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Pani-rosa [81]
4 years ago
11

Question 1 (1 point)

Physics
1 answer:
TiliK225 [7]4 years ago
3 0

Answer:

The resultant velocity of the plane relative to the ground is;

150 kh/h north

Explanation:

The flight speed of the plane = 210 km/h

The direction of flight of the plane = North

The speed at which the wind is blowing = 60 km/h

The direction of the wind = South

Therefore, representing the speed of the plane and the wind in vector format, we have;

The velocity vector of the plane = 210.\hat j

The velocity vector of the wind = -60.\hat j

Where, North is taken as the positive y or \hat j direction

The resultant velocity vector is found by summation of the two vectors as follows;

Resultant velocity vector = The velocity vector of the plane + The velocity vector of the wind

Resultant velocity vector = 210.\hat j  + (-60.\hat j) =  210.\hat j  - 60.\hat j = 150.\hat j

The resultant velocity vector = 150.\hat j

Therefore, the resultant velocity of the plane relative to the ground = 150 kh/h north.

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2 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.
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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.

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