Answer is b hope this helps
Answer:
57 %
Explanation:
input power = 16.4 kW = 16.4 x 10^3 W = 16400 W
Water pumped per second = 67 L/s
Mass of water pumped per second, m = Volume of water pumped epr second x density of water
m = 67 x 10^-3 x 1000 = 67 kg/s
height raised, h = 14 m
Output Power = m x g x h / t = 67 x 10 x 14 = 9380 W
efficiency = output power / input power = 9380 / 16400 = 0.57
% efficiency = 57 %
thus, the efficiency of the pump is 57 %.
Answer:
COMPLETE QUESTION
A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?
Explanation:
Given that,
Extension of spring
x = 0.0208m
Mass attached m = 3.39kg
Additional mass to have a frequency f
Let the additional mass be m
Using Hooke's law
F= kx
Where F = W = mg = 3.39 ×9.81
F = 33.26N
Then,
F = kx
k = F/x
k = 33.26/0.0208
k = 1598.84 N/m
The frequency is given as
f = ½π√k/m
Make m subject of formula
f² = ¼π² •(k/m
4π²f² = k/m
Then, m4π²f² = k
So, m = k/(4π²f²)
So, this is the general formula,
Then let use the frequency above
f = 3Hz
m = 1598.84/(4×π²×3²)
m = 4.5 kg
Answer:
The period of motion of new mass T = 0.637 sec
Explanation:
Given data
Mass of object (m) = 9 gm = 0.009 kg
Δx = 3.5 cm = 0.035 m
We know that spring force is given by
F = m g
F = 0.009 × 9.81 = 0.08829 N
Spring constant
k = 2.522 
New mass
= 26 gm = 0.026 kg
Now the period of motion is given by
T = 0.637 sec
This is the period of motion of new mass.
