A 44 kg block lies on a horizontal frictionless surface. A horizontal force of 110 N is applied to the block. (a) Determine the
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Answer:
The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.
Explanation:
From coulomb's law, F = Eq
Thus,
F = E₁q₁
F = E₂q₂
Then
E₂q₂ = E₁q₁
where;
E₂ is the external electric field due to second test charge = ?
E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C
q₁ is the first test charge = 13 mC
q₂ is the second test charge = 23 mC
Substitute in these values in the equation above and calculate E₂.
The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.
However, the direction of the external field is still to the right.
1) A) Object 1 has the greater momentum
The magnitude of the momentum of an object is given by:
where
m is the mass of the object
v is its speed
Object 1 has a mass of and a speed of
, so its momentum is
Object 2 has a mass of and a speed of
, so its momentum is
So we see that , so object 1 has the greater momentum.
2) The objects have the same kinetic energy.
The kinetic energy of an object is given by
where
m is the mass of the object
v is its speed
Object 1 has a mass of and a speed of
, so its kinetic energy is
Object 2 has a mass of and a speed of
, so its kinetic energy is
So we see that , so the objects have same kinetic energy