Answer:A
Explanation: number that shows the total atomic mass of the substance
The magnitude of the average emf induced in the loop is given by (we ignore the signs since we are interested only in the magnitude)
where
is the variation of magnetic flux through the area enclosed by the loop, and
is the time interval.
The magnetic flux is given by
where B is the intensity of the magnetic field, A is the area enclosed by the loop and
is the angle between the perpendicular to the area and the magnetic field. In our problem, this angle is zero because the loop is perpendicular to the magnetic field, so the cosine is 1. The area of the loop is fixed, and it is
where
is the radius of the loop. The only element which is variable in the formula is B, which changes from 0.069 T to -0.044 T (opposite direction). So we can rewrite the flux variation as
where
By using
, we can find the magnitude of the emf induced:
When a heavy football player and a light one run into each other, does the lighter player really exert as much force on the heavy player s the heavy player exerts on the light one. Yes. The interaction between the two players, the force each exerts on the other have equal strength.
Hope this helps you!!!
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Answer: 6.48m/s
Explanation:
First, we know that Impulse = change in momentum
Initial velocity, u = 19.8m/s
Let,
Velocity after first collision = x m/s
Velocity after second collision = y m/s
Also, we know that
Impulse = m(v - u). But then, the question said, the guard rail delivered a "resistive" impulse. Thus, our impulse would be m(u - v).
5700 = 1500(19.8 - x)
5700 = 29700 - 1500x
1500x = 29700 - 5700
1500x = 24000
x = 24000/1500
x = 16m/s
Also, at the second guard rail. impulse = ft, so that
Impulse = 79000 * 0.12
Impulse = 9480
This makes us have
Impulse = m(x - y)
9480 = 1500(16 -y)
9480 = 24000 - 1500y
1500y = 24000 - 9480
1500y = 14520
y = 14520 / 1500
y = 9.68
Then, the velocity decreases by 3.2, so that the final velocity of the car is
9.68 - 3.2 = 6.48m/s
Answer:
The particle comes to rest before reaching the position x=4m.
Explanation:
When the object is at x=0m, all of its energy is kinetic energy. Using the equation for kinetic energy yields KE=1/2mv^2=(12)(2)(3)^2=9J. Using the given equation for potential energy when the object is at x=4m yields U=4x^2=4(4)^2=64J. Since the system only has 9J of energy, the object comes to rest before reaching x=4m.