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AURORKA [14]
3 years ago
8

On the surface of planet y, which has a mass of 4.83x1024 kg, a 30.0 kg object weighs 50.0 n. what is the radius of the planet?

Physics
1 answer:
Aleksandr [31]3 years ago
5 0
We can solve the problem in two steps:

1) From the weight W=50.0 N of the object, we can find the value of the gravitational acceleration g of the planet. In fact, the weight is equal to
W=mg
where m=30 kg is the mass of the object. From this, we find g:
g= \frac{W}{m}= \frac{50.0 N}{30 kg}=1.67 m/s^2

2) The gravitational acceleration of a planet with mass M and radius r is given by
g= \frac{GM}{r^2}
where G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2} is the gravitational  constant. In our problem, the mass of the planet is 
M=4.83 \cdot 10^{24} kg, and we found g in step 1), g=1.67 m/s^2, so we have everything to solve and find the value of the radius r:
r= \sqrt{ \frac{GM}{g} }= \sqrt{ \frac{(6.67\cdot 10^{-11})(4.83 \cdot 10^{24})}{1.67} }=1.39\cdot 10^7 m
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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
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Answer:

a) -0.0934 kJ/kg. K

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

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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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