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:
64 J
Explanation:
The potential energy change of the spring ∆U = -W where W = work done by force, F.
Now W = ∫F.dx
So, ∆U = - ∫F.dx = - ∫Fdxcos180 (since the spring force and extension are in opposite directions)
∆U = - ∫-Fdx
= ∫F.dx
Since F = 40x - 6x² and x moves from x = 0 to x = 2 m, we integrate thus, ∆U = ∫₀²F.dx
= ∫₀²(40x - 6x²).dx
= ∫₀²(40xdx - 6x²dx)
= ∫₀²(40x²/2 - 6x³/3)
= ∫₀²(20x² - 2x³)
= [20x² - 2x³]₀²
= [(20(2)² - 2(2)³) - (20(0)² - 2(0)³)
= [(20(4) - 2(8)) - (0 - 0))
= [80 - 16 - 0]
= 64 J
Answer:
D.
Explanation:
8.1X10 to the power of 6 Joules.
Answer:
The specific latent heat (L) of a material is a measure of the heat energy (Q) per mass (m) released or absorbed during a phase change.
It's defined through the formula Q = mL.
Explanation:
