Answer:
22.15 N/m
Explanation:
As we know potential energy = m*g*h
Potential energy of spring = (1/2)kx^2
m*g*h = (1/2)kx^2
Substituting the given values, we get -
(400)*(9.8)*(10) = (0.5)*(k)*(2.0^2)
k = 39200/2.645
k = 19600 N/m
For safety reasons, this spring constant is increased by 13 % So the new spring constant is
k = 19600 * 1.13 = 22148 N/m = 22.15 N/m
Answer:
30.96 m
Explanation:
If the particle has a lifetime of 129 ns as measured by observer A, and has a speed of 0.8c as measured by observer A, the distance between the markers will be:
d = v * Δt
v = 0.8*c = 0.8 * 3e8 = 2.4e8
Δt = ζ = 129 ns = 1.29e-7 s
d = 2.4e8 * 1.29e-7 = 30.96 m
This is the distance as measured by observer A.
Answer:
5773.50269 Hz
23 A
Explanation:
= Inductance = 6 mH
= Capacitance = 5 μF
= Resistance = 3 Ω
= Maximum emf = 69 V
Resonant angular frequency is given by

The resonant angular frequency is 5773.50269 Hz
Current is given by

The current amplitude at the resonant angular frequency is 23 A
Answer:
I attached an image that should help.
Explanation:
Check it out.
This problem uses the relationships among current
I, current density
J, and drift speed
vd. We are given the total of electrons that pass through the wire in
t = 3s and the area
A, so we use the following equation to to find
vd, from
J and the known electron density
n,
so:

<span>The current
I is any motion of charge from one region to another, so this is given by:
</span>

The magnitude of the current density is:

Being:

<span>
Finally, for the drift velocity magnitude vd, we find:
</span>
Notice: The current I is very high for this wire. The given values of the variables are a little bit odd