If two objects move farther apart from one another, the force of gravity will
2 answers:
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The strength of the electric field is 5 N/C
Explanation:
The magnitude of the electric field produced by a single-point charge is given by:
where
is the Coulomb's constant
Q is the magnitude of the charge
r is the distance from the charge
In this problem, we have:
is the charge producing the field
r = 100 m is the distance from the charge at which we want to calculate the field
Substituting into the equation, we find the s trength of the electric field:
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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:
The minimum coefficient of friction is 0.27.
Explanation:
To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.
Centripetal force is written as
with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as
with ω denoting the angular velocity, which we are given. With that, the above becomes:
Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?
Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:
and so 45 rev/min = 4.71 rad/s.
A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.