Answer:14 m/s
Explanation:
Kinetic energy(ke)=175J
Momentum(M)=25kgm/s
Speed=v
Mass=m
Ke=(m x v x v)/2
175=(mv^2)/2
Cross multiply
175 x 2=mv^2
350=mv^2
Momentum=mass x velocity
25=mv
m=25/v
Substitute m=25/v in 350=mv^2
350=25/v x v^2
350=25v^2/v
v^2/v=v
350=25v
v=350/25
v=14 m/s
Answer: false
Explanation:
You have to make sure it doesn’t stay wet
we know the equation for the period of oscillation in SHM is as follows:
T = 2 * pi * sqrt(mass/k)
we know f = 1/T, so f = 1/(2 * pi) * sqrt(k/m).
since d = v*T, we can say v = d/t = d * f
the final equation, after combining everything, is as follows:
v = d/(2 * pi) * sqrt(k/m)
by plugging everything in
v = .75/(2 * pi) * sqrt((1 * 10^5)/(30))
We find our velocity to be:
v = 6.89 m/s
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
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