Here I come and we wanna go home!!!
I believe it's D. The centripetal force moves away from the center, but the marble stays there due to the string.
Answer:
F. 25.82 s
Explanation:
Given:
Δy = 90 m
v₀ = 0 m/s
a = 0.27 m/s²
Find: t
Δy = v₀ t + ½ at²
90 m = (0 m/s) t + ½ (0.27 m/s²) t²
t = 25.82 s
Answer:
0.00417 kW/K or 4.17 W/K
Second law is satisfied.
Explanation:
Parameters given:
Rate of heat transfer, Q = 2kW
Temperature of hot reservoir, Th = 800K
Temperature of cold reservoir, Tc = 300K
The rate of entropy change is given as:
ΔS = Q * [(1/Tc) - (1/Th)]
ΔS = 2 * (1/300 - 1/800)
ΔS = 2 * 0.002085
ΔS = 0.00417 kW/K or 4.17 W/K
Since ΔS is greater than 0, te the second law of thermodynamics is satisfied.
Given:
Height of tank = 8 ft
and we need to pump fuel weighing 52 lb/ 
to a height of 13 ft above the tank top
Solution:
Total height = 8+13 =21 ft
pumping dist = 21 - y
Area of cross-section = 
= 
=16
Now,
Work done required = 
= 
= 832
![[ 21y ]_{0}^{8} - [\frac{y^{2}}{2}]_{0}^{8}\\](https://tex.z-dn.net/?f=%5B%2021y%20%5D_%7B0%7D%5E%7B8%7D%20-%20%5B%5Cfrac%7By%5E%7B2%7D%7D%7B2%7D%5D_%7B0%7D%5E%7B8%7D%5C%5C)
)
= 113152 = 355477 ft-lb
Therefore work required to pump the fuel is 355477 ft-lb
