Physics Apparent or Real

Question

This is something new for me and I don't know how to solve it, if someone can help, thanks!

Answer 1

Ex 2

(1) 8 N represents real weight of the body Figure (a)

(2) 7 N represents apparent wt of the body in water Figure (b)

(3) The difference is due to upthrust = 1N ( 8N -7 N)

Volume of ball = 1/(1000*g) m³ 10^-4 m³

(4) Pressure due to water = P1 -P2 = 106360 Pa -103366 Pa

= 994 Pa = Height* Density of water*g

So height h= 2994/1000*10= 0.2994m

Answer 2

these problems seem to have been written by a non-english speaker so they're harder to understand, but

A.

1. the dynomometer is apparently a force measurement device and the eight newtons it's measuring is the weight of the ball when unsubmerged.
2. the condition for equilibrium is that all the forces on the ball sum (vector summation) to zero.
3. the force creating the difference between a) and b) is the buoyant force. it's 1 N
4. the volume of the ball can be found from the definition of the buoyant force, F_B = (\rho g V_{displaced}), where \rho is the density of water, g is the acceleration due to gravity, and V_{displaced} is the volume of the displaced water. in this case that displaced volume is the volume of the ball, so V = 1N/(1000 kg/m^3 * 9.8 m/s^2) = about 10^{-4} m^3 = 0.1 liters = 100 mL

B. this question is ambiguous about the pressure due to the water above the ball at the bottom of the tank. i believe they meant to say that the absolute pressure on the ball is 106360 pascals. the pressure due to the water by itself can be used to determine the depth of the ball once it hits the bottom. that depth is the distance over which the ball accelerates due to the 7 N force that pulls it to the bottom of the tank once the dynamometer is detached. the pressure is related to the depth by the equation P = \rho g h, where P is the pressure due to the water not the atmosphere: 106360-103360 = 3000 Pa. so h = P/(rho * g) = 3000 Pa/(1000 kg/m^3 * 9.8 m/s^2) = about 0.3 m. the ball accelerates according to newton's second law, so the acceleration is a = F/m = 7 N / 0.8 kg = 8.75 m/s^2. The mass of the ball is 0.8 kg and the force is 7 N. to find the velocity the ball reaches just before hitting the bottom of the tank i would use a projectile motion equation like v^2 = v_0^2 + 2ah. the initial velocity is zero, so v = sqrt(2ah) = sqrt(2 * 8.75 m/s^2 * 0.3 m) = about 2.3 m/s