Answer:
The part of the system that is considered the resistance force is;
B
Explanation:
The simple machine is a system of pulley that has two pulleys
The effort, which is the input force at A gives the value of the tension at C and D which are used to lift the load B
Therefore, we have;
A = C = D
B = C + D = C + C = 2·C
∴ C = B/2
We have;
C = B/2 = A
Therefore, with the pulley only a force, A equivalent to half the weight, B, of the load is required to lift the load, B
The resistance force is the constant force in the system that that requires an input force to overcome in order for work to be done
It is the force acting to oppose the sum of the other forces system, such as a force acting in opposition to an input force
Therefore, the resistance force is the load force, B, for which the input force, A, is required in order for the load to be lifted.
Answer:
a) The Net power developed in this air-standard Brayton cycle is 43.8MW
b) The rate of heat addition in the combustor is 84.2MW
c) The thermal efficiency of the cycle is 52%
Explanation:
To solve this cycle we need to determinate the enthalpy of each work point of it. If we consider the cycle starts in 1, the air is compressed until 2, is heated until 3 and go throw the turbine until 4.
Considering this:
Now we can calculate the enthalpy of each work point:
h₁=281.4KJ/Kg
h₂=695.41KJ/Kg
h₃=2105KJ/Kg
h₄=957.14KJ/Kg
The net power developed:
The rate of heat:
The thermal efficiency:
Answer:
a) v = +/- 0.323 m/s
b) x = -0.080134 m
c) v = +/- 1.004 m/s
Explanation:
Given:
a = - (0.1 + sin(x/b))
b = 0.8
v = 1 m/s @ x = 0
Find:
(a) the velocity of the particle when x = -1 m
(b) the position where the velocity is maximum
(c) the maximum velocity.
Solution:
- We will compute the velocity by integrating a by dt.
a = v*dv / dx = - (0.1 + sin(x/0.8))
- Separate variables:
v*dv = - (0.1 + sin(x/0.8)) . dx
-Integrate from v = 1 m/s @ x = 0:
0.5(v^2) = - (0.1x - 0.8cos(x/0.8)) - 0.8 + 0.5
0.5v^2 = 0.8cos(x/0.8) - 0.1x - 0.3
- Evaluate @ x = -1
0.5v^2 = 0.8 cos(-1/0.8) + 0.1 -0.3
v = sqrt (0.104516)
v = +/- 0.323 m/s
- v = v_max when a = 0:
-0.1 = sin(x/0.8)
x = -0.8*0.1002
x = -0.080134 m
- Hence,
v^2 = 1.6 cos(-0.080134/0.8) -0.6 -0.2*-0.080134
v = sqrt (0.504)
v = +/- 1.004 m/s
Print(‘Predictions are hard’)
print(‘Especially about the future’)
user_num = 5
print(‘user_num is:’+str(user_num))
There are many ways to do the last print. Look up string interpolation
