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KATRIN_1 [288]
3 years ago
15

The period of a sound wave is 0.002 seconds. The speed of sound is 344 m/s. whats the frequency

Physics
1 answer:
Archy [21]3 years ago
4 0

Answer:

= 500 Hz

Explanation:

A Period of a wave is the time taken for one complete oscillations by the wave particles. It is measured in seconds.

Frequency on the other hand is the number of complete oscillations covered by a wave in one second. It is measured in Hertz.

Period is the inverse of frequency.

That is; Period, T = 1/ frequency, f

T = 1/f and thus; f = 1/T

Therefore;

f = 1/0.002 seconds

 = 500 Hz

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7nadin3 [17]

Answer:

<h2>C. <u>0.55 m/s towards the right</u></h2>

Explanation:

Using the conservation of law of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.

Momentum = Mass (M) * Velocity(V)

BEFORE COLLISION

Momentum of 0.25kg body moving at 1.0m/s = 0.25*1 = 0.25kgm/s

Momentum of 0.15kg body moving at 0.0m/s(body at rest) = 0kgm/s

AFTER COLLISION

Momentum of 0.25kg body moving at x m/s = 0.25* x= 0.25x kgm/s

<u>x is the final velocity of the 0.25kg ball</u>

Momentum of 0.15kg body moving at 0.75m/s(body at rest) =

0.15 * 0.75kgm/s = 0.1125 kgm/s

Using the law of conservation of momentum;

0.25+0 = 0.25x + 0.1125

0.25x = 0.25-0.1125

0.25x = 0.1375

x = 0.1375/0.25

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Since the 0.15 kg ball moves off to the right after collision, the 0.25 kg ball will move at <u>0.55 m/s towards the right</u>

<u></u>

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3 years ago
A 5-kg projectile is fired over level ground with a velocity of 200 m/s at an angle of 25°above the horizontal. Just before it h
Nutka1998 [239]

Answer:

the change in thermal energy of the projectile is 43.8 kJ

Explanation:

Given;

mass of the object, m = 5kg

initial velocity of the projectile, v₁ = 200 m/s

final  velocity of the projectile, v₂ = 150 m/s

To determine the the change in the thermal energy of the projectile and air, we consider change in potential and kinetic enrgy of the projectile. Since the projectile was fired over level ground, change in potential energy is zero.

Then, change in thermal energy of the projectile, KE = Δ¹/₂mv²

KE = Δ¹/₂mv² = ¹/₂m(v₁²-v₂²)

KE =  ¹/₂ × 5(200²-150²) = 2.5(17500) = 43750 J = 43.8 kJ

Therefore, the change in thermal energy of the projectile is 43.8 kJ

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{\texttt{\huge{\red{ANSWER :-}}}}

<h3><u>L</u><u>A</u><u>P</u><u>L</u><u>A</u><u>C</u><u>E</u><u> </u><u>L</u><u>A</u><u>W</u><u> </u><u>:</u><u>-</u></h3>

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