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gregori [183]
2 years ago
15

The same force is applied to two skateboards. One rolls across the room and the other moves a few feet and comes to a stop. Wher

e was there more work done? The skateboard that traveled the shorter distance shows more work because there was more resistance. The skateboard that traveled the longer distance shows more work because it was lighter. The skateboard that traveled further shows more work because the distance was greater. The skateboard that traveled the shorter distance shows more work because the force was greater.
Physics
2 answers:
igor_vitrenko [27]2 years ago
8 0

The longer you spend reading and thinking about this question,
the more defective it appears.

-- In each case, the amount of work done is determined by the strength
of the force AND by  the distance the skateboard rolls <em><u>while you're still </u></em>
<em><u>applying the force</u>.   </em>Without some more or different information, the total
distance the skateboard rolls may or may not tell how much work was done
to it.<em>
</em>
-- We know that the forces are equal, but we don't know anything about
how far each one rolled <em>while the force continued</em>.  All we know is that
one force must have been removed.

-- If one skateboard moves a few feet and comes to a stop, then you
must have stopped pushing it at some time before it stopped, otherwise
it would have kept going. 

-- How far did that one roll while you were still pushing it ?

-- Did you also stop pushing the other skateboard at some point, or
did you stick with that one?

-- Did each skateboard both roll the same distance while you continued pushing it ?

I don't think we know enough about the experimental set-up and methods
to decide which skateboard had more work done to it.

Lyrx [107]2 years ago
5 0

Answer: More Work was done by the Skateboard that traveled farther.

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Amanda [17]

Answer:

If it isn't an ion it should have 35 elektrons to cancel the positivity of the nucleus.

7 0
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A car is moving with a velocity of45m/s. Is brought to rest in 5s.the distance travelled by car before it comes to rest is
ser-zykov [4K]

Answer:

The car travels the distance of 225m before coming to rest.

Explanation:

v = 45m/s

t = 5s

Therefore,

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= 45*5

= 225m

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2 years ago
The weight of an astronout is 60kg on earth, find the wight of the same object on the planet where the gravitational attraction
Marianna [84]

The astronaut's weight is not 60 kg anywhere, because kg is a unit of mass, not weight.

If the astronaut's mass is 60 kg, then his weight is (60 kg)x(acceleration of gravity).

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3 0
3 years ago
Time period of a simple pendulum is measured at Karachi. What change will occur in the time period, if it is measured on mount e
Reil [10]
The period of a pendulum is given by T=2 \pi  \sqrt{ \frac{l}{g} }
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5 0
3 years ago
2) [25 pts] The bob of mass m=5 kg shown in the figure is being held by a force F. If the angle is 0⃗
Colt1911 [192]

Complete Question

The complete question is shown on the first uploaded image  

Answer:

a

   The tension is  T =  48.255 \ N

b

   The time taken is  t =  0.226 \ s

c

   The position for maximum velocity is  

        S = 0  

d

     The maximum velocity is  V_{max} =0.384 \ m/s

Explanation:

The free body for this question is shown on the second uploaded image

From the question we are told that

    The mass of the bob  is m_b  =  5 \ kg

     The angle is  \theta = 10^o

     The length of the string is  L =  0.5 \ m

The tension on the string is mathematically represented as

              T  = mg cos \theta

substituting values

             T =  5 * 9.8 cos(10)

             T =  48.255 \ N

The motion of the bob is mathematically represented as

             S =  A sin (w t  + \frac{\pi }{2} )

     =>   S =  A sin (wt)

Where  w is the angular speed

and  \frac{\pi}{2} is the phase change

At initial position S =  0

   So   wt  = cos^{-1} (0)

          wt  = 1

Generally w can be mathematically represented as

          w =  \frac{2 \pi }{T}

Where T is the period of oscillation which i mathematically represented as

          T  =  2 \pi \sqrt{\frac{L}{g} }

So  

      t   = \frac{1}{w}

       t   = \frac{T}{2 \pi}

       t =  \sqrt{\frac{L}{g} }

substituting values

       t =  \sqrt{\frac{0.5}{9.8} }

       t =  0.226 \ s

Looking at the equation

         wt =  1

We see that maximum velocity of the bob will be at  S = 0  

i. e     w =  \frac{1}{t}

The maximum velocity is mathematically represented as

          V_{max}   =  w A

Where A is the amplitude which is mathematically represented as

         A = L sin \theta

So

      V_{max} = \frac{2 \pi }{T }  L sin \theta

      V_{max} = \frac{2 \pi }{2 \pi } \sqrt{\frac{g}{L} }   L sin \theta  Recall   T  =  2 \pi \sqrt{\frac{L}{g} }

     V_{max} = \sqrt{gL} sin \theta

substituting values        

       V_{max} = \sqrt{9.8 * 0.5 } sin (10)

       V_{max} =0.384 \ m/s

6 0
2 years ago
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