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 with Explanation:
We are given that
Weight of an ore sample=17.5 N
Tension in the cord=11.2 N
We have to find the total volume and the density of the sample.
We know that
Tension, T=
=buoyancy force
T=Tension force
W=Weight
By using the formula

N

Where
=Volume of object
=Density of water
=Acceleration due to gravity
Substitute the values then we get


Volume of sample=
Density of sample,
Where mass of ore sample=1.79 kg
Substitute the values then, we get

Density of the sample=
Answer:
A) 
B) F = 1632.65 N
Explanation:
Given details
outside air speed is given as 
since inside air is atmospheric , 
a) By using bernoulli equation between outside and inside of flight


![\Delta P = \frac{1}{2} \rho[ v_2^2 -v_1^2]](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Crho%5B%20v_2%5E2%20-v_1%5E2%5D)
![\Delta P = \frac{1}{2} 1.29 [ 150^2 - 0^2]](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%20%5Cfrac%7B1%7D%7B2%7D%201.29%20%5B%20150%5E2%20-%200%5E2%5D)

b) force exerted on window
Area of window 
We know that force is given as


F = 1632.65 N