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Anna [14]
3 years ago
13

A 15 Kilogram anvil is hung from a ceiling . While it is hanging there , it has been determined to have 65 Joules of gravitation

al potential energy relative to the floor . If the anvil is released , what would be its speed just before it hits the floor ?
Physics
1 answer:
tigry1 [53]3 years ago
3 0

Answer:

v = 2.94 m/s

Explanation:

Given that,

Mass of anvil, m = 15 kg

The gravitational potential energy relative to the floor, P = 65 J

We need to find the speed just before it hits the floor when the Anvil is released. Let it is v.

We can use the conservation of energy.

\dfrac{1}{2}mv^2=mgh\\\\65=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{65\times 2}{15}} \\\\v=2.94\ m/s

So, the required speed is 2.94 m/s.

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it depends on the appliance

Explanation:

bigger appliances will use more power, smaller appliances will use a lesser amount

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What does the person in front of the ambulance experience?
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A higher frequency. I just got this answer on usa test prep.

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Indicate whether the following statements are always true or can be false. An object's velocity will change if a net force acts
Sergio [31]

Answer:

1-An object's velocity will change if a net force acts on the object. (True)

2-In order not to slow down, a car moving at a constant velocity needs a small net force applied. (False)

3- The net force which acts on an object which remains at rest is zero. (True)

4- If an object's speed does not change, no net force is acting on the object.(True)

5- If two objects are under the influence of equal forces, they have the same acceleration. (False)

6- During the collision of a car with a large truck, the truck exerts a greater force on the car than the car exerts on the truck.(False) (Newton Third Law)

3 0
3 years ago
An object with a mass of 2 kg has a force of 4 N acting on it. What is the acceleration of the object?
STatiana [176]
Answer: 2 m/s^2

Explain

£F=ma
net force = mass x acceleration
4N is a force acting on it
4N=(2kg) a
2kg is the mass
you divide the 2 over
and A= 2 m/s^2

8 0
3 years ago
A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t
vladimir1956 [14]

a) \omega=\frac{-mvR}{I+MR^2}

b) v=\frac{-mvR^2}{I+MR^2}

Explanation:

a)

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

L_1=0

Later, after the girl throws the rock, the angular momentum will be:

L_2=(I_M+I_g)\omega +L_r

where:

I is the moment of inertia of the merry-go-round

I_g=MR^2 is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

\omega is the angular speed of the merry-go-round and the girl

L_r=mvR is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

L_1=L_2

So we find:

0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

b)

The linear speed of a body in rotational motion is given by

v=\omega r

where

\omega is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

\omega=\frac{-mvR}{I+MR^2} is the angular speed

r=R is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

v=\omega R=\frac{-mvR^2}{I+MR^2}

5 0
3 years ago
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