Answer:
5.571 sec
Explanation:
angular frequency = √ (k/m) = √ (49.3 / 5) = 3.14 rad/s
Period To = 2π / angular frequency
Period To = 2π/3.14 = 2 × 3.14 / 3.142 = 2.00 sec which you got
T measured by the observer = To / (√ (1 - (v²/c²))) = 2 / √( 1 - 0.871111) = 2 / 0.35901 = 5.571 sec
t=2.00/(1-√((2.80*10^8)^2/(3.00*10^8)^2))= should have been ( To / (√ (1 - (v²/c²))). where To = 2.00 sec
In series with the circuit, so for it pass the current to be mensured.
Letter A
If you notice any mistake in my english, please let me know, because i am not native.
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
Answer:
Explanation:
a ) Thermal efficiency = work output / heat input
= .38 MW / 1 MW = .38
OR 38%
Heat rejected at cold reservoir = heat input - work output
1 MW - .38 MW
= 0.62 MW.
b ) For reversible power output
efficiency = T₂ - T₁ / T₂ ; T₂ is temperature of hot reservoir and T₁ is temperature of cold reservoir.
= 1200 - 300 / 1200 = 900 / 1200
= .75
or 75%
rate at which heat is rejected
= 1 - .75 x 1
= .25 MW .
