A89.3 kg man sits on the stern of a 5.8 m
long boat. the prow of the boat touches the
pier, but the boat isn’t tied. the man notices his mistake, stands up and walks to the boat’s prow, but by the time he reaches the prow, it’s moved 4.17 m away from the pier.
assuming no water resistance to the boat’s
motion, calculate the boat’s mass (not counting the man).
answer in units of kg

34.9 kg

Explanation:

Since there are no net external forces on the system, the center of gravity does not move.

Let's say that m is the mass of the man, M is the mass of the boat, and L is the length of the boat.

When the man is at the stern, the distance between the center of gravity and the pier is:

CG = (m L + M (L/2)) / (m + M)

When the man reaches the prow, the distance between the center of gravity and the pier is:

CG = (m x + M (x + L/2)) / (m + M)

Since these are equal:

(m L + M (L/2)) / (m + M) = (m x + M (x + L/2)) / (m + M)

m L + M (L/2) = m x + M (x + L/2)

m L + M (L/2) = m x + M x + M (L/2)

m L = m x + M x

m L − m x = M x

m (L − x) = M x

M = m (L − x) / x

Plugging in values:

M = 89.3 kg (5.8 m − 4.17 m) / 4.17 m

M = 34.9 kg

an enlarged virtual image

c. rb

explanation:

the atomic radius of a chemical element is a measure of the size of its atoms, usually the mean or typical distance from the center of the nucleus to the boundary of the surrounding shells of electrons.