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DIA [1.3K]
3 years ago
12

A 79.7 kg base runner begins his slide into second base while moving at a speed of 4.77 m/s. The coefficient of friction between

his clothes and Earth is 0.635. He slides so that his speed is zero just as he reaches the base. The acceleration of gravity is 9.8 m/s 2 . What is the magnitude of the mechanical energy lost due to friction acting on the runner? Answer in units of J.
Physics
1 answer:
SashulF [63]3 years ago
6 0

To solve this problem we will apply the concept related to the kinetic energy theorem. Said theorem states that the work done by the net force (sum of all forces) applied to a particle is equal to the change experienced by the kinetic energy of that particle. This is:

\Delta W = \Delta KE

\Delta W = \frac{1}{2} mv^2

Here,

m = mass

v = Velocity

Our values are given as,

m = 79.7kg

v = 4.77m/s

Replacing,

\Delta W = \frac{1}{2} (79.7kg)(4.77m/s)^2

\Delta W = 907J

Therefore the mechanical energy lost due to friction acting on the runner is 907J

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Answer:

C. It is caused by flowing negatively charged particles.

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The force is equal at all points in the field as the field lines are parallel in a uniform field. So, negatively charged particles will move in the opposite direction to the current.

Hence, the correct answer is "C. It is caused by flowing negatively charged particles."

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2 years ago
The Milky Way Galaxy is (a) another name for our solar system; (b) a small group of stars visible in our night sky; (c) a collec
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2 years ago
Raindrops fall vertically at 7.5 m/s relative to the Earth. What does an observer in a car moving at 20.2 m/s in a straight line
Vilka [71]

Answer:

vDP = 21.7454 m/s

θ = 200.3693°

Explanation:

Given

vDE = 7.5 m/s

vPE = 20.2 m/s

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Assume that

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vPE to be in direction of i

According to relative motion concept the velocity vDP is given by

vDP = vDE - vPE     (I)

Substitute in (I) to get that

vDP = - 7.5 j - 20.2 i

The magnitude of vDP is given by

vDP = √((- 7.5)²+(- 20.2)²) m/s =  21.7454 m/s

θ = Arctan (- 7.5/- 20.2) = 20.3693°

θ is in 3rd quadrant so add 180°

θ = 20.3693° + 180° = 200.3693°

4 0
3 years ago
To accelerate, an object must *
Sonja [21]
Answer: Not 100% sure but I think it’s C.

Hope this helps! ^^
4 0
3 years ago
The critical angle between a medium and air is 430. What is the speed of light in that medium?
vagabundo [1.1K]

Answer:

The speed of light is that medium is 281907786.2 m/s.

Explanation:

since the critical angle is Фc = 430, we know that the refractive index is given by:

n = 1/sin(Фc)

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then if n is the refractive index of the medium and c is the speed of light, then the speed of light in the medium is given by:

v = c/n

  = (3×10^8)/(1.06)

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Therefore, the speed of light is that medium is 281907786.2 m/s.

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2 years ago
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