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Feliz [49]
3 years ago
11

A man stands by a train track to watch the train car go by. Inside, a man walks through the train car eating a hot dog (the hot

dog is moving 1 m/s towards his mouth). Determine the hot dog's velocity with respect to the man standing still outside of the train car.
A) -11 m/s
B) -9 m/s
C) 9 m/s
D) 11 m/s

Physics
2 answers:
antiseptic1488 [7]3 years ago
5 0

Answer:

D) 11 m/s

Explanation:

The problem asks us to calculate the velocity of the hot dog with respect to the observer stationary outside the train. This velocity is given by:

v=v_t + v_m + v_h

where

v_t=+10 m/s is the velocity of the train (towards right)

v_m=+2 m/s is the velocity of the man (towards right)

v_h=-1 m/s is the velocity of the hot-dog (towards left, so we put a negative sign)

Substituting the numbers into the equation, we find

v=+10 m/s+2 m/s-1 m/s=+11 m/s

and the positive sign means the velocity is toward right.

xenn [34]3 years ago
4 0

Answer:

D) 11 m/s

Explanation:

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Answer:

Explanation:

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To objective is to find the:

(i) required heat exchanger area.

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

Given that:

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

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Q_{supplied } = \dfrac{2}{0.03} \ MW

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

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\Delta T_1 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_1 = 300 -290 \\ \\ \Delta T_1 = 10 \ K

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66.667 \times 10^6 = 1200 \times A \times 4.97 \\ \\  A= \dfrac{66.667 \times 10^6}{1200 \times 4.97} \\ \\  \mathbf{A = 11178.236 \ m^2}

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Q_{H} = mC_p(T_{in} -T_{out} )  \\ \\  66.667 \times 10^6= m \times 4.18 (300 -292) \\ \\ m = \dfrac{  66.667 \times 10^6}{4.18 \times 8} \\ \\  \mathbf{m = 1993630.383 \ kg/s}

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what is potential difference?

The difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other.

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