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nadezda [96]
3 years ago
12

You are pulling a 80 kg box with a rope. The force you exert on the box is 90 N at 30 degrees from the ground. What would be the

coefficient of kinetic friction if the box is moving at constant velocity?
Physics
1 answer:
Feliz [49]3 years ago
7 0

Answer:

Coefficient of dynamic friction= md= 0.09931

Explanation:

To determine the coefficient of dynamic friction we must first match the friction force that is permendicular to the normal force of the block and opposite to the drag force, to the component of the drag force in this same direction. This component on the X axis of the drag force will be:

F= 90N × cos(30°) = 77.9423N

This component on the X axis of the drag force must be equal to the dynamic friction force that is equal to the coefficient of dynamic friction by the normal force of the block weight:

F= md × m × g= 77.9423N

m= mass of the block

md= coefficient of dynamic friction

g= gravity acceleration

F= md × 80kg× 9.81 (m/s²)= 77.9423(kg×m/s²)

md= (77.9423(kg×m/s²) / 784.8 (kg×m/s²)) = 0.09931

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An electron is initially moving at 1.4 x 107 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude 120 N
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Answer:

K.E = 15.57 x 10⁻¹⁷ J

Explanation:

First, we find the acceleration of the electron by using the formula of electric field:

E = F/q

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but, from Newton's 2nd Law:

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Comparing both equations, we get:

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K.E = (0.5)(9.1 x 10⁻³¹ kg)(1.85 x 10⁷ m/s)²

<u>K.E = 15.57 x 10⁻¹⁷ J</u>

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