Complete Question:
A hollow cylinder with an inner radius of 4.0mm and an outer radius of 30mm conducts a 3.0-A current flowing parallel to the axis of the cylinder. If the current density is uniform throughout the wire, what is the magnitude of the magnetic field at a point 12mm from its center?
Answer:
The magnitude of the magnetic field = 7.24 μT
Explanation:
Inner radius, a = 4.0 mm = 0.004 m
Outer radius, b = 30 mm = 0.03 m
Radius, r = 12 mm = 0.012 m
let h² = b² - a²
h² = 0.03² - 0.004²
h² = 0.000884
Let d² = r² - a²
d² = 0.012² - 0.004²
d² = 0.000128
Current I = 3A
μ = 4π * 10⁻⁷
The magnitude of the magnetic field is given by:
B = 7.24 * 10⁻⁶T
B = 7.24 μT
Usually nice weather , i dont know the answer to the second part
Answer:
4.25 m/s
Explanation:
Force, F = 22 N
Time, t = 0.029 s
mass, m = 0.15 kg
initial velocity of the cue ball, u = 0
Let v be the final velocity of the cue ball.
Use newton's second law
Force = rate of change on momentum
F = m (v - u) / t
22 = 0.15 ( v - 0) / 0.029
v = 4.25 m/s
Thus, the velocity of cue ball after being struck is 4.25 m/s.
T=s/v=>t=1500/1,5=1000s
1,5km=1500m
Answer:
<em>The velocity of the two cars is 10 m/s after the collision.</em>
Explanation:
<u>Law Of Conservation Of Linear Momentum
</u>
The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and velocity v is
P=m.v
If we have a system of bodies, then the total momentum is the sum of them all
If some collision occurs, the velocities change to v' and the final momentum is:
In a system of two masses, the law of conservation of linear momentum takes the form:
If both masses stick together after the collision at a common speed v', then:
The car of mass m1=1000 Kg travels at v1=25 m/s and collides with another car of m2=1500 Kg which is at rest (v2=0).
Knowing both cars stick and move together after the collision, their velocity is found solving for v':
v' = 10 m/s
The velocity of the two cars is 10 m/s after the collision.
