Answer:
(a) 8.117 (b) 0.742
Explanation:
We have given distance s =40 m time t=7.5 sec
Final velocity v =2.55 m/sec
From the first equation of motion (negative sign because there is retrdation as the truck speed is slowing down )
So --------------eqn 1
From the second equation of motion ( negative sign because there is retrdation as the truck speed is slowing down )
So
------------------eqn 2
On solving eqn1 and eqn 2
u=8.117 m/sec and a=-0.742
The range of potential energies of the wire-field system for different orientations of the circle are -
θ U
0° 375 π x
90° 0
180° - 375 π x
We have current carrying wire in a form of a circle placed in a uniform magnetic field.
We have to the range of potential energies of the wire-field system for different orientations of the circle.
<h3>What is the formula to calculate the Magnetic Potential Energy?</h3>The formula to calculate the magnetic potential energy is -
U = M.B = MB cos
where -
M is the Dipole Moment.
B is the Magnetic Field Intensity.
According to the question, we have -
U = M.B = MB cos
We can write M = IA (I is current and A is cross sectional Area)
U = IAB cos
U = IπB cos
For = 0° →
U(Max) = MB cos(0) = MB = IπB = 5 × π ×
× 3 ×
=
375 π x .
For = 90° →
U = MB cos (90) = 0
For = 180° →
U(Min) = MB cos(0) = - MB = - IπB = - 5 × π ×
× 3 ×
=
- 375 π x .
Hence, the range of potential energies of the wire-field system for different orientations of the circle are -
θ U
0° 375 π x
90° 0
180° - 375 π x
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Answer:
t = 0.0735 m
Explanation:
Angular acceleration of the flywheel is given as
now after t = 8 s the speed of the flywheel is given as
now rotational kinetic energy of the wheel is given as
now we have