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Georgia [21]
3 years ago
11

I need help on career prep

Physics
1 answer:
Tema [17]3 years ago
4 0
Here is some helpful information:

- Career prep is dedicating time and energy to deciding what you want to spend doing as life work. In other words it is choosing a career.

- Career prep involves preparation, research, resources, connections, decision making....

- Career prep is meant to be a process that helps you choose the field that you will pursue . It is meant to prepare you for the career you have chosen to commit to.

- Career prep involves you asking serious questions like what i am good at? what do I like doing? what do I so well I can get paid for doing it? what do I do that I don't mind working hard every day?

I hope this helped. please vote my answer branliest. Thanks.
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A music fan at a swimming pool is listening to a radio on a diving platform. The radio is playing a constant- frequency tone whe
joja [24]

Answer:

The Doppler Effect is given by the following relation;

f' = \left (\dfrac{v + v_0}{v - v_s} \right) \times f

Where;

f' = The frequency the observer hears

f = Actual frequency of the wave

v = The velocity of the sound wave

v_o = The velocity of the observer

v_s = The velocity of the source

Where the observer is stationary, we have;

(i) When the source is moving in the direction of the observer

f' = \left (\dfrac{v }{v - v_s} \right) \times f

(ii) When the source is receding from the observer, we have;

f' = \left (\dfrac{v }{v + v_s} \right) \times f

Therefore;

(a) A person left behind on the platform

For a person left behind on the platform, we have that the radio source is receding, therefore, we have;

f' = \left (\dfrac{v }{v + v_s} \right) \times f

(1) Given that (v + v_s) > v, therefore, v < (v + v_s), f' < f, the frequency heard by the person left on the platform, f', is smaller (lower) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively smaller

(b) A person down below floating on a rubber raft

For the the person down below on the rubber raft, the radio source is advancing

Therefore, the radio source is moving towards the person at rest down on the rubber raft, therefore, we have;

f' = \left (\dfrac{v }{v - v_s} \right) \times f

(1) Given that (v - v_s) < v, therefore, f' > f, the frequency heard by the person down below floating on the rubber raft, f', is greater (higher) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively greater (higher)

Explanation:

7 0
3 years ago
How many pounds is a 1 pound rock?
RideAnS [48]

Answer: 1 pound.

Explanation:

6 0
3 years ago
Read 2 more answers
Which of the following processes can produce either rain or snow
umka21 [38]

Bergeron–Findeisen Process.

3 0
3 years ago
Read 2 more answers
What is the law of variation of the period T of a simple pendulum
DIA [1.3K]
The period of a simple pendulum is given by:
T=2 \pi  \sqrt{ \frac{L}{g} }
where L is the length of the pendulum and g=9.81 m/s^2 is the gravitational acceleration. As we can see, the period of a simple pendulum depends only on its length.
3 0
3 years ago
A football player runs at 8m/s and plows into a 80kg referee standing on the field causing the referee to fly forward at 5m/s. I
madreJ [45]

The topic here is momentum.

When a collision is said to be elastic, it means that the colliding objects now travel at their own new, indivual and distinct velocities, often in different directions.

So we write that as,

(mass of football player x velocity of football player) + (mass of referee x velocity of referee) = (mass of football player x velocity of football player) + (mass of referee x velocity of referee)

(M × 8) + (80 × 0) = (M× 0) + (80 × 5)

8M = 400

M = 50 kg

4 0
3 years ago
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