Answer:
m 200 g , T 0.250 s,E 2.00 J
;
2 2 25.1 rad s
T 0.250
(a)
2 2
k m 0.200 kg 25.1 rad s 126 N m
(b)
2
2 2 2.00 0.178 mm 200 g , T 0.250 s,E 2.00 J
;
2 2 25.1 rad s
T 0.250
(a)
2 2
k m 0.200 kg 25.1 rad s 126 N m
(b)
2
2 2 2.00 0.178 m
Explanation:
That is a reason
In order to determine the acceleration of the block, use the following formula:
Moreover, remind that for an object attached to a spring the magnitude of the force acting over a mass is given by:
Then, you have:
by solving for a, you obtain:
In this case, you have:
k: spring constant = 100N/m
m: mass of the block = 200g = 0.2kg
x: distance related to the equilibrium position = 14cm - 12cm = 2cm = 0.02m
Replace the previous values of the parameters into the expression for a:
Hence, the acceleration of the block is 10 m/s^2
Answer:
Δμ = hΔf/B
Explanation:
If the photon energy , ΔE = hΔf where Δf = small frequency shift and since the potential energy change of the magnetic dipole moment μ in magnetic field B from parallel to anti-parallel state is ΔU = ΔμB. where Δμ = small shift in magnetic moment.
Since the magnetic energy change equals the photon energy,
ΔE = ΔU
hΔf = ΔμB
Δμ = hΔf/B
Assuming that the can is motionless, we can then assume that the vertical component of T = mg and that Fe = the horizontal component of T.
<span> Since T itself is larger than it's vertical or horizontal components separately, then T is greater than all the forces.</span>
Im not 100% sure you have to tell me if im wrong or not.
D
B
C
