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:
4km
Explanation:
15 minutes is 1/4 of an hour.
1/4 of 16 is 4.
Total thermal energy is the answer to your question.
Answer:
F = 0.1575 N
Explanation:
When the third sphere touches the first sphere, the charge is distributed between both spheres, then now the first sphere has only half of his original charge.
In this moment then
Sphere one has a charge = Q/2
Sphere three has a charge = Q/2
Now when the third sphere touches the second sphere again the charge is distributed in a manner that both sphere has the same charge.
How the total charge is Q = Q/2 + Q = 3/2Q, when the spheres are separated each one has 3/4Q
Sphere two has a charge = 3/4Q
Sphere three has a charge = 3/4Q
The electrostatic force that acts on sphere 2 due to sphere 1 is:
F = 
F= 
how 
= 0.42
Then
F = 
F = 0.1575 N
in situations involving equal masses, chemical reactions produce less energy than nuclear reactions.
