A small sphere hangs from a string attached to the ceiling of a uniformly accelerating train car. It is observed that the string
makes an angle of 37° with respect to the vertical. The magnitude of the acceleration a of the train car is most nearly?
1 answer:
Answer:
a=7.384 m/s^2
Explanation:
let T be the tension in the string, m= mass
and a= acceleration
from the FBD in the attachment we can write
Tcos37°= mg
Tsin37° =ma
dividing both the equations we get
tan37° =a/g
therefore a=g×tan37°
a= 9.81×0.7535 = 7.384 m/s^2
the magnitude of the acceleration a of the train = 7.384 m/s^2
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Answer:
V_{a} - V_{b} = 89.3
Explanation:
The electric potential is defined by
= - ∫ E .ds
In this case the electric field is in the direction and the points (ds) are also in the direction and therefore the angle is zero and the scalar product is reduced to the algebraic product.
V_{b} - V_{a} = - ∫ E ds
We substitute
V_{b} - V_{a} = - ∫ (α + β/ y²) dy
We integrate
V_{b} - V_{a} = - α y + β / y
We evaluate between the lower limit A 2 cm = 0.02 m and the upper limit B 3 cm = 0.03 m
V_{b} - V_{a} = - α (0.03 - 0.02) + β (1 / 0.03 - 1 / 0.02)
V_{b} - V_{a} = - 600 0.01 + 5 (-16.67) = -6 - 83.33
V_{b} - V_{a} = - 89.3 V
As they ask us the reverse case
V_{b} - V_{a} = - V_{b} - V_{a}
V_{a} - V_{b} = 89.3
A)
The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):
(1)
where k is the spring constant.
The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:
(2)
where x is the displacement, m the mass, and v the speed.
We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:
Using (2) we can rewrite this as
And using (1), we find
Substituting into the last equation, we find the value of x:
B)
In this case, the kinetic energy is 1/10 of the total energy:
Since we have
we can write
And so we find: