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Anit [1.1K]
4 years ago
14

A pendulum has 366 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom o

f its swing?
Physics
2 answers:
lana66690 [7]4 years ago
8 0
968 j or kg because it cannot lose any of it energy and cannot gain any extra mass from nothing
Tatiana [17]4 years ago
5 0
Energy can neither be created nor destroy, it can only be transformed from one form to another form.
The potential energy possessed by the pendulum can not be destroyed, neither can more energy be added to it, but it can be converted from potential energy to kinetic energy.
So, at the bottom of the swing, the kinetic energy of the pendulum is going to be 366 J.
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A 80 kg block starts from rest at the top of a 15m frictionless ramp. The block slides down the frictionless ramp onto a rough h
Nadusha1986 [10]

Answer:

W(fric) = -12000 J

Explanation:

Given that

mass ,m= 80 kg

Initial speed ,u= 0 m/s

Height ,h= 15 m

Final speed ,v= 0 m/s

We know that

Work done by all the forces = Change in the kinetic energy

W_{gravity}+W_{fric}=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2

W_{gravity}+W_{fric}=\dfrac{1}{2}m\times 0^2-\dfrac{1}{2}m\times 0^2

W_{gravity}+W_{fric}=0

W(fric)= - m g h

W(fric) = - 80 x 10 x 15     ( g=10 m/s²)

W(fric) = -12000 J

negative sign indicates that force and displacement is in opposite direction.

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3 years ago
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Answer:

<em>The distance of the light is 9.4608 x 10^25 m</em>

<em></em>

Explanation:

Time taken by the light = 10 billion years = 10 x 10^9 years

speed of light = 3 x 10^8 m/s

speed of light in m/years is = (3 x 10^8)/(60 x 60 x 24 x 365) = 9.4608 x 10^15 m/year

distance = speed x time

therefore, the distance of this light = 10 x 10^9 x 9.461 x 10^15 = <em>9.4608 x 10^25 m</em>

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3 years ago
At 0.0 degrees Celsius, a wire cable is 410.0000 meters in length. When the temperature increased to 30.0 degrees Celsius, the w
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Answer: The coefficient of expansion for the wire 0.000028 (^oC)^{-1}.

Explanation:

Original length of the wire = L= 410.0000 m

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Final Temperature = T_2=30.0^oC

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\alpha _L=\frac{\Delta L}{L\times (T_2-T_1)}=\frac{0.3444 m}{410.0000 m\times (30.0^oC-0.0^oC)}

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