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vladimir1956 [14]

2 years ago

11

A push on a 1-kg brick accelerates it. Neglecting friction, equally accelerating a 10-kg brick requires 10 times as much force.

Using your knowledge of Newton’s Second Law, explain why this is correct.

1 answer:

sergij07 [2.7K]2 years ago

6 0

It is CVM this is simple

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VLD [36.1K]

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How do you transfer kinetic energy to electrical energy using a generator system

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2 years ago

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A compact car, mass 660 kg, is moving at 1.00 ✕ 102 km/h toward the east. The driver of the compact car suddenly applies the bra

Maslowich

Answer: $59.43\times 10^3 kg-m/s$

Explanation:

Given

mass of car $m=660 kg$

Initial velocity of car $=102 km/h\approx 28.33 m/s$ towards east

Time taken to stop $t=2.1 s$

Force exerted $F_{avg}=4.8\times 10^3 N$

change in momentum is given by impulse imparted to the car

$Impulse(J)=-F\cdot t$

$J=-4.8\times 10^3\times 2.1$

$J=-59.49\times 10^3 kg-m/s$

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6 0

3 years ago

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A 25kg chair initially at rest on a horizontal floor requires 165 N force to set it in motion. Once the chair is in motion, a 12

bazaltina [42]

The coefficient of static friction between the chair and the floor is 0.67

Explanation:

Given:

Weight of the chair = 25kg

Force = 165 N (F_applied)

Force = 127 N (F_max)

To find: Coefficient of static friction

The “coefficient of static friction” between a chair and the floor is defined as the ration of maximum force to the normal force acting on the chair

μ_s= $F_{max}/F_{n}$

The F_n is equal to the weight multiplied by its gravity

∴ $F_{n}$ =mg

Thus the coefficient of static friction changes as

μ_s= $F_{max}/mg$

μ_{s} = $=165N/((25kg)\times(9.80 m/s^2 ) )$

= 0.67

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2 years ago

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Answer:

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Explanation:

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