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Morgarella [4.7K]
3 years ago
7

g Suppose you're on a hot air balloon ride, carrying a buzzer that emits a sound of frequency f. If you accidentally drop the bu

zzer over the side while the balloon is rising at constant speed, what can you conclude about the sound you hear as the buzzer falls toward the ground
Physics
1 answer:
dlinn [17]3 years ago
8 0

Answer:

the observed frequency will reduce but the wavelength will increase

Explanation:

As we know

fo = fs (v/(v-vs))

fo = observed frequency

vs = velocity of source

As per this equation,  

When an observer moves away from the stationary source, the observed frequency reduces. Since the observer in the balloon is moving away from the source which itself is moving in opposite direction, the observed frequency will reduce.  

Since wavelength = V/fs . The source frequency is unchanged but the velocity is increasing as it is moving in downward direction. Hence, the wavelength will increase

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How to tell how much work gravity does on something?
Grace [21]
By dropping a ball and seeing how long it takes to hit the ground or throw a ball up and time it as well
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3 years ago
Why do things become hot or cold?<br> Relate to Thermal Energy and The Law of Conservation of Energy
Mekhanik [1.2K]

<u>Answer:</u>

Things become hot and cold because of the transfer of energy.

<u>Explanation:</u>

The energy possessed by an object or system is called thermal energy and heat is the flow of this energy. While the law of conversation of energy states that energy is not destroyed or created, it just transfers from one object to another.  

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4 0
3 years ago
A factory worker pushes a crate of mass 31.0 kg a distance of 4.35 m along a level floor at constant velocity by pushing horizon
Debora [2.8K]

Answer:

a. 79.1 N

b. 344 J

c. 344 J

d. 0 J

e. 0 J

Explanation:

a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force F_p by the worker must be equal to the friction force F_f on the crate, which is the product of friction coefficient μ and normal force N:

Let g = 9.81 m/s2

F_p = F_f = \mu N = \mu mg = 0.26 * 31 * 9.81 = 79.1 N

b. The work is done on the crate by this force is the product of its force F_p and the distance traveled s = 4.35

W_p = F_ps = 79.1*4.35 = 344 J

c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35

W_f = F_fs = -79.1*4.35 = -344 J

This work is negative because the friction vector is in the opposite direction with the distance vector

d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.

e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction

W_p + W_f = 344 - 344 = 0 J

8 0
3 years ago
Read 2 more answers
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