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diamong [38]
3 years ago
5

The worlds fastest humans can reach speeds of about 11 m/s in order to increase his gravitational potential energy by an amount

equal to his Kinetic energy at full speed how high with the sprinter need to climb
Physics
1 answer:
Amanda [17]3 years ago
4 0
 
What a delightful little problem !

-- When he is running on level ground, his kinetic energy is

             KE = (1/2) x (mass) x (speed)² .

-- When he climbs up from the ground, his potential energy is

             PE = (mass) x (gravity) x (height above the ground).

We're looking for the height that makes these quantities of energy equal,
figuring that when he runs, his speed is  11 m/s.

The first time I looked at this, I thought we would need to know the runner's
mass.  But it turns out that we don't.

       <u>PE = KE</u>

      (mass) x (gravity) x (height) = (1/2) (mass) (11 m/s)²

Divide each side by (mass) : 

       (gravity) x (Height)  =  (1/2) (11 m/s)²

Divide each side by gravity:

                      Height = (1/2) (121 m²/s²) / (9.8 m/s²)

                                =  <em>6.173 meters</em>

                         (about  20.3 feet !)


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A 27.0-m steel wire and a 48.0-m copper wire are attached end to end and stretched to a tension of 145 N. Both wires have a radi
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Answer:

The time taken by the wave to travel  along the combination of two wires is 458 ms.

Explanation:

Given that,

Length of steel wire= 27.0 m

Length of copper wire = 48.0 m

Tension = 145 N

Radius of both wires = 0.450 mm

Density of steel wire \rho_{s}= 7.86\times10^{3}\ kg/m^{3}

Density of copper wire \rho_{c}=8.92\times10^{3}\ kg/m^3

We need to calculate the linear density of steel wire

Using formula of linear density

\mu_{s}=\rho_{s}A

\mu_{s}=\rho_{s}\times\pi r^2

Put the value into the formula

\mu_{s}=7.86\times10^{3}\times\pi\times(0.450\times10^{-3})^2

\mu_{s}=5.00\times10^{-3}\ kg/m

We need to calculate the linear density of copper wire

Using formula of linear density

\mu_{c}=\rho_{s}A

\mu_{c}=\rho_{s}\times\pi r^2

Put the value into the formula

\mu_{c}=8.92\times10^{3}\times\pi\times(0.450\times10^{-3})^2

\mu_{c}=5.67\times10^{-3}\ kg/m

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

v_{s}=\sqrt{\dfrac{T}{\mu_{s}}}

v_{s}=\sqrt{\dfrac{145}{5.00\times10^{-3}}}

v_{s}=170.3\ m/s

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

v_{c}=\sqrt{\dfrac{T}{\mu_{c}}}

v_{c}=\sqrt{\dfrac{145}{5.67\times10^{-3}}}

v_{c}=159.9\ m/s

We need to calculate the time taken by the wave to travel  along the combination of two wires

t=t_{s}+t_{c}

t=\dfrac{l_{s}}{v_{s}}+\dfrac{l_{c}}{v_{c}}

Put the value into the formula

t=\dfrac{27.0}{170.3}+\dfrac{48.0}{159.9}

t=0.458\ sec

t=458\ ms

Hence, The time taken by the wave to travel  along the combination of two wires is 458 ms.

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