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son4ous [18]
3 years ago
15

Define work done against friction(helpp)​

Physics
1 answer:
77julia77 [94]3 years ago
7 0

Answer:

The work done against friction is the work done on an object that overcomes this frictional force allowing the object to move - it doesn't include any extra work used to accelerate an object for example - only the work used to beat the frictional force.

Explanation:

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A center-seeking force related to acceleration is _______ force.
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<span>The question is 'a centre seeking force related to acceleration is ............... force. The answer is centripetal force. Motion in a curved path is an accelerated motion and it requires a force that will direct the moving object towards the centre of curvature of the path of motion. This centre seeking force is known as centripetal force.</span>
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· List the atomic numbers of the elements in Period 2.
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Answer:

3,4,5,6,7,8,9,10

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4. beryllium

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6. carbon

7. nitrogen

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Explanation:

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a stone is dropped from top of a tower of 50m high ,simltaneously another stone is thrown upward with a speed of 20m/s find the
aleksklad [387]
At the time that I'll call ' Q ', the height of the stone that was
dropped from the tower is

             H = 50 - (1/2 G Q²) ,

and the height of the stone that was tossed straight up
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             H  =  20Q - (1/2 G Q²) .

The stones meet when them's heights are equal,
so that's the time when

                         <span>50 - (1/2 G Q²)  =  20Q - (1/2 G Q²) .

This is looking like it's going to be easy.

Add  </span><span>(1/2 G Q²)  to each side.
Then it says
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Divide each side by 20:          2.5  =  Q .

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5 0
3 years ago
A ball is thrown up at a speed of 20 m/s.
ioda

Given:

(Initial velocity)u=20 m/s

At the maximum height the final velocity of the ball is 0.

Also since it is a free falling object the acceleration acting on the ball is due to gravity g.

Thus a=- 9.8 m/s^2

Now consider the equation

v^2-u^2= 2as

Where v is the final velocity which is measured in m/s

Where u is the initial velocity which is measured in m/s

a is the acceleration due to gravity measured in m/s^2

s is the displacement of the ball in this case it is the maximum height attained by the ball which is measured in m.

Substituting the given values in the above formula we get

0-(20x20)= 2 x- 9.8 x s

s= 400/19.6= 20.41m

Thus the maximum height attained is 20.41 m by the ball

6 0
3 years ago
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