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iragen [17]
2 years ago
11

A 110-pound person pulls herself up 4.0 feet. This took her 2.5 seconds. How much power was developed?

Physics
2 answers:
mrs_skeptik [129]2 years ago
5 0

Answer:

0.32 Hp is the answer. right

lesya [120]2 years ago
4 0

Answer: Power developed by her is <u><em>0.32 hp</em></u>

 

Explanation:

Work done , W= Force(F) X Displacement (d) in the direction of force

Thus work done , W=Fd=110\times 4.0 ft-lbf = 440 ft-lbf

Now 1.0 ft-lbf = 1.355 joules of energy

Therefore , work done, W= 440x 1.355 J = 596.2 J

Average Power , P=\frac{work done(W)}{time (t)}

=> P=\frac{596.2}{2.5} watts = 238.5 watts

We know 746 hp = 1.0 watts

=> P=\frac{238.5}{746} hp=0.32 hp

Thus average power developed by her is 0.32 hp

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(a) If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.5
Tju [1.3M]

Answer:

The fraction of its energy that it radiates every second is 3.02\times10^{-11}.

Explanation:

Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}

Given that,

Kinetic energy = 6.2 MeV

Radius = 0.500 m

We need to calculate the acceleration

Using formula of acceleration

a=\dfrac{v^2}{r}

Put the value into the formula

a=\dfrac{\dfrac{1}{2}mv^2}{\dfrac{1}{2}mr}

Put the value into the formula

a=\dfrac{6.2\times10^{6}\times1.6\times10^{-19}}{\dfrac{1}{2}\times1.67\times10^{-27}\times0.51}

a=2.32\times10^{15}\ m/s^2

We need to calculate the rate at which it emits energy because of its acceleration is

\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}

Put the value into the formula

\dfrac{dE}{dt}=\dfrac{(1.6\times10^{-19})^2\times(2.3\times10^{15})^2}{6\pi\times8.85\times10^{-12}\times(3\times10^{8})^3}

\dfrac{dE}{dt}=3.00\times10^{-23}\ J/s

The energy in ev/s

\dfrac{dE}{dt}=\dfrac{3.00\times10^{-23}}{1.6\times10^{-19}}\ J/s

\dfrac{dE}{dt}=1.875\times10^{-4}\ ev/s

We need to calculate the fraction of its energy that it radiates every second

\dfrac{\dfrac{dE}{dt}}{E}=\dfrac{1.875\times10^{-4}}{6.2\times10^{6}}

\dfrac{\dfrac{dE}{dt}}{E}=3.02\times10^{-11}

Hence, The fraction of its energy that it radiates every second is 3.02\times10^{-11}.

5 0
2 years ago
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