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Maurinko [17]
3 years ago
15

Which statement is true regarding the ping-pong ball at the bottom of the ramp in the experiment, “Rolling Along”? Question 5 op

tions: Only potential energy from the ball coming down the ramp will be given to the ping-pong ball at the bottom of the ramp. All kinetic energy remains with the ball that is coming down the ramp- none of it transfers to the ping-pong ball at the bottom. No energy was involved in this experiment. The kinetic energy of the ball coming down the ramp will be given to the ping-pong ball at the bottom of the ramp.
Physics
1 answer:
Gwar [14]3 years ago
7 0

Hi there!

The answer to your question is A

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A hunter on level ground fires a bullet at an angle of 2 degrees below the horizontal while simultaneously dropping another bull
Marizza181 [45]

If hunter fires a bullet 2 degree below the horizontal

then the components of its speed is given as

v_x = v_o cos2

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While the other bullet is dropped downwards so its initial speed will be zero

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3 years ago
A metal ring 5.00 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpend
labwork [276]

Answer:

Ein: 2.75*10^-3 N/C

Explanation:

The induced electric field can be calculated by using the following path integral:

\int E_{in} dl=-\frac{\Phi_B}{dt}

Where:

dl: diferencial of circumference of the ring

circumference of the ring = 2πr = 2π(5.00/2)=15.70cm = 0.157 m

ФB: magnetic flux = AB (A: area of the loop = πr^2 = 1.96*10^-3 m^2)

The electric field is always parallel to the dl vector. Then you have:

E_{in}\int dl=E_{in}(2\pi r)=E_{in}(0.157m)

Next, you take into account that the area of the ring is constant and that dB/dt = - 0.220T/s. Thus, you obtain:

E_{in}(0.157m)=-A\frac{dB}{dt}=-(1.96*10^{-3}m^2)(-0.220T/s)=4.31*10^{-4}m^2T/s\\\\E_{in}=\frac{4.31*10^{-4}m^2T/s}{0.157m}=2.75*10^{-3}\frac{N}{C}

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