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Sergeu [11.5K]
3 years ago
10

A boy jumps from a wall 3m high. What is an estimate of the change in momentum of the boy when he lands without rebounding?

Physics
1 answer:
EastWind [94]3 years ago
8 0
The answer is D. Because the boy jump 3 high
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Which of the following is a<br> fossil fuel?<br> A. Coal<br> B. Wind energy<br> C. Biomass energy
almond37 [142]

A. Coal

is the correct answer

5 0
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A factory emits pollutants at a rate of 25 g/s. The factory is located between two mountain ranges resulting in an effective val
Lostsunrise [7]

Answer:

1.25\ \mu\text{g/m}^3

Explanation:

v = Velocity of the breeze = 4 m/s

w = Width of the valley = 5000 m

h = Height of the valley = 1000 m

Volumetric flow rate is given by

\dot{V}=vwh\\\Rightarrow \dot{V}=4\times 5000\times 1000\\\Rightarrow \dot{V}=2\times10^{7}\ \text{m}^3/\text{s}

\dot{m} = Mass flow rate of pollutant = 25 g/s = 25\times 10^6\ \mu\text{g/s}

Concentration is given by

C=\dfrac{\dot{m}}{\dot{V}}\\\Rightarrow C=\dfrac{25\times 10^6}{2\times 10^7}\\\Rightarrow C=1.25\ \mu\text{g/m}^3

The steady state concentration of pollutants in the valley, is 1.25\ \mu\text{g/m}^3.

6 0
3 years ago
An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea
melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, v=1.37\times10^7\ \rm m/s

Explanation:

<u>Given:</u>

  • Linear charge density of the ring=0.1\ \rm \mu C/m
  • Radius of the ring R=0.2 m
  • Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C

Potential due the ring at a distance x from the centre of the rings is given by

V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V

Let\Delta U be the change in potential Energy given by

\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s

So the electron will be moving with v=1.37\times10^7\ \rm m/s

5 0
3 years ago
A particle is moving along the x-axis. Its position as a function of time is given as x=bt-ct^2a) What must be the units of the
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Answer:

We are given x= bt +ct²

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A. bxt= m

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So b= m/s and c= m/s²

B.

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So at x=0 t=0

x=0 t= 2

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C. At rest wen V= 0

We have t= b/2c

D. To find acceleration we use

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3 years ago
A DC generator is the same as a(n)
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Answer:

A DC generator is the same as a(n)

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