A 0.2-stone is attached to a string and swung in a circle of radius 0.6 m on a horizontal and frictionless surface. If the stone
1 answer:
You might be interested in
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.
Answer:
Explanation:
As we know that mass is the product of volume and density
so we will have
here we have
so we will have
so we will have
now let the position of Pluto is at origin so we have