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Anna11 [10]
2 years ago
11

Anyone know how to do this?

Physics
1 answer:
MrMuchimi2 years ago
7 0

The voltage from one side of the battery all the way around to the other side of the battery is 12v .

If 4 of those volts show up across the circle-thing, then the rest of the 12v ... 8v ... Must show up across the set of parallel rectangles.

To get that answer, I subtracted the 4 from the 12.

Just like it says in choice-C.

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A wave traveling in water has a frequency of 250 Hz and a wavelength of 6.0m. What is the speed of the wave?
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Assuming a 8 kilogram bowling ball moving at 2 m/s bounces off a spring at the same speed that had before bouncing what is the a
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a) 32 kg m/s

Assuming the spring is initially at rest, the total momentum of the system before the collision is given only by the momentum of the bowling ball:

p_i = m u = (8 kg)(2 m/s)=16 kg m/s

The ball bounces off at the same speed had before, but the new velocity has a negative sign (since the direction is opposite to the initial direction). So, the new momentum of the ball is:

p_{fB}=m v_b =(8 kg)(-2 m/s)=-16 kg m/s

The final momentum after the collision is the sum of the momenta of the ball and off the spring:

p_f = p_{fB}+p_{fS}

where p_{fS} is the momentum of the spring. For the conservation of momentum,

p_i = p_f\\p_i = p_{fB}+p_{fS}\\p_{fS}=p_i -p_{fB}=16 kg m/s -(-16 kg m/s)=32 kg m/s


b) -32 kg m/s

The change in momentum of bowling ball is given by the difference between its final momentum and initial momentum:

\Delta p=p_{fb}-p_i=-16 kg m/s - 16 kg m/s=-32 kg m/s


c) 64 N

The change in momentum is equal to the product between the average force and the time of the interaction:

\Delta p=F \Delta t

Since we know \Delta t=0.5 s, we can find the magnitude of the force:

F=\frac{\Delta p}{\Delta t}=\frac{-32 kg m/s}{0.5 s}=-64 N

The negative sign simply means that the direction of the force is opposite to the initial direction of the ball.


d) The force calculated in the previous step (64 N) is larger than the force of 32 N.

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Chiefly in science fiction) an invisible barrier of exerted strength or impetus.
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