A comet experiences uniform acceleration from 60 m/s to 183 m/s in 10.0 s. What is its rate of acceleration?
1 answer:
You might be interested in
The distance should be 4m from the wire in order to get the magnetic field of 0.100μ .
- The magnitude and direction of the magnetic field due to a straight wire carrying current can be calculated using the previously mentioned Biot-Savart law. Let "I" be the current flowing in a straight line and "r" be the distance. Then the magnetic field produced by the wire at that particular point is given by
...(1)
- Since the wire is assumed to be very long, the magnitude of the magnetic field depends on the distance of the point from the wire rather than the position along the wire.
It is given that magnetic field 40.0 cm away from a straight wire is 1.00μT having current 2.00 A .
From equation (1) magnetic field 40.0 cm = 0.4m away from a straight wire is 1.00μT which is given by .....(2)
From equation (1) magnetic field 'r' m away from a straight wire is 0.100μT which is given by ...(3)
On dividing equation (2) by (3) , we get
Learn more about magnetic field here :
#SPJ4
Answer:
a) At 0.20 m, the magnitude of the field is 675.0 kV
The direction of the field is acting outwards from the center of the charged spheres
b) At 0.10 m, the magnitude of the field is 135 kV
The direction is acting outwards from the center of the charged spheres
c) At 0.025 m
The magnitude of the field, V = -270.0 kV
The direction of the field is inwards, towards the center of the charged spheres
Explanation:
The charged spherical shell parameters are;
The charge on the inner sphere, q₁ = -1.50 × 10⁻⁶ C
The radius of the inner shell, R₁ = 0.050 m
The charge on the outer sphere, q₂ = +4.50 × 10⁻⁶ C
The radius of the outer shell, R₂ = 0.15 m
Let 'r', represent the distance at which the electric field is measured, the following relationships can be obtained;
When r < R₁ < R₂,
When R₁ < r < R₂,
When R₁ < R₂ < r,
a) When r = 0.20 m, we have;
R₁ < R₂ < r, therefore
By plugging in the values, we get;
Therefore, the magnitude of the field, V = 675.0 kV
The direction of the field is outwards
b) When r = 0.10 m, we have;
When R₁ < r < R₂, therefore;
By plugging in the values, we get;
Therefore, the magnitude of the field, V = 135 kV
The direction of the field is outwards from the center
c) When r = 0.025 m, we have;
When r < R₁ < R₂, therefore;
By plugging in the values, we get;
Therefore, the magnitude of the field, V = -270.0 kV
The direction of the field is inwards, towards the center of the charged spheres.