A point charge is placed 3m from a 4uc charge what is the strength of the electric field on the point charge at this distance ro
1 answer:
The strength of the electric field on the point charge at this distance will be 4000 V/m.
<h3>What is the strength of the electric field?</h3>
The strength of the electric field is the ratio of electric force per unit charge.
The given data in the problem is;
Qis the unit charge = 4.0 × 10⁻⁶ C
E is the strength of the electric field
R is the distance from point charge = 3 m
The strength of the electric field is;
Hence, the strength of the electric field on the point charge at this distance will be 4000 V/m.
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Answer:
Spring's displacement, x = -0.04 meters.
Explanation:
Let the spring's displacement be x.
Given the following data;
Mass of each shrew, m = 2.0 g to kilograms = 2/1000 = 0.002 kg
Number of shrews, n = 49
Spring constant, k = 24 N/m
We know that acceleration due to gravity, g is equal to 9.8 m/s².
To find the spring's displacement;
At equilibrium position:
Fnet = Felastic + Fg = 0
But, Felastic = -kx
Total mass, Mt = nm
Fg = -Mt = -nmg
-kx -nmg = 0
Rearranging, we have;
kx = -nmg
Making x the subject of formula, we have;
Substituting into the formula, we have;
x = -0.04 m
Therefore, the spring's displacement is -0.04 meters.
A) The velocity of sphere A after the collision is 1.00 m/s to the right
B) The collision is elastic
C) The velocity of sphere C is 2.68 m/s at a direction of
D) The impulse exerted on C is 4.29 kg m/s at a direction of
E) The collision is inelastic
F) The velocity of the center of mass of the system is 4.00 m/s to the right
Explanation:
A)
We can solve this part by using the principle of conservation of momentum. The total momentum of the system must be conserved before and after the collision:
is the mass of sphere A
is the initial velocity of the sphere A (taking the right as positive direction)
is the final velocity of sphere A
is the mass of sphere B
is the initial velocity of the sphere B
is the final velocity of the sphere B
Solving for vA:
The sign is positive, so the direction is to the right.
B)
To verify if the collision is elastic, we have to check if the total kinetic energy is conserved or not.
Before the collision:
After the collision:
The total kinetic energy is conserved: therefore, the collision is elastic.
C)
Now we analyze the collision between sphere B and C. Again, we apply the law of conservation of momentum, but in two dimensions: so, the total momentum must be conserved both on the x- and on the y- direction.
Taking the initial direction of sphere B as positive x-direction, the total momentum before the collision along the x-axis is:
While the total momentum along the y-axis is zero:
We can now write the equations of conservation of momentum along the two directions as follows:
(1)
We also know the components of the momentum of B after the collision:
So substituting into (1), we find the components of the momentum of C after the collision:
So the magnitude of the momentum of C is
Dividing by the mass of C (1.60 kg), we find the magnitude of the velocity:
And the direction is
D)
The impulse imparted by B to C is equal to the change in momentum of C.
The initial momentum of C is zero, since it was at rest:
While the final momentum is:
So the magnitude of the impulse exerted on C is
And the direction is the angle between the direction of the final momentum and the direction of the initial momentum: since the initial momentum is zero, the angle is simply equal to the angle of the final momentum, therefore .
E)
To check if the collision is elastic, we have to check if the total kinetic energy is conserved or not.
The total kinetic energy before the collision is just the kinetic energy of B, since C was at rest:
The total kinetic energy after the collision is the sum of the kinetic energies of B and C:
Since the total kinetic energy is not conserved, the collision is inelastic.
F)
Here we notice that the system is isolated: so there are no external forces acting on the system, and this means the system has no acceleration, according to Newton's second law:
Since F = 0, then a = 0, and so the center of mass of the system moves at constant velocity.
Therefore, the centre of mass after the 2nd collision must be equal to the velocity of the centre of mass before the 1st collision: which is the velocity of the sphere A before the 1st collision (because the other 2 spheres were at rest), so it is simply 4.00 m/s to the right.
Learn more about momentum and collisions:
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