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grigory [225]
3 years ago
9

What is the kinetie energy of a 3-kilogram ball that is rolling at 2 meters per second?

Physics
1 answer:
Sedaia [141]2 years ago
7 0
KE=1/2m v2= 1/2 (3kg) (2m/s)2 = 6J (6joules)
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SCIENCE<3:: will give brainliest! :)
Alex777 [14]

Answer:

Option two

Explanation:

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2 years ago
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What causes a objects to move or stay still? claim and evidence
jek_recluse [69]

Answer:

A force

Explanation:

A push or a pull is an example of a force and can cause an object to speed up, slow down, etc..  Newton's laws tell us that 1- an object will not change its motion unless a force acts on it 2- the force on an object is equal to its mass times its acceleration. 3- The third law states that for every action (force) in nature there is an equal and opposite reaction.. However, forces like gravity and friction can resist movement.

4 0
3 years ago
You are driving your motorcycle in a circle of radius 75 m on wet pavement. what is the fastest you can go before you lose tract
stira [4]

On driving your motorcycle in a circle of radius 75 m on wet pavement, the fastest you can go before you lose traction, assuming the coefficient of static friction is 0.20 is 147m/s

Friction helps to maintain the slipping of the vehicle on the road hence lays a very important role.

Maximum velocity of a road with friction is given by the formula,

v = μRg

where, v is the maximum velocity

μ is the coefficient of static friction

R is the radius of the circle road

g is the acceleration due to gravity

Given,

μ = 0.20

R = 75m

g = 9.8m/s²

On substituting the given values in the above formula,

v = 0.20× 75 ×9.8

v = 147m/s

So, the Maximum velocity of the wet road is 147m/s.

Learn more about Velocity here, brainly.com/question/18084516

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3 0
2 years ago
Calculate the net force on particle q1. First, find the direction of the force particle q2 is exerting on particle q1. Is it pus
ValentinkaMS [17]

The net force on particle particle q1 is 13.06 N towards the left.

<h3>Force on q1 due to q2</h3>

F(12) = kq₁q₂/r₂

F(12) = (9 x 10⁹ x 13 x 10⁻⁶ x 7.7 x 10⁻⁶)/(0.25²)

F(12) = -14.41 N  (towards left)

<h3>Force on q1 due to q3</h3>

F(13) = (9 x 10⁹ x 7.7 x 10⁻⁶ x 5.9 x 10⁻⁶)/(0.55²)

F(13) = 1.352 N (towards right)

<h3>Net force on q1</h3>

F(net) = 1.352 N - 14.41 N

F(net) = -13.06 N

Thus, the net force on particle particle q1 is 13.06 N towards the left.

Learn more about force here: brainly.com/question/12970081

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8 0
1 year ago
An 100 kg object traveling at 50 m/s collides (perfectly inelastic) with a 50 kg object initially at rest.
qaws [65]

Answer:

Option C. 5,000 kg m/s

Explanation:

<u>Linear Momentum on a System of Particles </u>

Is defined as the sum of the momenta of each particles in a determined moment. The individual momentum is the product of the mass of the particle by its speed

P=mv

The question refers to an 100 kg object traveling at 50 m/s who collides with another object of 50 kg object initially at rest. We compute the moments of each object

m_1=(100\ kg)(50\ m/s)=5,000\ kg\ m/s

m_2=(50\ kg)(0\ m/s) = 0

The sum of the momenta of both objects prior to the collision is

P=5,000\ kg\ m/s+0\ kg\ m/s

\boxed{ P=5,000\ kg\ m/s}

7 0
3 years ago
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