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 answer is 35 degrees Celsius. Hope I helped :) Please vote brainliest.
Answer:
<em>A. Statistics addresses gaps in knowledge.</em>
Explanation:
The following statements that does not describe a limitation of statistics is <em>statistics addresses gaps in knowledge.</em>
It will travel slowest through gases.
Explanation:
1. Force applied on an object is given by :
F = W = mg
(a) A 160 lb human being, F = 160 lb
g = acceleration due to gravity, g = 32 ft/s²
m = 5 kg
(b) A 1.9 lb cockatoo, F = 1.9 lb
m = 0.059 kg
2. (a) A 2300 kg rhinoceros, m = 2300 kg
(b) A 22 g song sparrow, m = 22 g = 0.022 kg
Hence, this is the required solution.
