The ration of output work to input work expressed as a percentage is called <u>Efficiency</u>.
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
The launch velocity of the marble launcher is 34.65 m/s
Given that the launch velocity of marble launcher, launches a 25g marble to a distance of 73 cm (0.73 m) and the marble roll up to 6.2 meters before stopping. The launch height is 20 cm (0.2 m).
The time for landing can be calculated by the second equation of motion formula:
h = ut +
g
Let u = 0
0.2 = 0×t +
× 9.8 × 
= 
= 0.04
t = 0.2s
Now, the launch velocity of the marble launcher can be calculated by:
Speed = Distance / Time
Speed = 
Speed = 
Speed = 34.65 m/s
Therefore, the launch velocity of the marble launcher is 34.65 m/s
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