right amount of energy to cause polar molecules, like water, to rotate, thereby transferring kinetic energy into thermal energy. The dipole nature of water molecules allows for efficient energy transfer, specifically heating the areas where these molecules are present, such as in food.
To find the separation between the hotspots, we use the relationship between the speed of light (c), frequency (f), and wavelength (λ) given by c = λ * f. With c being approximately 3 x 108 meters per second and f being 2.45 GHz, we can calculate the wavelength and thus the distance between hotspots. The speed of the microwaves is the speed of light, and so the wavelength (λ) for 2.45 GHz radiation would be about 0.122 meters (12.2 cm), hence the hotspots would be separated by approximately half a wavelength, around 6.1 cm.
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A step-down transformer providing electricity for a residential neighborhood has exactly 2750 turns in its primary. When the potential difference across the primary is 5670 V, the potential difference at the secondary is 240 V. How many turns are in the secondary? Show all work.
Answer:
116
Explanation:
The transformer equation states that:
[tex]\frac{V_p}{V_s}=\frac{N_p}{N_s}[/tex]
where
Vp is the voltage in the primary coil
Vs is the voltage in the secondary coil
Np is the number of turns in the primary coil
Ns is the number of turns in the secondary coil
Here we have:
Vp = 5670 V
Np = 2750
Vs = 240 V
So we can solve the formula to find Ns:
[tex]N_s = N_p \frac{V_s}{V_p}=(2750)\frac{240 V}{5670 V}\sim 116[/tex]
So, the secondary coil has 116 turns.
The wavelengths of light emitted by a firefly span the visible spectrum but have maximum intensity near 550 nm. A typical flash lasts for 100 ms and has a power of 1.2 mW. If we assume that all of the light is emitted at the peak-intensity wavelength of 550 nm, how many photons are emitted in one flash?
To calculate the number of photons emitted in one flash, use the formula: Number of photons = Energy of the flash / Energy of one photon. By using the given power and duration of the flash, you can calculate the energy of the flash and then calculate the energy of one photon using the wavelength. Finally, divide the energy of the flash by the energy of one photon to find the number of photons emitted.
Explanation:To calculate the number of photons emitted in one flash, we need to use the formula:
Number of photons = Energy of the flash / Energy of one photon
First, let's calculate the energy of the flash. The power of the flash is given as 1.2 mW, and it lasts for 100 ms. Using the formula P = E/t, we can find the energy of the flash as 1.2 mJ.
Next, we need to calculate the energy of one photon. The energy of a photon can be calculated using the formula E = hc/λ, where E is the energy, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength.
The wavelength is given as 550 nm, which is equivalent to 550 x 10^-9 m. Plugging in the values, we can calculate the energy of one photon as 3.606 x 10^-19 J.
Now, we can use the formula to calculate the number of photons:
Number of photons = 1.2 x 10^-3 J / 3.606 x 10^-19 J = 3.328 x 10^15 photons
Therefore, approximately 3.328 x 10^15 photons are emitted in one flash.
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To calculate the number of photons emitted in one flash, we can use the relationship between power and energy. The energy of a single photon can be determined using the wavelength. The number of photons emitted in one flash is approximately 3.33 x 10^14 photons.
Explanation:To calculate the number of photons emitted in one flash, we need to use the relationship between power and energy of a photon. Power is given as 1.2 mW and since the flash lasts for 100 ms, we can calculate the energy of the flash as:
E = P x t = (1.2 mW) x (100 ms) = 0.12 mJ
To find the number of photons, we can use the formula:
Number of photons = Energy of flash / Energy of a single photon
The energy of a single photon can be determined using the relationship between energy and wavelength:
Energy = hc / λ
Where h is Planck's constant, c is the speed of light, and λ is the wavelength. Plugging in the values:
Energy = (6.63 x 10^-34 J.s) x (3 x 10^8 m/s) / (550 x 10^-9 m) ≈ 3.6 x 10^-19 J
Finally, plugging the values into the formula:
Number of photons = (0.12 x 10^-3 J) / (3.6 x 10^-19 J) ≈ 3.33 x 10^14 photons
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