The latent heat of vaporization of h₂o at body temperature (37°c) is 2.42 x 10⁶ j/kg. to cool the body of a 60.4-kg jogger [average specific heat capacity = 3500 j/(kg c°)] by 1.08 c°, how many kilograms of water in the form of sweat have to be evaporated?

Explanation:

        Latent heat of vaporization of at 37°C is .

        When the sweat on our body evaporates, it absorbs energy from our body to overcome it's Latent heat of vaporisation. Thus our body cools down when sweat evaporates.

       So, Energy absorbed by sweat to evaporate = Energy lost by body

Specific heat capacity of human body = . Jogger weights 60.4 kg. Body temperature decreases by

       Energy absorbed from body =

∴ 0.094 kg of sweat has evaporated from the body.

wavelegth as in spectrum red oygbiv

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft