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Mumz [18]
3 years ago
5

In the desert air, sound travels at 358 meters per second (m/s). In the polar air, sound travels at 330 meters per

Physics
1 answer:
Sati [7]3 years ago
7 0

Answer:

The correct option is;

C. The temperature of the air affects the speed of sound

Explanation:

The information given are;

The speed with which sound travels in the desert air = 358 m/s

The speed with which sound travels in the polar air = 330 m/s

The major difference between the desert air and the polar air is that the temperature of the desert air is hotter than the temperature of the air in the polar region. Therefore the speed of sound is affected by the air temperature.

The equation that gives the relationship of the speed of sound in air, v_{air}, to temperature is presented as follows;

v_{air} = 331\frac{m}{s} \times \sqrt{\dfrac{T_K}{273 K} } =  331\frac{m}{s} \times \sqrt{1 + \dfrac{T_{^{\circ}C}}{273 ^{\circ} C} }

Which shows that the speed of sound in air, v_{air}, rises as the temperature of the air rises.

Where;

T_{^{\circ}C} = The temperature of the air in degrees Celsius

T_K = The temperature of the air in degrees Kelvin.

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gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

u(x) = \frac{-2}{x} + \frac{4}{x^2}

At x = 1.00 m

u(1) = \frac{-2}{1} + \frac{4}{1^2}

= 4J

Kinetic energy = (1/2)mv²

= \frac{1}{2} * 4(3)^2

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

u(5) = \frac{-2}{5} + \frac{4}{5^2}

= \frac{4-10}{25} = \frac{-6}{25} J

= -0.24J

Kinetic energy =

\frac{1}{2} * 4Vf^2

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

Vf = \sqrt{frac{22.4}{2}

= 3.33 m/s

b) magnitude of force when x = 5.0m

u(x) = \frac{-2}{x} + \frac{4}{x^2}

\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}

= \frac{2}{x^2} - \frac{8}{x^3}

At x = 5.0 m

\frac{2}{5^2} - \frac{8}{5^3}

F = \frac{2}{25} - \frac{8}{125}

= 0.016N

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3 years ago
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Answer:

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Two flat glass plates are separated at one end by a wire of diameter 0.20 mm; at the other end they touch. Thus, the air gap bet
stepan [7]

Answer:

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4 years ago
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