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bezimeni [28]
3 years ago
13

Label and describe what is happening in this picture

Physics
1 answer:
SOVA2 [1]3 years ago
7 0
Something is reproducing.
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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} }

\sf{ \underline{ \bold{  {proved}}}}

Hope I helped!

Best regards!!

5 0
3 years ago
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