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3 years ago

I need help in this small question:

Physics

1 answer:

Natali5045456 [20]3 years ago

5 0

Answer:

1) The energy released during nuclear fission or fusion, especially when used to generate electricity is called nuclear energy.

2) It is not renewable because it is an element that has no way whatsoever to regenerate or replicate itself, nor gets created by any natural terrestrial means, neither makes itself available by arriving from outer space (like sunlight). There is a limited amount of it available on the Earth, and every bit you use is a bit you’ll never have available again (as Uranium atoms get destroyed by the fission process).

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If a person climbed Mt. Everest has a mass of 105 kg and a weight of 625 N what would be the acceleration due to gravity?

DanielleElmas [232]

The acceleration due to gravity would be 5.95 m/s²

A force is known to be a push or pull and it is the change in momentum per time. It can be expressed by using the relation.

Force = mass × acceleration.

From the parameters given:

Mass = 105 kg
Force = 625 N

By replacing the given values into the above equation, we can determine the acceleration.

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625 N = 105 kg × acceleration.

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2 years ago

Show that (a)KE=1/2mv2

evablogger [386]

Answer:

$\boxed{ \bold{ \huge{ \boxed{ \sf{see \: below}}}}}$

Explanation:

$\underline{ \bold{ \sf{To \: prove \: that \: kinetic \: energy = \frac{1}{2} m {v}^{2} }}}$

Let us consider, a body of mass ' m ' is lying at rest ( initial velocity = 0 ) on a smooth surface. Let a constant force F displaces this body in its own direction by a displacement ' d '. Let 'v' be it's final velocity. The work done ' W ' by the force is given by :

$\sf{W = FD}$

⇒ $\sf{W = m \: \times a \: \times s} \: \: \: \: \: \: \: \: \: ( \: ∴ \: f \: = \: ma \: ; \: s \: = d)$

⇒ $\sf{W = m \: \times \frac{v - u}{t} \times \frac{u + v}{2} \times t \: \: \: \: \: \: \: \: \: (∴ \: a = \frac{v - u}{t} and \: s = \frac{u + v}{2} \times t}$

⇒ $\sf{W = m \times \frac{ {v}^{2} - {u}^{2} }{2} }$

⇒ $\sf{W = \frac{1}{2} m {v}^{2} \: \: \: \: \: \: \: \: \: \: \: \: (since, \: initial \: velocity(u) = 0)}$

The work done becomes the kinetic energy of the body. Thus, the kinetic energy of a body of mass ' m : moving with the velocity equal to 'v ' is 1 / 2 mv²

∴ $\sf{KE= \frac{1}{2} m {v}^{2} }$