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2 years ago

11

A pump lifts water from a lake to a large tank 20 m above the lake. How much work against gravity does the pump do as it transfe

rs 5.0 m^3 if the density is ?=1000kg/m^3?

2 answers:

Aleonysh [2.5K]2 years ago

3 0

Answer:

980 kJ

Explanation:

Work = change in energy

W = mgh

W = (1000 kg/m³ × 5.0 m³) (9.8 m/s²) (20 m)

W = 980,000 J

W = 980 kJ

The pump does 980 kJ of work.

irina1246 [14]2 years ago

3 0

Answer:

980 kJ

Explanation:

It takes the work against gravity does the pump do as it transfers from 5.0 m^3 the density is 980kj.

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2 years ago

Read 2 more answers

29. 2072 Set C Q.No. 10c

Annette [7]

Answer:

$90.2^{\circ}C$

Explanation:

Considering the thermal conductivity of aluminium and brass as $k_{al}=205 W/mK$ and $k_{br}=109 W/mk$ respectively

The temperature at the end of aluminium and brass are given as $T_{al}=150^{\circ}C$ and $T_{br}=20^{\circ}C$ respectively with length of rod L=1.3 m , Length of aluminium $L_{al}=0.8 m$ , length of brass $L_{br}=0.5 m$ and letting temperature at steady state be T

At steady state, thermal conductivity of both aluminium and brass are same hence

$H_{br}=H_{al}$

$k_{al}A\frac {T_H-T}{L_{al}}= k_{br}A\frac {T-T_H}{L_{br}}$

Upon re-arranging

$T=\frac {k_{al}L_{al}T_{br}+k_{al}L_{br}T_{al}}{k_{br}L_{al}+k_{al}L_{br}}$

$(205)\frac {150-T}{0.8}=109\frac {T-20}{0.5}$

$T=\frac {(109*0.8*20)+(205*0.5*150)}{(109*0.8)+(205*0.5)}$

$T=90.2^{\circ}C$

Therefore, the temperatures at which the metals are joined is $90.2^{\circ}C$

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2 years ago

A 0.28-kg stone you throw rises 34.3 m in the air. The magnitude of the impulse the stone received from your hand while being th

goldfiish [28.3K]

Answer:

7.3 kg m/s

Explanation:

First of all, let's calculate the gravitational potential energy of the stone as it reaches its highest point:

$U=mgh=(0.28 kg)(9.8 m/s^2)(34.3 m)=94.1 J$

For the law of conservation of energy, this is equal to the initial kinetic energy of the stone at ground level (where the potential energy is zero), just after the stone leaves your hand:

$K=\frac{1}{2}mv^2=94.1 J$

From this equation we can find the velocity of the stone as it leaves your hand:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(94.1 J)}{0.28 kg}}=25.9 m/s$

The initial velocity of the stone (before leaving your hand) is zero:

$u=0$

The impulse received by the stone is equal to its change in momentum, so:

$I=\Delta p=m\Delta v=m(v-u)=(0.28 kg)(25.9 m/s-0)=7.3 kg m/s$

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2 years ago

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AleksandrR [38]

Answer:

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2 years ago

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aksik [14]

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2 years ago