If the air temperature is the same as the temperature of your skin (about 30 ∘C), your body cannot get rid of heat by transferring it to the air. In that case, it gets rid of the heat by evaporating water (sweat). During bicycling, a typical 67.0 kg person’s body produces energy at a rate of about 501 W due to metabolism, 81.0 % of which is converted to heat. [Recall that the normal internal body temperature is 98.6 ∘F and the specific heat capacity of the body is 3480 J/(kg⋅∘C) .]

m_w = 0.7453 kg

1 bottle of water

Explanation:

Given:

- T_1 = 30 degrees

- mass of the cyclist m_c = 67 kg

- Heat energy rate produced W = 501 W

- Metabolism rate = 81 %

- Heat of vaporization of water L_v = 2.42 * 10^6 J/kg

Find:

How many kilograms of water must the person’s body evaporate in an hour to get rid of this heat?

The evaporated water must, of course, be replenished, or the person will dehydrate. How many 750-mL bottles of water must the bicyclist drink per hour to replenish the lost water?

Solution:

- We first need to calculate the mass of vaporized water. It consists of two phases. Phase 1 is associated with increase in temperature, while phase 2 is transition from liquid to vapor. The total heat thermal equilibrium would be:

                                        Q_body = Q_water

Where,

                                        Q_body = W*t

                                        Q_water = m_w*L_v

Equating the two Eqs we have:

                                        m_w*L_v = W*t

                                        m_w = W*t / L_v

Plug in values:

                                        m_w = 501*60*60 / 2.42*10^6

                                         m_w = 0.7453 kg

- The amount of water evaporated in mL = 745 mL

Since a bottle of water has 750 mL, Hence, we require only 1 bottle.

b.transmitters send radio waves, and receivers capture radio waves.

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards