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zvonat [6]
2 years ago
9

A 0.5 kg rock is thrown straight downward off a cliff. If the rock only experience a gravitation force for 4.3 seconds, and ends

with a downward velocity of 46.1 m/s, what was rock initial velocity? What is the gravitational force?
Physics
1 answer:
pogonyaev2 years ago
3 0

Answer:Answer:

See diagram for answers. The force and acceleration vectors are directed inwards towards the center of the circle and the velocity vector is directed tangent to the circle.

Explanation:

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4 years ago
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What would MOST LIKELY happen if the amount of incoming solar energy decreased without a change in the amount of reflection or o
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Is power a vector quantity​
Morgarella [4.7K]

Answer:

Power is a scalar quantity and has a unit,a magnitude(a numerical value) but no direction.

Explanation:

7 0
3 years ago
When you walk at an average speed (constant speed, no acceleration) of 20.7 m/s in 75.8 sec you will cover a distance of______?
insens350 [35]

Answer:

You will cover a distance of 1569.06 metres. Or you could round down to 1,569m.

Explanation:

20.7*75.8=1563.06

7 0
4 years ago
Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.
Sliva [168]

To solve the problem it is necessary to take into account the concepts related to energy efficiency in the engines, the work done, the heat input in the systems, the exchange and loss of heat in the soupy the radius between the work done the lost heat ( efficiency).

By definition the efficiency of the heat engine is

\epsilon = 1- \frac{T_c}{T_h}

Where,

T_c = Temperature at the room

T_h  =Temperature of the soup

The work done is defined as,

dW = \epsilon(dQ_h)

Where Q_h represents the input heat and at the same time is defined as

dQ_h =c_v (dT_h)

Where,

c_V =Specific Heat

The change at the work would be defined then as

dW = \epsilon(dQ_h)

dW = \epsilon c_v (dT_h)

dW = (1-\frac{T_c}{T_h})c_v (dT_h)

W = \int dW = \int (1-\frac{T_c}{T_h})c_v (dT_h)

W = c_v (T_h-T_c)-c_v T_c ln(\frac{T_h}{T_c})

On the other hand we have that the heat lost by the soup is equal to

dQ_h =c_v (dT_h)

Q_h =c_v (T_h-T_c)

The ratio between both would be,

\frac{W}{Q_h} = \frac{c_v (T_h-T_c)-c_v T_c ln(\frac{T_h}{T_c})}{c_v (T_h-T_c)}

\frac{W}{Q_h} = \frac{1+ln(\frac{T_h}{T_c})}{1-\frac{T_h}{T_c}}

Replacing with our values we have,

\frac{W}{Q_h} = 1+\frac{ln(\frac{340K}{300K})}{1-\frac{340K}{300K}}

\frac{W}{Q_h} = 0.0613

Therefore the fraction of heat lost by the soup that can be turned into useable work by the engine is 0.0613.

5 0
3 years ago
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