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:
I'm pretty sure it's 3.
Explanation:
Because if you look at your options the only that would be relevant to tick marks would be either 4 or 3. And it said in the question that we're looking for the one for the dependent variable. And the dependent variable is on the Y- Axis and the 3 is the tick marks for the y-axis. So your answer is 3.
Answer: The amount of energy consisted in the molecules determines the state of matter.
Explanation:
Data is inappropriate
here, we need gauge of the wire i.e., diameter of the wire, so that we calculate the resistance by using the formula
R = ρl/A
where R= resistance ; Ω
l = length of wire ; m
A = area of wire ; m²
ρ = resistivity ; Ω-m
But in general ohms law is
V = I R
R = V/I ;
but here we also calculate "R" from length of wire in which the current is flowing.
I hope it is helpful to you.
