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alexgriva [62]
2 years ago
6

Please help!!!!!!!!!!!!! i will give brainly

Physics
1 answer:
liq [111]2 years ago
8 0

Answer:

1.) strength or energy to do an action or movement

2.)an object will jot change its motion unless acted on by an unbalanced force

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An automobile engine delivers 47.4 hp. how much time will it take for the engine to do 6.82 × 105 j of work? one horsepower is e
kogti [31]
1 horsepower is equal to 746 W, so the power of the engine is
P=47.4 hp \cdot 746  \frac{W}{hp}=35360 W
The power is also defined as the energy E per unit of time t:
P= \frac{E}{t}
Where the energy corresponds to the work done by the engine, which is E=6.82 \cdot 10^5 J. Re-arranging the formula, we can calculate the time t needed to do this amount of work:
t =  \frac{E}{P}= \frac{6.82 \cdot 10^5 J}{35360 W}=19.3 s
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1 because the the mid night summer is dark
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marta [7]
Hope this helps you.

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3 years ago
Dumbo was picked on for having big. what might be an advantage to havin large ears
Aleonysh [2.5K]

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3 years ago
A river flows down towards a lake along an incline. initially the river is 90 m above the lake flowing at a rate of 3 m/s. at th
lubasha [3.4K]

Total energy =kinetic energy +potential energy

Change in energy =change in (kinetic energy +potential energy)

potential energy, P.E=mgh

where m is the mass, g is acceleration due to gravity and h is the height.

potential energy per unit mass =gh

change in potential energy per unit mass = \Delta P.E.=g\Delta h

where, h is the height.

kinetic energy= K.E. =\frac{1}{2}mv^2

change in kinetic energy per unit mass,\Delta K.E. =\frac{1}{2}\Delta v^2

In the given question:

Height varies from 90 m to zero as river flows from 90 m height to lake at 0 m

Velocity varies from 3m/s at top to o m/s at bottom.

Therefore,

\Delta E =\Delta K.E.+\Delta P.E.\\ \Rightarrow \Delta E/m=\frac{1}{2}\Delta v^2+g\Delta h=\frac{1}{2}(0^2-3^2) m^2s^{-2}+9.8 (0-9)m^2s^{-2}=(-\frac{9}{2}-88.2)m^2s^{-2}=-92.7 m^2s^{-2}

Here, it was mentioned in the question internal energy of the water is constant and there is no change in the pressure at the inlet and outlet.

4 0
3 years ago
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