How far would you travel in 10 seconds if you are moving at 15 m/s?
2 answers:
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The formula for the rotational kinetic energy is
where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.
ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s
Then,
where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.
ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s
Then,
Answer:
T = 676 N
Explanation:
Given that: f = 65 Hz, L = 2.0 m, and ρ = 5.0 g = 0.005 kg
A stationary wave that is set up in the string has a frequency of;
f =
⇒ T = 4M
Where: t is the tension in the wire, L is the length of the wire, f is the frequency of the waves produced by the wire and M is the mass per unit length of the wire.
But M = L × ρ = (2 × 0.005) = 0.01 kg/m
T = 4 × ×
× 0.01
= 4 × 4 ×4225 × 0.01
= 676 N
Tension of the wire is 676 N.
The correct answer to the question is: 4) 74.9 N/m.
EXPLANATION:
As per the question, the stretched length of the spring is given as x = 0.250 m.
The potential energy gained by the spring is given as 2.34 joules.
We are asked to calculate the spring constant of the spring.
The potential energy gained by the spring is nothing else than the elastic potential energy .
The elastic potential energy of the spring is calculated as -
Potential energy P.E =
⇒k =
=
= 74.88 N/m
= 74.9 N/m. [ans]
Hence, the force constant of the spring is 74.9 N/m.