What is the period and frequency of the second hand on a clock?
1 answer:
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1) The most potential energy is at position A
2) The most kinetic energy is at position C
Explanation:
1)
The gravitational potential energy is the energy possessed by an object due to its position in a gravitational field, and it is given by the equation
where
m is the mass of the object
g is the acceleration of gravity
h is the height of the object relative to the ground
From the equation, we see that the potential energy is directly proportional to the heigth of the object: therefore, the roller coaster in this problem will have the most potential energy at its highest postion, so at position A.
2)
The total mechanical energy of the roller coaster at any point along the track is given by
where
PE is the potential energy
KE is the kinetic energy
Assuming there is no friction, the mechanical energy E is constant. This means that when PE increases, KE decreases, and when PE increases, KE decreases.
Therefore, the cart will have maximum kinetic energy when the potential energy is at minimum: and since the potential energy is directly proportional to the height of the track, this will occur at the lowest position, so at position C.
Learn more about kinetic and potential energy:
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Answer:
8.91 J
Explanation:
mass, m = 8.20 kg
radius, r = 0.22 m
Moment of inertia of the shell, I = 2/3 mr^2
= 2/3 x 8.2 x 0.22 x 0.22 = 0.265 kgm^2
n = 6 revolutions
Angular displacement, θ = 6 x 2 x π = 37.68 rad
angular acceleration, α = 0.890 rad/s^2
initial angular velocity, ωo = 0 rad/s
Let the final angular velocity is ω.
Use third equation of motion
ω² = ωo² + 2αθ
ω² = 0 + 2 x 0.890 x 37.68
ω = 8.2 rad/s
Kinetic energy,
K = 0.5 x 0.265 x 8.2 x 8.2
K = 8.91 J
Answer:
a) 2.933 m
b) 4.534 m
Explanation:
We're given the equation
v(t) = -0.4t² + 2t
If we're to find the distance, then we'd have to integrate the velocity, since integration of velocity gives distance, just as differentiation of distance gives velocity.
See attachment for the calculations
The conclusion of the attachment will be
7.467 - 2.933 and that is 4.534 m
Thus, The distance it travels in the second 2 sec is 4.534 m
