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3 years ago

Measurements of two electric currents are shown in the chart.

Physics

2 answers:

Pani-rosa [81]3 years ago

6 0

Answer

Correct answer is B. Current Y has a greater potential difference, and the charges flow at a slower rate.

Explanation

Current Y has a potential difference of 9 volts and current X has a potential difference of 1.5 volts. This tell us that current Y has a greater potential difference.

A current is the rate of low of charge. Current X is 7.8 A which is bigger than current Y which is 0.5 A. This tell us that current X flows faster than current Y.

From these statements, we conclude that Current Y has greater potential difference than current X and the charges in current Y flow at a slower rate. The correct answer is B.

lana [24]3 years ago

4 0

The correct answer is:

B. Current Y has a greater potential difference, and the charges flow at a slower rate.

The statement best compares the two currents.

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4 years ago

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22

alexgriva [62]

1) The law of motion of the projectile is
$h(t) = 2+22.5 t-4.9 t^2$
To find the velocity, we should compute the derivative of h(t):
$v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t$
So now we can calculate the speed at t=2 s and t=4 s:
$v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s$
$v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s$
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.

2) The projectile reaches its maximum height when the speed is equal to zero:
$v(t)=0$
So we have
$22.5-9.8 t=0$
And solving this we find
$t=2.30 s$

3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
$h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m$

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
$2+22.5t -4.9 t^2 = 0$
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
$t=4.68 s$
And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
$v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36 m/s$
with negative sign, because it is directed downward.

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4 years ago