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zhenek [66]
3 years ago
10

A car travels 10.0 m/s. What is its velocity in km/h? A. 0.6 km/h B. 36,000 km/h C. 36 km/h D. 360 km/h

Physics
1 answer:
Mrrafil [7]3 years ago
5 0

Answer:

36km/h C.

Explanation:

10* 3600/1000= 10*18/5= 36km/h

C. 36km/h

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A 16 kg object cuts 4 km in 25 minutes, find the applied force on the object?
GenaCL600 [577]

Answer:

14min i think im not quite sure

Explanation:

7 0
2 years ago
A single insulated duct flow experiment using air operating at steady-state is performed in a lab. One measurement location (Sta
weqwewe [10]

Answer:

a) -0.0934 kJ/kg. K

b) The direction of flow is from right to left.

Explanation:

A free flow diagram of the horizontal insulated duct is as shown below.

NOW,

Let assume that the direction of flow is from left to right and consider the following relation for the entropy rate balance equation for a control volume as:

\frac{\sigma_{cv}}{m}= (s_2-s_1) \geq  0 \ \ \ -------> \ \ \ 1

Now; if the value for this relation is greater than zero; then we conclude that our assumption is correct.

If the value is less than zero; then we conclude that the assumption is wrong.

Then, the flow is said to be  in the opposite direction

Formula for the change in specific entropy can be calculated as:

s_2-s_1 = s^0(T_2) - s^0(T_1)-R \ In ( \frac{P_2}{P-1}) \ \ \  ------->  \ \ \ 2

where;

s_1, s_2 , s^0(T_2), s^0(T1) are specific entropies

R = universal gas constant

P_1 = pressure at location 1

P_2 = pressure at location 2

We obtain the specific properties of air at temperature at T_1 = (67°C + 273)K = 340 K from the table A-22 ( Ideal gas properties of air)

s^0(T1) = 1.8279 kJ/kg.K

We also obtain the specific properties of air at temperature T_2 = 22°C + 273) K = 295 K

From the table A- 22

s^0(T_2) = 1.68515 kJ/kg . K

R = \frac{8.314 kJ}{28.97 kg.K}

P_1 = 0.95 bar

P_2 = 0.8 bar

Now replacing our values  into equation (2) from above; we have;

s_2-s_1 = s^0(T_2) -s^0(T_1)-R \ In (\frac{P_2}{P_1} )

s_2-s_1 = 1.68515 -1.8279-\frac{8.314}{28.97}  \ In (\frac{0.8}{0.95} )

s_2-s_1 = 1.68515 -1.8279+ 0.0493

s_2-s_1 =-0.0934 \  kJ/kg.K

Equating our result to equation (1)

s_2-s_1 \geq 0\\-0.0934 \leq 0

Therefore , our assumption is wrong and the direction of flow is said to be from right to left.

We therefore conclude that the direction of flow is from right to left.

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3 years ago
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Here are the 2 reasons:

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A car travels 2155.0m in 195.9s. What is the car's average speed?
inna [77]

Average speed of the car is 11 m/s

Explanation:

  • Speed is calculated by the rate of change of displacement.
  • It is given by the formula, Speed = Distance/Time
  • Here, distance = 2155 m and time = 195.9 s

Speed of the car = 2155/195.9 = 11 m/s

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<span>Work: W = Fd. 50(distance) multiplied by 90(force) would equal 4500 J or, answer D</span>
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