Invader Zim’s spaceship is sitting at rest in outer space. The ship then accelerates at a uniform rate of 12 m/s2 for 10 seconds
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Answer:
a) 0.28 m or 28 cm is the minimum height above ground the fish reaches.
b) at the height of 0.484 m height , the pufferfish will eventually come to rest.
c) There exists two types of energy remain at the equilibrium point in the system. These are :
Gravitational potential energy = 23.72J
Spring potential energy = 0.384 J
Explanation:
Given that :
Mass of the pufferfish m =5kg
initial height of the fish h =5.5m
length of the spring l =0.5m
Spring constant K =3000N/m
a)
Assuming no energy loss to friction, what is the minimum height above the ground that the pufferfish reaches?
Lets assume that the minimum height the fish reaches is = x meters
Now by using the conservation of energy; we realize that :
Initial total energy = final total energy
Gravitational potential energy =
Gravitational potential energy' + Spring potential energy (kinetic energy is zero in both cases)
Replacing our given values into the above equation; we have :
269.5 = 47.5 x + 1500(0.5 -x )²
269.5 = 47.5 x + 1500(0.25 - x²)
269.5 = 47.5 x + 375 - 1500 x²
269.5 - 375 = 47.5 x - 1500 x²
-105.5 = 47.5 x - 1500 x²
-105.5 + 1500 x² - 47.5 x = 0
1500 x² - 47.5 x - 105.5 = 0
By using quadratic equation and taking the positive value;
x = 0.28 m or 28 cm is the minimum height above ground the fish reaches.
b)
At the equilibrium position the weight of fish will be equal to the force applied by the spring thus
mg = kx
substituting our given values ; we have:
(5)(9.8) = 3000x
x = 61.22
x = 0.016m : so this is the compression in the spring
Now; to determine the height the pufferfish gets to before it eventually come to rest; we have
(0.5-0.016) m = 0.484m
therefore, at the height of 0.484 m height , the pufferfish will eventually come to rest.
c)
There exists two types of energy remain at the equilibrium point in the system. These are :
Gravitational potential energy = mgh' = (5)(9.8)(0.484)
= 23.72J
and spring potential energy
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.
