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2 years ago

15

A baseball leaves a bat with a horizontal velocity of 20.0 m/s. When it has left the bat for a time of 0.250 s (Assume air resis

tance is negligible). How far will it have moved horizontally?

1 answer:

miss Akunina [59]2 years ago

5 0

Answer:

The distance is 5 m.

Explanation:

Given that,

Horizontal velocity = 20.0 m/s

Time t = 0.250 s

We need to calculate the horizontal distance

The distance traveled by the baseball with horizontal velocity in time t is given by

Using formula for horizontal distance

$d = v\times t$

$d = 20.0\times0.250$

$d=5\ m$

Hence, The distance is 5 m.

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Explanation:

Given

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mass flow rate of geothermal water $\dot{m_h}=2 kg/s$

diameter of inner wall $d_i=1.5 cm$

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specific heat of water $c=4.18 kJ/kg-K$

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

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$LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}$

$LMTD=\frac{80-104}{\ln \frac{80}{104}}$

$LMTD=91.49^{\circ}C$

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$A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2$

$A=\pi DL=5.144$

$L=\frac{5.144}{\pi \times 0.015}$

$L=109.16 m$

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