What is the final velocity of a dragster that travels 400 meters from the start line
1 answer:
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Answer:
a) ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ ) N.m
b) ΔU = -0.000747871 J
c) w = 47.97 rad / s
Explanation:
Given:-
- The area of the circular ring, A = 4.45 cm^2
- The current carried by circular ring, I = 13.5 Amps
- The magnetic field strength, vec ( B ) = (1.05×10−2T).(12i^+3j^−4k^)
- The magnetic moment initial orientation, vec ( μi ) = μ.(−0.8i^+0.6j^)
- The magnetic moment final orientation, vec ( μf ) = -μ k^
- The inertia of ring, T = 6.50×10^−7 kg⋅m2
Solution:-
- First we will determine the magnitude of magnetic moment ( μ ) from the following relation:
μ = N*I*A
Where,
N: The number of turns
I : Current in coil
A: the cross sectional area of coil
- Use the given values and determine the magnitude ( μ ) for a single coil i.e ( N = 1 ):
μ = 1*( 13.5 ) * ( 4.45 / 100^2 )
μ = 0.0060075 A-m^2
- From definition the torque on the ring is the determined from cross product of the magnetic moment vec ( μ ) and magnetic field strength vec ( B ). The torque on the ring in initial position:
vec ( τi ) = vec ( μi ) x vec ( B )
= 0.0060075*( -0.8 i^ + 0.6 j^ ) x 0.0105*( 12 i^ + 3 j^ -4 k^ )
= ( -0.004806 i^ + 0.0036045 j^ ) x ( 0.126 i^ + 0.0315 j^ -0.042 k^ )
- Perform cross product:
- The initial torque ( τi ) is written as follows:
vec ( τi ) = ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ )
- The magnetic potential energy ( U ) is the dot product of magnetic moment vec ( μ ) and magnetic field strength vec ( B ):
- The initial potential energy stored in the circular ring ( Ui ) is:
Ui = - vec ( μi ) . vec ( B )
Ui =- ( -0.004806 i^ + 0.0036045 j^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )
Ui = -[( -0.004806*0.126 ) + ( 0.0036045*0.0315 ) + ( 0*-0.042 )]
Ui = - [(-0.000605556 + 0.00011)]
Ui = 0.000495556 J
- The final potential energy stored in the circular ring ( Uf ) is determined in the similar manner after the ring is rotated by 90 degrees with a new magnetic moment orientation ( μf ) :
Uf = - vec ( μf ) . vec ( B )
Uf = - ( -0.0060075 k^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )
Uf = - [( 0*0.126 ) + ( 0*0.0315 ) + ( -0.0060075*-0.042 ) ]
Uf = -0.000252315 J
- The decrease in magnetic potential energy of the ring is arithmetically determined:
ΔU = Uf - Ui
ΔU = -0.000252315 - 0.000495556
ΔU = -0.000747871 J
Answer: There was a decrease of ΔU = -0.000747871 J of potential energy stored in the ring.
- We will consider the system to be isolated from any fictitious forces and gravitational effects are negligible on the current carrying ring.
- The conservation of magnetic potential ( U ) energy in the form of Kinetic energy ( Ek ) is valid for the given application:
Ui + Eki = Uf + Ekf
Where,
Eki : The initial kinetic energy ( initially at rest ) = 0
Ekf : The final kinetic energy at second position
- The loss in potential energy stored is due to the conversion of potential energy into rotational kinetic energy of current carrying ring.
-ΔU = Ekf
0.5*T*w^2 = -ΔU
w^2 = -ΔU*2 / T
Where,
w: The angular speed at second position
w = √(0.000747871*2 / 6.50×10^−7)
w = 47.97 rad / s
Answer:
5.7 x 10^12 C
Explanation:
Let the charge on earth and moon is q.
mass of earth, Me = 5.972 x 10^24 kg
mass of moon, Mm = 7.35 x 10^22 kg
Let d be the distance between earth and moon.
the gravitational force between them is
The electrostatic force between them is
According to the question
1 % of Fg = Fe
q = 5.7 x 10^12 C
Thus, the charge on earth and the moon is 5.7 x 10^12 C.