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
We can use the equation for Newton's Law of Gravitation
Fg = (Gm₁m₂)/r²
Where gravitational constant = G = 6.674 x 10⁻¹¹ N · m²/kg²
mass m₁ = 0.145 kg
mass m₂ = 6.8 kg
distance between centers of masses = r = 0.5 m
Substitute these values into...
Fg = (Gm₁m₂)/r²
Fg = ((6.674 x 10⁻¹¹)(0.145)(6.8)) / (0.5)²
Fg = 2.63 x 10⁻¹⁰ N
Therefore, your answer should be <span>2.6 × 10–10</span>
Answer:
(I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.
(II). The torque is 84.87 N-m.
Explanation:
Given that,
Initial spinning = 50.0 rad/s
Time = 20.0
Distance = 2.5 m
Mass of pole = 4 kg
Angle = 60°
We need to calculate the angular acceleration
Using formula of angular velocity




The angular acceleration is -2.5 rad/s²
We need to calculate the number of revolution
Using angular equation of motion

Put the value into the formula


The number of revolution is 500 rad.
(II). We need to calculate the torque
Using formula of torque


Put the value into the formula


Hence, (I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.
(II). The torque is 84.87 N-m.
Answer:
The correct option is;
Still constant
Explanation:
The relative refractive index ₁n₂ between the two medium can be as follows;

Therefore, given that the speed of light in medium 1 is constant and the speed of light on medium 2 is also constant, the relative refractive index ₁n₂ = c₁/c₂ is always constant.