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amm1812
3 years ago
13

An object is (starting at rest) begins to plummet towards the earth. Assuming it is in a vacuum, how fast is the object travelin

g after 5s?

Physics
1 answer:
Eva8 [605]3 years ago
8 0

Answer:

–50.96

Explanation:

The following data were obtained from the question:

Initial velocity (Vᵢ) = 0 m/s

Acceleration (a) = – 9.8 m/s²

Time (t) = 5.2 s

Final velocity (Vբ) =.?

Acceleration is simply defined as the change of velocity with time. Mathematically, it is expressed as:

Acceleration (a) = Final velocity (Vբ) – Initial velocity (Vᵢ) /Time (t) =

a = (Vբ – Vᵢ) / t

With the above formula, we can determine how fast the object is traveling after 5 s as follow:

Initial velocity (Vᵢ) = 0 m/s

Acceleration (a) = – 9.8 m/s²

Time (t) = 5.2 s

Final velocity (Vբ) =.?

a = (Vբ – Vᵢ) / t

– 9.8 = (Vբ – 0) / 5.2

– 9.8 = Vբ / 5.2

Cross multiply

Vբ = –9.8 × 5.2

Vբ = –50.96 m/s

Therefore, the object is traveling at

–50.96 m/s

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In order to calculate the time taken by the snowball to reach the highest point in its journey, we need to consider the variables along the y-direction.

Let us list out what we know from the question so that we can decide on the equation to be used.

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Acceleration in the Y direction a_{y} = -9.8 m/s^{2}, since the acceleration due to gravity points in the downward direction.

Final Y Velocity V_{fy} = 0 because at the highest point in its path, an object comes to rest momentarily before falling down.

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From the list above, it is easy to see that the equation that best suits our purpose here is V_{fy} = V_{iy} + a_{y}t

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Explanation:

Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.

To objective is to find the:

(i) required heat exchanger area.

(ii) flow rate to be maintained in the evaporator.

Given that:

water temperature = 300 K

At a reasonable depth, the water is cold and its temperature = 280 K

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where;

\zeta = \dfrac{W_{out}}{Q_{supplied }}

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However, from the evaporator, the heat transfer Q can be determined by using the formula:

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