A canoe floats in a lake. Inside the canoe is a 25 kg steel solid ball. If the ball is thrown into the lake, does the level of the lake go up, go down or remain the same? You will do a calculation using the 3.0 cm diameter ball bearing to simulate the situation. And you will answer whether the level in the lake changes when the 25 kg steel ball is thrown from the canoe into the water.

Explanation:

volume of ball bearing = 4/3 π r³

= 4/3 x 3.14 x 1.5³

= 14.13 cc.

if D  be the density of steel , weight of the ball bearing

= 14.13 x D x g

For the first case , water will be displaced to keep it floating

weight of displaced water will be equal to weight of steel

weight of displaced water = 14.13 Dg

mass of displaced water = 14.13 D

volume of displaced water = mass / density of water

= 14.13 D / d                             ; (where d is density of water) .

Now when the steel ball bearing is dipped in water , it will displace water equal to its volume only and it will be drowned

its volume = 14 .13 cc

14.13 D / d  >  14.13  ( because D/d is more than one , since D > d )

volume of water displaced in first case is greater

water level will go up higher in first case .

Hence in the second case water level will go down .

Same will happen in case of 25 kg steel .

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.