Answer:
The rate at which this vacuum cleaner supply energy to the room when running is 285.71 Watts.
Explanation:
power efficiency of electric motor = 70% = 0.70
The power output of the vacuum cleaner =
= 200 W
The power output of the vacuum cleaner = 



The rate at which this vacuum cleaner supply energy to the room when running is 285.71 Watts.
Answer:
Below see details
Explanation:
A) It is attached. Please see the picture
B) First to calculate the overall mean,
μ=65∗25/75+80∗25/75+95∗25/75
μ=65∗25/75+80∗25/75+95∗25/75 = 80
Next to calculate E(MSTR) = σ2+(1/r−1) ∑ni(μi−μ)^2 = 5634
And E(MSE) = σ^2= 9
C) Yes, it is substantially large than E(MSE) in this case.
D) If we sampled 25 employees from each group, we are likely to get a F statistics to indicate differences of job satisfactions among three types of length of service of employees.
Answer:
T=151 K, U=-1.848*10^6J
Explanation:
The given process occurs when the pressure is constant. Given gas follows the Ideal Gas Law:
pV=nRT
For the given scenario, we operate with the amount of the gas- n- calculated in moles. To find n, we use molar mass: M=102 g/mol.
Using the given mass m, molar mass M, we can get the following equation:
pV=mRT/M
To calculate change in the internal energy, we need to know initial and final temperatures. We can calculate both temperatures as:
T=pVM/(Rm); so initial T=302.61K and final T=151.289K
Now we can calculate change of U:
U=3/2 mRT/M using T- difference in temperatures
U=-1.848*10^6 J
Note, that the energy was taken away from the system.