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2 years ago

Which of the following describes sound waves?

Physics

2 answers:

lianna [129]2 years ago

8 0

The answer is c) electromagnetic sawed in which the vibrations are perpendicular to the motion of the sound

hoa [83]2 years ago

3 0

Answer:

D. Mechanical waves in which the vibrations are parallel to the

motion of the sound

Explanation:

Ap3x

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Drink water at least every<br> weather.<br> minutes while exercising in hot

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3 years ago

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As a projectile falls, what happens to the components of velocity?

netineya [11]

Answer:

Option (c).

Explanation:

An object when when projected at an angle, will have some horizontal velocity and vertical velocity such that,

$v_x=v\cos\theta\ \text{and}\ v_y=v\sin\theta$

$\theta$ is the angle of projection

The horizontal component of the projectile remains the same because there is no horizontal motion. Vertical component changes at every point.

As a projectile falls, vertical velocity increases in magnitude, horizontal velocity stays the same .

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3 years ago

Organizing and interpreting information from the senses is called

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I believe the answer you are looking for is perception.

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3 years ago

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A 3.0 kg rifle is held firmly by a 50.0 kg woman, initially standing still. A 0.06 kg bullet leaves the rifle muzzle with a velo

ohaa [14]

Answer:

48kg

Explanation:

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2 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

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

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

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

6 0

3 years ago