A cube, 1 m on each edge, is held 0.75 m below the surface of a pool of water by a rope. The mass of the cube is 700 kg, and the weight distribution is such that one face remains parallel to the surface of the water. a. Find the tension in the rope. b. Assuming the rope is cut, how far will the cube protrude from the water

Explanation:

The mass of cube = 700 kg

volume = 1 m³

density = 700 kg / m²

Its density is less than that of water so it will try to float on the surface .

Tension in rope will be equal to net upward force

upthrust = volume x density of water x g

= 1 x 10³ x 9.8

= 9800 N

weight of cube = mass x g

= 700 x 9.8

= 6860 N .

Net upward force = 9800 - 6860

= 2940 N.

Tension in the rope = 2940 N.

Rope will hold the cube inside and not allow it to go outside water .

b )

If rope is cut , cube being lighter , will float on surface of water .

Part of cube inside water while floating

= 6860 / 9800

= .7

.7 m will remain inside water

part floating outside

= 1 - 0.7

= 0.3 m .

the answer to this is proletariats.

hope this : )

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2