Answer:
0.025V + (0.000218V/s³) t³
Explanation:
Parameters given:
Radius of coil, r = 3.85 cm = 0.0385 m
Number of turns, N = 450
Magnetic field, B = ( 1.20×10^(−2) T/s )t + (2.60×10^(−5) T/s4 )t^4.
The magnitude of Induced EMF is given as:
E = N * A * dB/dt
Where A is the area of the coil
First, we differentiate the magnetic field with respect to time:
dB/dt = 0.012 + 0.000104t³
Therefore, EMF will be:
E = 450 * 3.142 * (0.012 + 0.000104t³)
E = 2.096(0.012 + 0.000104t³)
E = 0.025V + (0.000218V/s³)t³
Answer:
F t = m Δv impulse delivered = change in momentum
Δv = 100 * .1 / .5 = 20 m/s original speed of puck
KE = 1/2 m v^2 = .5 * 20^2 / 2 = 100 J initial KE of puck
E = μ m g d energy lost by puck
Ff = μ m g = m a deceleration of puck due to friction
a = μ g = 9.8 * .2 = 1.96 m/s^2
v2 = a t + v1 = -1.96 * 4 + 20 = 12.2 m/s speed of puck on striking box
m v2 = M V conservation of momentum when puck strikes box
V = m v2 / M = 12.2 * .5 / .8 = 7.63 m/s speed of box after collision
KE = 1/2 M V^2 = .8 * 7.63^2 / 2 = 23.3 J KE of box after collision
KE = μ M g d energy lost by box in sliding distance d
d = 23.3 / (.3 * .8 * 9.8) = 9.91 m distance box slides
Explanation:
The magnetic force acting on the charged particle in magnetic field is given by :
Here,
q is the charged particle
v is the speed of particle
B is magnetic field
is the angle between the v and B.
The correct statements for the magnetic force are :
- An electric charge moving perpendicular to a magnetic field experiences a magnetic force. In this case,
. It will experience maximum force. - The direction of the magnetic force acting on a moving electric charge in a magnetic field is perpendicular to the direction of motion.
- A magnetic force is exerted on an electric charge moving through a uniform magnetic field.
- The direction of the magnetic force acting on a moving charge in a magnetic field is perpendicular to the direction of the magnetic field.
Answer:
The wedge balances propelled force.
Explanation:
According to the laws of motion, when appropriate weight is employed properly, a ball can move a truck. The same basic functions apply in this scenario. You often find a woodcutter raising the axe and letting it fall into the wood. This allows force to equally generate at both sides of the axe, splitting the wood apart.
