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Mkey [24]
2 years ago
7

Calculate the power output of a 1.5 g fly as it walks straight up a windowpane at 2.4 cm/s .

Physics
1 answer:
Pepsi [2]2 years ago
6 0

Answer:

Power output, P = 0.35 watts

Explanation:

It is given that,

Mass of the fly, m = 1.5 g = 0.0015 kg

Speed of the fly, v = 2.4 cm/s = 0.024 m/s

We have to find the power output of the fly. Power output is given by :

P=\dfrac{W}{t}

And W = F.d

d = distance traveled

P=\dfrac{F.d}{t}

Since, v = d /t

So, P = F . v

P=mg\times v

P=0.0015\ kg\times 9.8\ m/s^2\times 0.024\ m/s

P = 0.00035 watts

or P = 0.35 watts

Hence, this is the required solution.

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X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to
lys-0071 [83]

Answer:

a) \Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

b) \lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

c) E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

Explanation

Part a

For this case we can use the Compton shift equation given by:

\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

Part b

For this cas we can calculate the wavelength of the phton with this formula:

\lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

3 0
2 years ago
Read 2 more answers
The free-body diagram of a crate is shown. What is the net force acting on the crate? 352 N to the left 176 N to the left 528 N
Umnica [9.8K]

As per given conditions there are two directions along which forces are acting

1. Net force along left direction is given as

F_{left} = 352N + 176 N = 528 N

2. Net force towards right direction is given as

F_{right} = 528N + 440 N = 968 N

now since the two forces here in opposite direction so here we will have net force given as

F_{net} = F_{right} - F_{left}

F_{net} = 968 - 528

F_{net} = 440 N

so here net forces must be 440 N towards right

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An ice dancer with her arms stretched out starts into a spin with an angular velocity of 1.00 rad/s. Her moment of inertia with
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Answer:

A) 0.957 J

Explanation:detailed explanation and calculation is shown in the image below

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Sara and Saba are identical twins who are the same in every way, including their weights. One day, Sara and Saba decided to go f
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As Saba was wearing high heels they are long from the bottom so they sank however Sana was wearing snow boots which means they were flat and so she didn’t sink.

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Which occurrence would lead you to conclude that lights are connected in a
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