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djyliett [7]
2 years ago
10

Which of the following objects is in static equilibrium?

Physics
1 answer:
Likurg_2 [28]2 years ago
4 0
<h3><u>Answer:- </u></h3>

\longrightarrow \textsf{D. A bicycle sitting on the ground}

<h3><u>Solution</u>:-</h3>

<u>Let us check all options -</u>

  • A) An apple falling from a tree is under the influence of gravity. Thus some force is acting on the apple and it can not said to be in state of equilibrium.

  • B) A car moving at a constant velocity may said to be in state of equilibrium but the since the car is in motion ,equilibrium is dynamic in nature .Thus , it is not in static equilibrium.

  • C) A motorcycle speeding up has some acceleration due to some forces. Thus it can also not be said in state of static equilibrium.

  • D) Since the force acting on bicycle sitting down in zero as well as it is in state of rest , thus bicycle sitting on the ground may said to be in static equilibrium.

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Answer:

D

Explanation:

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3 years ago
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The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .
vlabodo [156]

Answer: 90\°

Explanation:

The Compton Shift \Delta \lambda in wavelength when the photons are scattered is given by the following equation:

\Delta \lambda=\lambda_{c}(1-cos\theta)     (1)

Where:

\lambda_{c}=2.43(10)^{-12} m is a constant whose value is given by \frac{h}{m_{e}c}, being h the Planck constant, m_{e} the mass of the electron and c the speed of light in vacuum.

\theta) the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through 180\°:

\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))     (2)

\Delta \lambda_{max}=\lambda_{c}(1-(-1))    

\Delta \lambda_{max}=2\lambda_{c}     (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)      (4)

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)      (5)

If we want \frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}, 1-cos\theta   must be equal to 1:

1-cos\theta=1   (6)

Finding \theta:

1-1=cos\theta

0=cos\theta  

\theta=cos^{-1} (0)  

Finally:

\theta=90\°    This is the scattering angle that will produce \frac{1}{4}\Delta \lambda_{max}      

7 0
3 years ago
represents the space-time, speed-time and acceleration-time graphs for a ball pulled upwards with a speed of 10 m / s from 1 met
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3 years ago
Jupiter's semimajor axis is 7.78×1011 m. The mass of the Sun is 1.99×1030 kg. (a) What is the period of Jupiter's orbit in secon
dedylja [7]

Explanation:

It is given that,

Semi major axis of the Jupiter, a=7.78\times 10^{11}\ m

Mass of the sun, M=1.99\times 10^{30}\ kg

(a) Let T is the period of Jupiter's orbit. It is given by :

T^2\propto a^3

T^2=\dfrac{4\pi^2}{GM}a^3

T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.99\times 10^{30}}\times (7.78\times 10^{11})^3

T=3.74\times 10^8\ s

(b) We know that,

1\ year=3.154\times 10^7\ s

or

1\ s=3.171\times 10^{-8}\ year

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T = 11.859 earth years

Hence, this is the required solution.

7 0
4 years ago
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At what depth of lake water is the pressure equal to 201kpa?
Alex787 [66]

Answer:

20m

Explanation:

Pressure = pgh

p = density of water 1000

kg/m^3

g = acceleration due to gravity 9.81 m/s^2

h is the depth of water

Pressure = 201 kPa = 201 x 10^3 Pa

201 x 10^3 = 1000 x 9.81 x h

201 x 10^3 = 9810h

h = 20.49 m

Approximately 20 m

5 0
3 years ago
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