A simple harmonic oscillator consists of a block of mass 1.80 kg attached to a spring of spring constant 210 N/m. When t = 1.70
1 answer:
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Answer:
The compression in the spring is 5.88 meters.
Explanation:
Given that,
Mass of the car, m = 39000 kg
Height of the car, h = 19 m
Spring constant of the spring,
We need to find the compression in the spring in stopping the ore car. It can be done by balancing loss in gravitational potential energy and the increase in elastic energy. So,
x is the compression in spring
So, the compression in the spring is 5.88 meters.
Answer:
t=15.68 s
Explanation:
Given that
m = 1520 kg
P =75 KW
We know that
Power ,P = F .v
F=force
v=velocity
v= 100 km/h
v=27.77 m/s
75 x 1000 = F x 27.77
F= 2700.75 N
F= m a
m=mass
a=acceleration
2700.75 = 1520 x a
a=1.77 m/s²
time t given as
v= u + a t
27.77 = 0 + 1.77 x t
t=15.68 s
Answer:
(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:
With been the vacuum permittivity
(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:
Explanation:
(A) Considering a uniform linear density on the ring, then:
(1)⇒
(2)⇒
(3)
Applying the technique of charge integration for finite charges:
(4)
Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.
Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:
(5)
Using the expressions (1),(4) and (5) you obtain:
Integrating results:
(S_a)
(B) For the expression of the magnitude of the field E(z), is important to remember:
(6)
But in this case you only work in the z variable, soo the expression (6) can be rewritten as:
(7)
Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):
(S_b)