Answer:
22.2 m/s
Explanation:
First, we need to convert km to m by multiplying by 1000. This means that the car traveled 320 000 meters.
Next, we convert hours to minutes by multiplying by 3600 (the number of seconds in an hour). This means that overall, the car traveled 320 000 m in 14 400 seconds.
The average speed can be found by using the equation . After substitution, this gives the fraction
, which reduces to 22
m/s, or about 22.2 m/s.
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:
Explanation:
From the second law of Newton movement laws, we have:
, and we know that a is the acceleration, which definition is:
, so:
The next step is separate variables and integrate (the limits are at this way because at t=0 the block was at rest (v=0):
(This is the indefinite integral), the definite one is: