Answer:
B
Explanation:
This is a two sample t-test and not a matched pair t-test
null hypothesis(H0) will be that mean energy consumed by copper rotor motors is greater than or equal to mean energy consumed by aluminium rotor motors
alternate hypothesis(H1) will be that mean energy consumed by copper rotor motors is less than or equal to mean energy consumed by aluminium rotor motors.
So, option D is rejected
The hypothesis will not compare mean of differences of values of energy consumed by copper rotor motor and aluminium rotor motor.
Option A and C are also rejected
Answer:
The answer is "+9.05 kw"
Explanation:
In the given question some information is missing which can be given in the following attachment.
The solution to this question can be defined as follows:
let assume that flow is from 1 to 2 then
Q= 1kw
m=0.1 kg/s
From the steady flow energy equation is:
![m\{n_1+ \frac{v^2_1}{z}+ gz_1 \}+Q= m \{h_2+ \frac{v^2_2}{2}+ gz_2\}+w\\\\\ change \ energy\\\\0.1[1.005 \times 800]-1= 0.01[1.005\times 700]+w\\\\w= +9.05 \ kw\\\\](https://tex.z-dn.net/?f=m%5C%7Bn_1%2B%20%5Cfrac%7Bv%5E2_1%7D%7Bz%7D%2B%20gz_1%20%5C%7D%2BQ%3D%20m%20%5C%7Bh_2%2B%20%5Cfrac%7Bv%5E2_2%7D%7B2%7D%2B%20gz_2%5C%7D%2Bw%5C%5C%5C%5C%5C%20change%20%5C%20energy%5C%5C%5C%5C0.1%5B1.005%20%5Ctimes%20800%5D-1%3D%200.01%5B1.005%5Ctimes%20700%5D%2Bw%5C%5C%5C%5Cw%3D%20%2B9.05%20%5C%20kw%5C%5C%5C%5C)
If the sign of the work performed is positive, it means the work is done on the surrounding so, that the expected direction of the flow is right.
Answer:
The exit temperature is 293.74 K.
Explanation:
Given that
At inlet condition(1)
P =80 KPa
V=150 m/s
T=10 C
Exit area is 5 times the inlet area
Now

If consider that density of air is not changing from inlet to exit then by using continuity equation

So 
m/s
Now from first law for open system

Here Q=0 and w=0

When air is treating as ideal gas

Noe by putting the values



So the exit temperature is 293.74 K.
Answer:
0 m/s , 3 m/s , 2 m/s^2
Explanation:
Given : s(t) = ( t^2 - 6t + 5)
v(t) = ds / dt = 2t - 6
s(0) = 5 m
s(6) = (6)^2 - 6*6 + 5 = 5 m
Vavg = ( s(6) - s(0) ) / 2 = 0 m\s
Find the turning point of particle:
ds/dt = 0 = 2t - 6
t = 3 sec
s(3) = 3^2 -6*3 + 5 = - 4
Total distance = 5 - (-4) + (5 - (-4)) = 18 m
Total time = 6s
Average speed = Total distance / Total time = 18 / 6 = 3 m/s
Taking derivative of v(t) to obtain a(t)
a (t) = dv(t) / dt = 2 m/s^2