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bekas [8.4K]
3 years ago
10

If you start rolling down this hill, your potential energy will be converted to kinetic energy. At the bottom of the hill ypu KE

will be equal to your PE at the top, what is my speed at the bottom of the hill?
Physics
1 answer:
Serjik [45]3 years ago
8 0

Answer:

v=\sqrt{(2x/m)}\  \ m/s

Explanation:

Potential Energy= Kinetic Energy

Let x be the value of Kinetic Energy.

We know that

PE=KE=x\\x=\frac{1}{2}mv^2\\

Make v the subject of the formula to get speed at the bottom of the hill.

v^2=2x/m\\v=\sqrt{2x/m}

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The resistivity of a metal increases slightly with increased temperature. This can be expressed as rho= rho0[1+α(T−T0)] , where
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Answer:

At 81. 52 Deg C its resistance will be 0.31 Ω.

Explanation:

The resistance of wire =R_T =\frac{\rho_T \ l}{A}

Where R_T =Resistance of wire at Temperature T

\rho_T = Resistivity at temperature T =\rho_0 \ [1 \ + \alpha\ (T-T_0\ )]

Where T_0 =20\ Deg\ C , \  \rho_0 = Constant,  \alpha =3.9 \times 10^-^3 DegC^-1 \ (Given)

l=Length of the wire

& A = Area of cross section of wire

For long and thin wire the resistance & resistivity relation will be as follows

\frac{R_T_1}{R_T_2}=\frac{\rho_0(1+\alpha \cdot(T_1-2 0 )}{\rho_0(1+\alpha \cdot (T_2 -20 )}

\frac{0.25}{0.31}=\frac{1}{[1+\alpha(T-20)]}

1.24=1+\alpha (T-20)

0.24=\alpha(\ T -20 )

Putting\ the\ value\ of \alpha = 3.9 \times 10^-^3 DegC^-1

T = 81.52 Deg C

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