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sasho [114]
3 years ago
10

When light is scattered while spreading out through the atmosphere, the light rays A) converge and follow the same direction. El

iminate B) deflect randomly in multiple directions. C) bounce back toward their initial direction. D) undergo regular reflection from a particle.
Physics
2 answers:
boyakko [2]3 years ago
7 0

Answer:

I believe the answere is B)

Explanation:

Well, as the light enters the atmosphere it meets resestance.And since the atmosphere is essentially a thick layer of chemicals in the form of gasses under pressure it isn't very smooth.So as a result, the light mainly scatters towards different dirrections.I am happy to help.Contact me if you have any further questions.

Yours sincerely,

Manos.

rjkz [21]3 years ago
7 0
I believe the answer to be B
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True [87]

Explanation:

Hooke's law is a law of physics that states that the force needed to extend or compress a spring by some distance scales linearly with respect to that distance—that is, Fₛ = kx, where k is a constant factor characteristic of the spring, and x is small compared to the total possible deformation of the spring.

8 0
3 years ago
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Suppose you spray your sister with water from a garden hose. The water is supplied to the hose at a rate of 0.401 × 10 − 3 m 3 /
Maurinko [17]

To solve this problem it is necessary to apply the concepts related to the Flow Rate. It is understood as the volumetric amount transported during a certain time and can be calculated as

Q = AV

Where

A = Area

V = Velocity

Our values are given as

r = 5.27*10^{-3}m \rightarrow A = \pi r^2 = \pi ( 5.27*10^{-3})^2 = 8.72*10^{-5}m^2

Q = 0.401*10^{-3}m^3/s

Replacing and rearranging to find the velocity

V = \frac{Q}{A}

V = \frac{0.401*10^{-3}}{8.72*10^{-5}}

V= 4.59m/s

Therefore the speed of the water at the end of the Nozzle is 4.59m/s

6 0
3 years ago
A sphere of radius r = 5cm carries a uniform volume charge density rho = 400 nC/m^3. Q. What is the total charge Q of the sphere
Tanzania [10]

Answer:

The total charge Q of the sphere is 2.094\times10^{-10}\ C.

Explanation:

Given that,

Radius = 5 cm

Charge density J= 400\ nC/m^3

We need to calculate the total charge Q of the sphere

Using formula of charge

q=\rho V

Where, \rho = charge density

V = volume

Put the value into the formula

q=\rho\times(\dfrac{4}{3}\pi r^3)

Put the value into the formula

q=\dfrac{4}{3}\times\pi\times400\times10^{-9}\times(5\times10^{-2})^3

q=2.094\times10^{-10}\ C

Hence, The total charge Q of the sphere is 2.094\times10^{-10}\ C.

6 0
3 years ago
 A car whose initial speed is 30 m/s slows uniformly to 10 m/s in 5 seconds. Determine the acceleration of the car. Sketch a gra
Anton [14]
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  • Time=5s

\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}

\\ \sf\longmapsto Acceleration=\dfrac{10-30}{5}

\\ \sf\longmapsto Acceleration=\dfrac{-20}{5}

\\ \sf\longmapsto Acceleration=-4m/s^2

Now

For the second question

  • Time=3s

Using 2nd equation of motion

\\ \sf\longmapsto s=ut+\dfrac{1}{2}at^2

\\ \sf\longmapsto s=30(3)+\dfrac{1}{2}(-4)(3)^2

\\ \sf\longmapsto s=90-2(9)

\\ \sf\longmapsto s=90-18

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5 0
3 years ago
Atoms in a solid are in continuous vibrational motion due to thermal energy. At room temperature, the amplitude of these atomic
Bingel [31]

Answer:

10 m/s

Explanation:

Given:

Amplitude of atomic vibrations (λ) = 10⁻⁹ cm = 10⁻⁹ × 10⁻² m = 10⁻¹¹ m [1 cm = 10⁻² m]

Frequency of the vibrations (f) = 10¹² Hz

In order to find the atom's maximum speed, we need to make use of the formula that relates speed, frequency and wavelength of the vibration.

Therefore, the formula for maximum speed is given as:

v=f\lambda

Now, plug in the values given and solve for speed 'v'. This gives,

v=(10^{12}\ Hz)(10^{-11}\ m)\\\\v=10^{12-11}\ m/s\\v=10\ m/s

Therefore, the atom's maximum speed due to thermal energy provided is 10 m/s.

6 0
4 years ago
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