Mac and Tosh are arguing in the cafeteria. Mac says that if he flings the Jell-O with a greater speed it will have a greater ine
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Answer:
a) M = 4,997 10²⁰ kg , b) T = 1.43 10³ s
Explanation:
a) This exercise should be solved in several parts, let's start by calculating the acceleration of gravity of this planet from kinematics
v = v₀ - a t
As it indicates that there is no atmosphere, the friction force is zero and the initial and final velocity have the same module, but the opposite direction
a = (v₀ - v) / t
a = (15 - (-15)) /9.00 = 30/9
a = 3.33 m / s²
Now we use Newton's second law where force is the force of universal attraction
F = m a
G m M / r² = m a
M = a r² / G
Let's calculate
M = 3.33 (1.00 10⁵)² / 6.67 10⁻¹¹
M = 4,997 10²⁰ kg
b) The period of the ship's orbit
In this case we have a centripetal acceleration
The radius of the orbit is the radius of the plant plus the height of the ship from the surface
R = + h
R = 1 10⁵ + 2.00 10⁴
R = 12 10⁴ m
F = m a
G m M / R² = m a
Centripetal acceleration is
a = v² / R
The orbit is circular therefore the velocity module is constant, so we can use the equation of uniform motion, where the distance is the length of the orbit, for a circle
d = 2π R
v = d / t
v = 2π R / T
Let's replace
G m M / R² = m (2π R / T)² / R
G M = R³ 4π² / T²
T² = 4π² R³ / G M
T² = (4π² (12 10⁴)³ / (6.67 10⁻¹¹ 4,997 10²⁰)
T² = 6.82 10¹⁶ / 3.33 10¹⁰
T = √ (2,048 10⁶)
T = 1.43 10³ s
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:
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:
where;
are specific entropies
R = universal gas constant
= pressure at location 1
= pressure at location 2
We obtain the specific properties of air at temperature at = (67°C + 273)K = 340 K from the table A-22 ( Ideal gas properties of air)
= 1.8279 kJ/kg.K
We also obtain the specific properties of air at temperature = 22°C + 273) K = 295 K
From the table A- 22
= 1.68515 kJ/kg . K
R =
0.95 bar
0.8 bar
Now replacing our values into equation (2) from above; we have;
Equating our result to equation (1)
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.
