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pentagon [3]
3 years ago
6

When a ball rolls downhill, the rolling motion results in the ball ____________.

Physics
2 answers:
Flura [38]3 years ago
8 0

Answer:

the answer is b!!

Explanation: i got it wrong grrr. but it told me the correct answer

emmainna [20.7K]3 years ago
7 0

I am pretty sure it is A Becoming warm

Since it’s moving and causing friction which makes it warm


Hope this helps

Mark me brainliest

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The speed of sound in air on a certain stormy day is about 1130 feet per second. you see a flash of lightning striking some dist
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Answer  4,520 feet   Explanation The light and the sound were produced at the same time but sound took long to reach since is speed is much lower than the speed of light. Speed is the rate of change of distance.  So; Distance = speed × time                 = 1130 × 4                  = 4,520 feet.

6 0
3 years ago
(HELP QUICK!!) Air pressure is a result of _____
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I’m so sorry if it’s wrong, but I think it’s
A. Winds
7 0
3 years ago
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Determine the force p required to (a) raise and (b) lower the 40-kg cylinder at a slow, steady speed. the coefficient of frictio
Norma-Jean [14]

Answer:

T - M g - Ff = M a     describes the acceleration of the object

T = M g + Ff      if the object moves at constant speed  (a = zero)

Ff = μ M g = .3 * 40 * 9.8 = 118 N     the force of friction

Ff will be in a direction opposite to the motion

M g = 40 kg * 9.8 m/s = 392 N

a) T = M g + Ff   for an object moving upwards

T = 392 + 118 = 510 N

b) T = M g - Ff     for an object moving downwards

T = 392 - 118 = 274 N

8 0
2 years ago
A hockey puck with mass 0.3 kg is shot across an ice-covered pond. Before the hockey puck was hit, the puck was at rest. After t
JulsSmile [24]

Answer:

The net friction force is 8.01 N

Explanation:

Net friction force = mass of hockey puck × acceleration

From the equations of motion

v^2 = u^2 + 2as

v = 40 m/s

u = 0 m/s (puck was initially at rest)

s = 30 m

40^2 = 0^2 + 2×a×30

60a = 1600

a = 1600/60 = 26.7 m/s^2

The acceleration of the puck is 26.7 m/s^2

Net friction force = 0.3 × 26.7 = 8.01 N

3 0
4 years ago
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If the magnitude of the magnetic force on a proton is F when it is moving at 14.0 o with respect to the field, what is the magni
sergeinik [125]

Answer:

The magnitude of the magnetic force at 32.5⁰ to the field, is 2.22 F

Explanation:

Given;

magnitude of the magnetic force is F at 14.0⁰ with respect to the field.

To determine the magnitude of the magnetic force at 32.5⁰ to the field, we apply the following formula and solve with proportion;

f*Sin(14.0⁰) = F

f*Sin(32.5⁰) = ?

= \frac{f*Sin(32.5^o)}{f*Sin(14.0^o)}.F\\\\ =2.22 \ F

Therefore, the magnitude of the magnetic force at 32.5⁰ to the field, is 2.22 F

6 0
4 years ago
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