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Kruka [31]
3 years ago
5

1. A pumpkin with a mass of 2 kg accelerates 2 m/s/s when an unknown force is applied to it. What is the amount of the force? ​

Physics
1 answer:
Andreyy893 years ago
8 0

Answer:

<h3>The answer is 4 N</h3>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 2 × 2

We have the final answer as

<h3>4 N</h3>

Hope this helps you

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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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