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Alex17521 [72]
4 years ago
12

the ball has joules of potential energy at point B. At position A all of the energy changes to kinetic energy. The velocity of t

he ball at position A is meters/seconds. Assume theres no air resistance. use g=9.8m/s2
Physics
1 answer:
pochemuha4 years ago
7 0
As I don’t see unit values we can do this with algebra gravitational potential energy is equal to mgh so you find that and then convert it to 1/2mv^2
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How much energy is stored in a spring with a spring constant of 390 N/m if the spring is compressed a distance of 0.45 m from it
masya89 [10]
F=-ks
F=-(390)(.45)
F=-175.5 N

Work=force x displacement
Work= 175.5(0.45)
Work= 78.98 J

Work = ∆E =78.98 J

Answer=79J (first option)
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after 2 hours the displacement of a car is 160km north the initial displacement of the car is 0. what is the average velocity of
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D

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300N effort is applied to lift the load of 900N by using a first class lever. If the distance between the load and fulcrum is 20
Klio2033 [76]

Answer:

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2 years ago
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How much work is done if you push a 200 N box across a floor with a force of 50 N for a distance of 20m
vovikov84 [41]
<h2>Answer: 1000 J</h2>

The Work W done by a Force F refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.  

It should be noted that it is a scalar magnitude, and its unit in the International System of Units is the Joule (like energy). Therefore, 1 Joule is the work done by a force of 1 Newton when moving an object, in the direction of the force, along 1 meter:  

1J=(1N)(1m)=Nm  

Now, when the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:  

W=(F)(d) (1)

When they are not parallel, both directions form an angle, let's call it \alpha. In that case the expression to calculate the Work is:  

W=Fdcos{\alpha} (2)

For example, in order to push the 200 N box across the floor, you have to apply a force along the distance d to overcome the resistance of the weight of the box (its 200 N).  

In this case both <u>(the force and the distance in the path) are parallel</u>, so the work W performed is the product of the force exerted to push the box F=50N by the distance traveled d. as shown in equation (1).

Hence:

W=(50N)(20m)  

W=1000Nm=1000J >>>>This is the work  

8 0
3 years ago
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