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Svet_ta [14]
3 years ago
15

A 5.00-kg box slides 6.50 m across the floor before coming to rest. what is the coefficient of kinetic friction between the floo

r and the box if the box had an initial speed of 3.00 m/s?
Physics
1 answer:
antoniya [11.8K]3 years ago
7 0
Work=Fd and is equal to the change in energy, friction does work to stop the box. The box's kinetic energy is initially (K=.5m(v^2)) 22.5J and ends up at 0J. Friction does 22.5J of work over 6.5m. 22.5J=(F)(6.5m), F=3.46N. The force of friction is equal to the coefficient of friction times the normal force of an object, F=μFⁿ. We can determine the normal force by multiplying mass and gravity, Fⁿ=50N. Bringing this back to the force applied by friction, 3.46N/50N=μ=0.0692.
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The acceleration of an object is constant when its velocity is :
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A sealed test tube traps 25.0 cm3 of air at a pressure of 1.00 atm and temperature of 18°C. The test tube’s stopper has a diamet
puteri [66]

Answer:

180° C

Explanation:

First we start by finding the area of the stopper.

A = πd²/4, where d = 1.5 cm = 0.015 m

A = 3.142 * 0.015² * ¼

A = 1.767*10^-4 m²

Next we find the force on the stopper

F = (P - P•)A, where

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10 = (P - 101325) * 1.767*10^-4

P - 101325 = 10/1.767*10^-4

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P1/T1 = P2/T2, and when we substitute the values, we have

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3 0
4 years ago
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