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Paul [167]
3 years ago
15

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.

|Huntrw6|

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A soccer ball is kicked and left
Vedmedyk [2.9K]

Answer:

Explanation:

Considering that this is parabolic motion, we know that the time the ball is in the air begins the instant it leaves the ground, reaches up to its max height, and then begins falling until it reaches the ground. Duh, right? Some important things happen during this trip. There are a few things we need to know in order to even begin the problem. Parabolic motion has x and y coordinates because it is 2-dimmensional; the acceleration in the x dimension is not the same as the acceleration in the y dimension; the velocity of an object at its max height is always 0; the time it takes to reach its max height (where the max height is half the distance the object travels) is half the time it takes to make the whole trip. Yikes. That's a lot to know and much to remember! Don't you just LOVE physics!?

For a. the hang time is the time the ball was in the air. Some of that stuff we talked about above is pertinent to solving this problem. We know that the velocity of the ball is 0 at its max height, and we also know that if we find the time it takes to reach its max height, we can double that number to find how long it was in the air for the whole trip. Use the one-dimensional equation

v=v_0+at to find out how long it took to reach the max height. Even though we don't yet know the max height, we DO know that the velocity at that point is 0. BUT before we do that, since we are working in the y-dimension only, it would behoove us (benefit us) to find the velocity particular to this dimension. We are going to answer c. first, then backtrack.

c. wants the initial vertical velocity. That is found in the magnitude of the "blanket" or generic velocity times the sin of the angle, namely:

V_y=25sin(45) so

V_y= 18 m/s Now we can use that as the initial upwards velocity in part a:

v=v_0+at and filling in:

0 = 18 + (-9.8)t and

-18 = -9.8t so

t = 1.8 seconds. But remember, this is only half the time it was in the air. The whole trip, then, takes 2(1.8) which is

t = 3.6 seconds

That's a and c. Now for b:

b. asks for the x component of the velocity:

V_x=Vcos\theta which works out to be the same as the vertical velocity, since the sin and cos of 45 degrees is the same:

V_x=25cos45 and

V_x= 18 m/s

Onto d:

d. wants the max height. Remember, it took 1.8 seconds to get to the max height, so using yet another one-dimensional equation:

Δx = v₀t + \frac{1}{2}at^2 where Δx is the displacement, v₀ is the initial upwards velocity, a is the pull of gravity, and t is the time it takes to reach that max height (Δx, our unknown). Filling in:

Δx = 18(1.8)+\frac{1}{2}(-9.8)(1.8)^2 and if you do the rounding correctly, you'll end up with this:

Δx = 32 - 16 so

the max height, Δx, is 16 meters.

e. wants the range. That translates to the distance the ball traveled. This is found in a glorified version of d = rt, where d is displacement, r is velocity, and t is...well, time (that doesn't change):

Δx = vt so

Δx = 18(3.6) remember that the ball was in the air for a total of 3.6 seconds, so

Δx = 65 meters.

Phew!!!!! That's a lot! I suggest you learn your physics or this will make you insane by the end of the course!

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