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hichkok12 [17]
2 years ago
10

All objects on earth, whether moving or stationary, are acted upon by which force?

Physics
1 answer:
frutty [35]2 years ago
5 0

Answer:

air resistance

Explanation:

hope it will help

Brainlists please

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Which has more kinetic energy a bowling ball or a soccer ball
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A bowling ball because it is heavier  and it has more air force going against it<span />
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You have an initial velocity of -3.0 m/s. You then experience an acceleration of 2.5 m/s2 for 9.0s; what
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27.9 idkkkk look it up on photomath
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A copper sphere was moving at 40 m/s when it hit another object. This caused all of the KE to be converted into thermal energy f
USPshnik [31]

Answer:

Temperature increase = 2.1 [C]

Explanation:

We need to identify the initial data of the problem.

v = velocity of the copper sphere = 40 [m/s]

Cp = heat capacity = 387 [J/kg*C]

The most important data given is the fact that when the shock occurs kinetic energy is transformed into thermal energy, therefore it will have to be:

E_{k}=Q\\ E_{k}= kinetic energy [J]\\Q=thermal energy [J]\\Re-employment values and equalizing equations\\\\\frac{1}{2} *m*v^{2}=m*C_{p}*dT  \\The masses are canceled \\\\dT=\frac{v^{2}}{C_{p} *2} \\dT=2.1 [C]

8 0
3 years ago
Calculate the distance travel if it accelerates from 0 to 27.8 meters per second in 2.5 seconds
luda_lava [24]

The distance travel is 69.5 meters.

<u>Explanation:</u>

Given datas are as follows

Speed = 27.8 meters / second

Time = 2.5 seconds

The formula to calculate the speed using distance and time is

Speed = Distance ÷ Time (units)

Then Distance = Speed × Time (units)

Distance = (27.8 × 2.5) meters          

Distance = 69.50 meters

Therefore the distance travelled is 69.50 meters.

8 0
3 years ago
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!

6 0
3 years ago
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