A man got 100 Kcal heat from its lunch. Its efficiency is only 25% and mass of man is 60 Kg. Calculate the height he can acquire

A worker eats a dish of icecream which funishes 200 kcal of heat. If the efficiency of the body is 25%, then calculate the height to which the worker can carry a weight of 56 kg using the energy of icecream. Worker's own weight is 70 kg. Given J = 4.2 J cal^(-1) .

The weight of a person is 60 kg . If he gets 10 calories of heat through food and the efficiency of his body is 28% , then upto what height he can climb? Take g = 10m s^(-2)

The weight of a person is 60 kg . If he gets 10 calories of heat through food and the efficiency of his body is 28% , then upto what height he can climb? Take g = 10m s^(-2)

The weight of a person is 60 kg . If he gets 10 calories of heat through food and the efficiency of his body is 28% , then upto what height he can climb? Take g = 10m s^(-2)

The weight of a person is 60 kg . If he gets 105 calories heat through food and the efficiency of his body is 28%, then upto how much height he can climb (approximately)

A man of 60 kg gains 1000 cal of heat by eating 5 mangoes. His efficiency is 29% . To what height he can jump by using this energy?

A man of 60kg gains 1000 cal of heat by eating 5 mangoes. His efficiency is 56%. To what height he can jump by using this energy?

A man of 60 kg gains 1000 cal of heat by eating 5 mangoes. His efficiency is 56%. To what height he can jump by using this energy?