Answer:

Explanation:
Given data

To find
Magnitude of the net magnetic field B
Solution
The magnitude of the net magnetic field can be find as:

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
different because joules law talks about heat produce in an electric whiles ohm' law talks about potential difference
By definition we have that
force=dP/dt,
where
p is momentum
so
<span>momentum is force*time
p= 15*3 = 45 Ns , west.
</span><span>the change in momentum of the object is 45 N.s</span>