Answer:
The value of charge q₃ is 40.46 μC.
Explanation:
Given that.
Magnitude of net force 
Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.
We need to calculate the distance
Using Pythagorean theorem

Put the value into the formula


We need to calculate the magnitude of the charge q₃
Using formula of net force

Put the value into the formula






Hence, The value of charge q₃ is 40.46 μC.
The addition of 24 kJ of energy will allow all of the mercury and lead to change from solid to liquid. The temperature of each substance will also increase.
Answer: Physical changes only change the appearance of a substance, not its chemical composition. Chemical changes cause a substance to change into an entirely substance with a new chemical formula. Chemical changes are also known as chemical reactions.
Explanation:
I think number 1 is incorrect I believe that answer is D. Number 6 I believe would be B. The rest seem to be correct.
Answer:

Explanation:
given,
length of the ship = 120 m
length of model of the ship = 4 m
Speed at which the ship travels = 70 km/h
speed of model = ?
by using froude's law

for dynamic similarities




hence, the velocity of model will be 12.78 km/h