1.t
2.f
3.f
4.t
5.t
6.t
7.f
8.f
9.t
10.t
11.f
12.t
13.t
14.f
15.t
16.f
17.t
18.t
19.f
20.t
21.f
22.t
23.f
24t
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
Judging on the fact that this is middle school physics, I would say that it is a beneficial relationship.
Answer:
20573.67N
Explanation:
Given;
mass (m) of the car = 2130kg
angle of inclination Θ = 15⁰
The normal force (F) on the car is given by
F = mgcosΘ
where g is the acceleration due to gravity.
Taking g as 10
and substituting the values of m and Θ into the equation. We have;
F = 2130 x 10 x cos 15⁰
F = 2130 x 10 x 0.9659
F = 20573.67N
Therefore the normal force on the car is 20573.67N
