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il63 [147K]
2 years ago
14

. An unbalanced force of 500 N is applied to a 75 kg object. What is the acceleration of the object?

Physics
1 answer:
Hoochie [10]2 years ago
5 0

The acceleration of the object is 6.7 m/s^2

Explanation:

We can solve the problem by using Newton's second law, which states that the net force exerted on an object is equal to the product between the mass of the object and its acceleration:

F=ma

where

F is the net force

m is the mass of the object

a is its acceleration

For the object in this problem,

F = 500 N is the applied force

m = 75 kg is the force

Solving the equation for a, we find the acceleration:

a=\frac{F}{m}=\frac{500}{75}=6.7 m/s^2

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

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marshall27 [118]

Answer:

65 m/s

Explanation:

v=v0+at <=> v = 11 + 12 t ∧ t = 4.5 s <=> v = 11 + 12×4.5 <=> v = 65 m/s

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A 52.0-kg person, running horizontally with a velocity of +3.63 m/s, jumps onto a 15.2-kg sled that is initially at rest. (a) Ig
trasher [3.6K]

Answer:

The coefficient of kinetic friction between the sled and the snow is 0.0134

Explanation:

Given that:

M = mass of person = 52 kg

m = mass of sled = 15.2 kg

U = initial velocity of person = 3.63 m/s

u = initial velocity of sled = 0 m/s

After collision, the person and the sled would move with the same velocity V.

a) According to law of momentum conservation:

Total momentum before collision = Total momentum after collision

MU + mu = (M + m)V

V=\frac{MU+mu}{M+m}

Substituting values:

V=\frac{MU+mu}{M+m}=\frac{52(3.63)+15.2(0)}{52+15.2} =2.81m/s

The velocity of the sled and person as they move away is 2.81 m/s

b) acceleration due to gravity (g) = 9.8 m/s²

d = 30 m

Using the formula:

V^2=2\mu(gd)\\\mu=\frac{V^2}{2gd} \\\mu=\frac{2.81^2}{2*9.8*30} =0.0134

The coefficient of kinetic friction between the sled and the snow is 0.0134

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Answer:

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