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2 years ago

15

Which three applications use technologies that apply the Doppler effect?

1 answer:

ivolga24 [154]2 years ago

8 0

Answer:

A, B & C are correct.

Explanation:

In physics, Doppler effect is a term used to describe the change in frequency of a wave during the relative motion existing between the source of the wave the person acting a the observer.

Looking at the options given, this Doppler effect is applicable to: air traffic control where traffic controllers attempt to direct the traffic on the ground; weather forecasting where the conditions of the atmosphere at a particular time and place is being predicted; and in building inspection where we can use a non destructive test equipment known as Schmidt hammer to career out min destructive tests on a building.

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A mass weight of 120N is hung from two strings. what is the tension?

kramer

The weight should be shared between the two string equally. Therefore, tension in each string, T is;

T = 120 N/2 = 60 N

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3 years ago

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Light is a form of what

mart [117]

Light is a form of energy.
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3 years ago

61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A

Mademuasel [1]

Answer:

Part a)

$percentage = 21.3$ %

Part b)

$percentage = 2.13 \times 10^{-5}$ %

Explanation:

As we know that total power used in the room is given as

$P = P_1 + P_2 + P_3 + P_4$

here we have

$P_1 = (110)(3) = 330 W$

$P_2 = 100 W$

$P_3 = 60 W$

$P_4 = 3 W$

$P = 330 + 100 + 60 + 3$

$P = 493 W$

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

$110\times i = 493$

$i = 4.48 A$

now resistance of transmission line

$R = \frac{\rho L}{A}$

$R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}$

$R = 5.23 \ohm$

now power loss in line is given as

$P = i^2 R$

$P = (4.48)^2(5.23)$

$P = 105 W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{105}{493} \times 100$

$percentage = 21.3$ %

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

$110 \times 10^3 (i ) = 493$

$i = 4.48\times 10^{-3} A$

now power loss in line is given as

$P = i^2 R$

$P = (4.48 \times 10^{-3})^2(5.23)$

$P = 1.05 \times 10^{-4} W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{1.05 \times 10^{-4}}{493} \times 100$

$percentage = 2.13 \times 10^{-5}$ %

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3 years ago

A piece of silicon sample has a resistivity of 0.1 ω.Cm. Its thickness is 100µm. The electron mobility is 1350cm 2 v -1sec-1. Wh

lawyer [7]

The answer is Rh = 135 cm^3 and B = 0.05185 wh/m^2

Explanation:

Resitivity of silicon = 0.1

thickness = 100um

so, I = ma

Required to find out concentration of electron , we know that

Rh = up

By putting in the values,

Rh = 1350 x 0.1

Rh = 135 cm^3

Now consider,

Rh = 1 / Rh.q

= 1 / Rh . q

= 1 / 135 x1.609 x10^-19

= 4.6037 x 10^16 / cm^3

Vh = BIRh / w

B = Vh w/ IRh

B = -70 x10^-6 x 100 x10^-6 / 1x 10^-3 x 135 x 10^-6

B = 0.05185 wh / m^2

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3 years ago

Oduvanchick [21]

Answer:

the answer is different

Explanation:

i took the test

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3 years ago