Answer:
q_poly = 14.55 KJ/kg
Explanation:
Given:
Initial State:
P_i = 550 KPa
T_i = 400 K
Final State:
T_f = 350 K
Constants:
R = 0.189 KJ/kgK
k = 1.289 = c_p / c_v
n = 1.2 (poly-tropic index)
Find:
Determine the heat transfer per kg in the process.
Solution:
-The heat transfer per kg of poly-tropic process is given by the expression:
q_poly = w_poly*(k - n)/(k-1)
- Evaluate w_poly:
w_poly = R*(T_f - T_i)/(1-n)
w_poly = 0.189*(350 - 400)/(1-1.2)
w_poly = 47.25 KJ/kg
-Hence,
q_poly = 47.25*(1.289 - 1.2)/(1.289-1)
q_poly = 14.55 KJ/kg
Answer:
<h3>The answer is 45 J</h3>
Explanation:
The work done by an object can be found by using the formula
<h3>workdone = force × distance</h3>
From the question
distance = 3 meters
force = 15 newtons
We have
workdone = 15 × 3
We have the final answer as
<h3>45 J</h3>
Hope this helps you
Answer: 1010.92 m/s
Explanation:
According to Newton's law of universal gravitation:
(1)
Where:
is the gravitational force between Earth and Moon
is the Gravitational Constant
is the mass of the Earth
is the mass of the Moon
is the distance between the Earth and Moon
Asuming the orbit of the Moon around the Earth is a circular orbit, the Earth exerts a centripetal force on the moon, which is equal to :
(2)
Where 
is the centripetal acceleration given by:
(3)
Being 
the orbital velocity of the moon
Making (1)=(2):
(4)
Simplifying:
(5)
Making (5)=(3):
(6)
Finding :
(7)
(8)
Finally:
The weight of the box is (mass) x (gravity) = (50 kg) x (9.8m/s²) = 490 newtons.
If the box is sliding at constant speed, and not speeding up or slowing down,
that means that the horizontal forces on it add up to zero.
Since you're pushing on it with 53N in <em><u>that</u></em> direction, friction must be pulling
on it with 53N in the <u><em>other</em></u> direction.
The 53N of friction is (the weight) x (the coefficient of kinetic friction).
53N = (490N) x (coefficient).
Divide each side by 490N : Coefficient = (53N) / (490N) = 0.1082 .
Rounded to the nearest hundredth, that's <em>0.11 </em>. (choice 'd')
Answer:
155.38424 K
2.2721 kg/m³
Explanation:
= Pressure at reservoir = 10 atm
= Temperature at reservoir = 300 K
= Pressure at exit = 1 atm
= Temperature at exit
= Mass-specific gas constant = 287 J/kgK
= Specific heat ratio = 1.4 for air
For isentropic flow
The temperature of the flow at the exit is 155.38424 K
From the ideal equation density is given by
The density of the flow at the exit is 2.2721 kg/m³
