Aglass ball of radius 2.52 cm sits at the bottom of a container of milk that has a density of 1.01 g/cm3. the normal force on the ball from the container's lower surface has magnitude 10.1 x 10-2 n. what is the mass of the ball?

77.97 grams

Explanation:

Given:

Radius of the ball, r = 2.52 cm = 0.0252 m

Density of the milk, d = 1.01 g/cm³ = 1010 kg/m³

Normal force, F = 10.1 × 10⁻² N

Now,

For the equilibrium

Bouyant Force + Normal Force - Weight of the ball = 0

thus,

ρVg + 10.1 × 10⁻² - mg = 0

here, m is the mass of the ball

V is the volume of the ball

V = = 6.70 × 10⁻⁵ m³

substituting the values in the above equation we get

1010 × 6.70 × 10⁻⁵ × 9.8 + 10.1 × 10⁻² - m × 9.8 = 0

or

m × 9.8 = 0.764

or

m = 0.0779 kg = 77.97 grams

ice to water 1 water to vapour 2 latent heat

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.