Physics, 28.11.2019 05:31 hallkayleel0lozgxtq

Aperson can generate about 300 w of power on a treadmill. if the treadmill is inclined at 3% and a 70-kg man runs at 3 m/s for 45 min, consider the following.(a) calculate the percentage of the power output that goes into heating up his body. %(b) calculate the percentage that keeps him moving on the treadmill. %(c) how much water would that heat evaporate? assume the water starts at body temperature, 37 °c.

Answers: 3

Show answers

Answers

Answer from: kitttimothy55

(a) 79.42 %

(b) 20.58 %

Solution:

As per the question:

Power generated by a person, P = 300 W

Inclination of the treadmill from the horizontal, $sin\theta$ = 3% = 0.03

Mass of man, m = 70 kg

Speed, v = 3 m/s

Time taken, t = 45 min = $45\times 60\ s$ = 2700 s

Now,

Weight of the man, W = mg = $70\times 9.8 = 686\ N$

Man's weight component along the inclination of the treadmill, F = $mgsin\theta = 686\times 0.03 = 20.58\ N$

In order to keep moving the power lost by the man in 1 s is:

P' = Fv = $20.58\times 3 = 61.74\ W$

Therefore, the power that is utilized in raising the body temperature of the man:

P'' = P - P' = 300 - 61.74 = 238.26 W

(a) The percentage of the power that is utilized in heating up the body:

%P = $\frac{P''}{P}\times 100 = \frac{238.26}{300}\times 100$ = 79.42%

(b) Percentage power required to keep the man moving:

%P = $\frac{P'}{P}\times 100 = \frac{61.74}{300}\times 100$ = 20.58%

(c) Water evaporated by the heat equals the product of the power utilized in raising the body temperature of the man and time:

Heat required to evaporate water, Q = $238.26\times 2700 = 643.302\ kJ$

At $37^{\circ}C$ , heat required for the evaporation of 1 kg of water:

⇒ Heat required to raise the temperature of 1 kg of water at $100^{\circ}C$ + Heat required for conversion of 1 kg of water into steam at $100^{\circ}C$

⇒ $mc\Delta T + mL_{v}= 1\times 4186(100 - 37) + 1\times 2256254 = 2519972\ J$

where

$L_{v}$ = Latent heat of vaporization

Therefore, mass of water evaporated, $m_{w} = \frac{Q}{2519972} = \frac{643302}{2519972} = 0.255\ kg$

Answer from: Quest

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm. the wire diameter is d = 0.9 mm. the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2) in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation:

Answer from: Quest

reserve is the answer