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3 years ago

Explain briefly work energy and power

Physics

1 answer:

Rufina [12.5K]3 years ago

4 0

Answer:

Work can be defined as transfer of energy.

Energy can be defined as the capacity for doing work

Power is the work done in a unit of time.

Explanation:

Work can be defined as transfer of energy. In physics we say that work is done on an object when you transfer energy to that object.

Power is the work done in a unit of time. In other words, power is a measure of how quickly work can be done.

Energy can be defined as the capacity for doing work. The simplest case of mechanical work is when an object is standing still and we force it to move

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An automobile and train move together along parallel paths at 23.2 m/s. The automobile then undergoes a uniform acceleration of

lara [203]

Answer:

Train will 588.215 m ahead of automobile

Explanation:

We have given initially speed of both train and automobile is same as 23 m/sec

After that acceleration of automobile $a=-4m/sec^2$

Initial velocity u = 23 m/sec

Final velocity v = 0 ( as it comes to rest )

According to first law of motion

v=u+at

$0=23-4\times t$

t = 5.75 sec

Now according to third law of motion $v^2=u^2+2as$

$0^2=23^2-2\times 4\times s$

s = 66.125 m

Now automobile accelerates at with acceleration $a=2.09m/sec^2$

Initial velocity u = 23 m/sec

Final velocity v = 23.2 m/sec

So $time\ t =\frac{v-u}{a}=\frac{23.2-23}{2.09}=0.0956sec$

Distance traveled by automobile in 0.0956 sec

$s=ut+\frac{1}{2}at^2=23\times 0.0956+\frac{1}{2}\times 2.09\times 0.0956^2=2.208m$

So total distance traveled by automobile = 66.125+2.208 = 68.333 m

Total time = 5.75 +22.7+0.0956=28.545 sec

Sp distance traveled by train in 28.545 sec

= 28.545×23 = 656.548 m

So distance between automobile and train = 656.548 - 68.33=588.215 m

So train will 588.215 m ahead of automobile

3 0

3 years ago

After a long night of cramming for a test, your college room-mate hits his head forcefully against the wall of your room in desp

zhannawk [14.2K]

Answer: the wall contracts the force exerted by his head. The wall produces the opposite force which is equal to the force his head bangs the wall with.

Explanation: if his head exerts a much greater force than the wall can counteract the wall will be destroyed, if the wall exerts a much greater force than his head exerts he will be pushed far back and might even suffer a broken head.

The wall in this case provides the opposite reactive force.

6 0

4 years ago

Which is the currrent SI unit to use in measuring the amount of material contained in a solid sphere

olya-2409 [2.1K]

Hey

SI unit for measure the amount of material contained in a solid sphere metre cube(m^3).

mean it have Volume .

3 0

4 years ago

Could you guys tell me whether the photo represents a balanced or unbalanced equation

NARA [144]

Answer:

that is preatty balenced m8

Explanation:

6 0

4 years ago

Ne W2

levacccp [35]

Answer:

3.82746e+26 watts

Explanation:

There are two ways to solve this problem. One way is to use the equation

L = 4πσR²T⁴

where

L = the sun's bolometric (all-spectrum) luminous power

σ = 5.670374419e-8 W m⁻² K⁻⁴ = the Stefan-Boltzmann constant

R = 6.957e+8 meters = the sun's radius

T = 5771.8 K = the sun's effective temperature

You find that

L = 3.82746e+26 watts

The other way to solve the problem is to use the Planck integral for radiant flux.

L = 4π²R ∫(v₁,v₂) 2hv³/{c² exp[hv/(kT)]−1} dv

where

h = 6.62607015e-34 J sec

c = 299792458 m sec⁻¹

k = 1.380649e-23 J K⁻¹

v₁ = 0 = frequency band lower bound, in Hz

v₂ = ∞ = frequency band upper bound, in Hz

You find, once again, that

L = 3.82746e+26 watts

The advantage of using the Planck integral becomes clear when you want to calculate the sun's luminous power only in a specific band, rather than across the entire spectrum. For example, if we do the calculation again, except that we use

v₁ = 4.1e+14 = frequency band lower bound, in Hz

v₂ = 7.7e+14 Hz = frequency band upper bound, in Hz

restricting ourselves to the visible spectrum. We find that

L (visible) = 1.56799e+26 watts

So the fraction of the sun's luminosity that is in the visible spectrum is

L (visible) / L = 0.4096686

5 0

4 years ago