Let me know if you need help figuring out how to make something in little alchemy 2.
2 answers:
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(a) The mass of the bullet is 
and its speed is 
, so its momentum is
The pitcher claims that he can throw the ball with the same momentum p of the ball. Since the mass of the ball is m=0.148 kg, this means that the velocity of the ball must be:
(b) The kinetic energy of the bullet is:
while the kinetic energy of the ball is:
So, the bullet has greater kinetic energy than the ball.
The pitcher claims that he can throw the ball with the same momentum p of the ball. Since the mass of the ball is m=0.148 kg, this means that the velocity of the ball must be:
(b) The kinetic energy of the bullet is:
while the kinetic energy of the ball is:
So, the bullet has greater kinetic energy than the ball.
Answer:
A) d. B) a. C) E*ε₀
Explanation:
A) In electrostatic conditions, no electric field can exist within a conductor, as any existing field should produce an instantaneous charge redistribution (as charges can move freely inside a conductor) in such a way to make it 0.
B) In electrostatic conditions, as charge can move freely in a conductor, in order to remove any field inside the conductor, any excess charge must build on the surface on the conductor, so the charge density inside the conductor is 0.
C) If there is an electric field with a magnitude E, and directed towards the surface of the conductor (which means that there exists a net negative charge on the surface), we can apply Gauss´s Law to a rectangular surface (like a pill box), parallel to the surface, half inside the conductor, and half outside it.
We know that the total flux of the electric field through this surface (which it has four sides), must be equal to the enclosed charge, divided by ε₀.
As no electric field can exist inside the conductor, the flux through the bottom side is 0.
For the sides perpendicular to the surface, the flux should be produced by any tangential electric field.
In electrostatic conditions, should any tangential field existed, this would produce a charge movement in order to make it 0, so the flux through the sides is 0 also.
We have only one face with a non-zero flux, which can be calculated as follows (assuming E constant along the surface):
E*A = Q/ε₀ (1)
If the net charge is evenly distributed along the surface, we can define a surface charge density, σ, as follows:
σ =
Replacing this value in (1):
E = σ / ε₀
⇒ σ = E*ε₀