A 1.2-kg mass suspended from a spring of spring constant 22 N.m-1 executes simple harmonic motion of amplitude 5 cm. What is the
1 answer:
You might be interested in
Explanation:
Given that,
Mass if the rock, m = 1 kg
It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.
We need to find the mass of the meter stick. The force acting by the stone is
F = 1 × 9.8 = 9.8 N
Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.
W = 3.266 N
The mass of the meters stick is :
So, the mass of the meter stick is 0.333 kg.
Answer:
The question clearly describes the circular motion.
The circular motion equation is
The path of the particle is circular.
Explanation:
In circular motion, the radial acceleration is always towards the center and constant in magnitude. Furthermore, the velocity of the circular motion is always tangential to the circle, that is it is always perpendicular to the radius, hence the acceleration.
The related concept to solve this exercise is given in the expressions that the magnetic field has both as a function of the number of loops, current and length, as well as inductance and permeability. The first expression could be given as,
The magnetic field H is given as,
Here,
n = Number of turns of the coil
I = Current that flows in the coil
l = Length of the coil
From the above equation, the number of turns of the coil is,
The magnetic field is again given by,
Where the minimum inductance produced by the solenoid coil is B.
We have to obtain n, that
Replacing with our values we have that,
Therefore the number of turn required is 28Truns