Answer:
a) 5 N b) 225 N c) 5 N
Explanation:
a) Per Coulomb's Law the repulsive force between 2 equal sign charges, is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them, acting along the line that joins the charges, as follows:
F₁₂ = K Q₁ Q₂ / r₁₂²
So, if we make Q1 = Q1/5, the net effect will be to reduce the force in the same factor, i.e. F₁₂ = 25 N / 5 = 5 N
b) If we reduce the distance, from r, to r/3, as the factor is squared, the net effect will be to increase the force in a factor equal to 3² = 9.
So, we will have F₁₂ = 9. 25 N = 225 N
c) If we make Q2 = 5Q2, the force would be increased 5 times, but if at the same , we increase the distance 5 times, as the factor is squared, the net factor will be 5/25 = 1/5, so we will have:
F₁₂ = 25 N .1/5 = 5 N
Answer:
0.12 K
Explanation:
height, h = 51 m
let the mass of water is m.
Specific heat of water, c = 4190 J/kg K
According to the transformation of energy
Potential energy of water = thermal energy of water
m x g x h = m x c x ΔT
Where, ΔT is the rise in temperature
g x h = c x ΔT
9.8 x 51 = 4190 x ΔT
ΔT = 0.12 K
Thus, the rise in temperature is 0.12 K.
Answer:
<em>Thus, the object is accelerating to the left</em>
Explanation:
<u>The Net Force</u>
The net force is the result of adding all the forces as vectors acting on a body.
Each vector can be expressed in its rectangular components Fx and Fy, and the sum is the sum of the rectangular components separately.
Second Newton's law gives the relation between the net force and the acceleration of the body:
We can see the acceleration is a vector with the same direction as the net force.
The diagram shows two vertical forces and two horizontal forces.
The vertical forces are acting in opposite directions and with the same magnitude, thus they cancel out, leaving zero net force in the y-axis.
The horizontal forces are opposite and with different magnitudes. Since the force acting to the left (F3) has a greater magnitude than the force acting to the right (F4), there is a net force directed to the left with a magnitude of 60 N - 20 N = 40 N
Thus, the object is accelerating to the left
The charge will most likely leave the electric field near C) Y
