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kati45 [8]
3 years ago
14

During bicycling, a 70 kg person's body produces energy at a rate of about 500 W due to metabolism, 80% of which is converted to

heat. How many
Kilograms of water must the person's body evaporate in an hour to get rid of this heat? The heat of vaporization of water at body temperature is 24
* 109 1/kg
Physics
1 answer:
Nataly_w [17]3 years ago
7 0

Explanation:

It is given that the energy produced by a person’s body due to metabolism is at a rate of 500W.

Energy produced by him in 1 hour

= 1800000 J

Of this, 80 % is converted into heat.

Heat energy produced

= 1440000 J =1.4x10^6

1.44x10^6/2.42x10^6=0.595kg

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An object is propelled along a straight-line path by a force. If the new force were doubled, is acceleration would
Aneli [31]
Still go straight but would obviously go up in speed!!





Hope this helps plz mark as brainlist and 5 star
6 0
3 years ago
Read 2 more answers
A uniformly charged solid disk of radius R = 0.45 m carries a uniform charge density of σ = 175 μC/m². A point P is located a di
siniylev [52]

Answer:

1408.685 KN/C

Explanation:

Given:

R = 0.45 m

σ = 175 μC/m²

P is located a distance a = 0.75 m

k = 8.99*10^9

  • The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:

                                  E = 2*pi*k*o * (1 - \frac{a}{\sqrt{a^2 + R^2} })\\

part a)

Electric Field strength at point P: a = 0.75 m

E = 2*pi*8.99*10^9*175*10^-6 * (1 - \frac{0.75}{\sqrt{0.75^2 + 0.45^2} })\\\\E = 9885021.285*(0.1425070743)\\\\E = 1408.685 KN/C

part b)

Since, R >> a, we can approximate a / R = 0 ,

Hence, E simplified relation becomes:

E = 2*pi*k*o * (1 - \frac{a/R}{\sqrt{a^2/R^2 + 1} })\\\\E = 2*pi*k*o * (1 - \frac{0}{\sqrt{0 + 1} })\\\\E = 2*pi*k*o

E = σ / 2*e_o

part c)

Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:

Electric Field strength due to point charge is:

E = k*δ*pi*R^2 / a^2  

Since, R << a, Surface area = δ*pi

Hence,

E = (k*δ*pi/a^2)

 

6 0
3 years ago
4. Assume a multiple level queue system with a variable time quantum per queue, and that the incoming job needs 50ms to run to c
Troyanec [42]

Answer:

Explanation:

For the completion of incoming job it will take 50ms

First queue takes 5ms quantum time and the subsequent queue takes double of the previous question

So,

First queue T_1 = 5ms

Second queue T_2 = 2 × T_1 = 2 × 5 = 10ms

Third queue T_3 = 2 × T_2 = 2 × 10 = 20ms

Fourth queue T_4 = 2 × T_3 = 2 × 20 = 40ms

Fifth queue T_5= 2 × T_4 = 2 × 40 = 80ms.

Now, the job will be done after the fifth queue.

So, after the first queue, the job is not completed, so, we have first interruption

After the second queue, the job is not completed, so, we have second interruption

After the third queue, the job is not completed, so we have third interruption.

After the fourth queue, the job is not yet completed, so we have the fourth interruption

And in the fifth queue the job is completed, so we don't have any interruption here.

So, the job will be interrupted 4 times and it will finished on the fifth queue.

6 0
3 years ago
A person jogs 4.o km in 32 minutes then 2.0 km in 22 minutes and finally 1.0 km in 16 minutes. What is the joggers average speed
AlladinOne [14]
Average speed = distance travelled / time used
6 0
3 years ago
4.72 A full-wave bridge-rectifier circuit with a 1-k load operates from a 120-V (rms) 60-Hz household supply through a 12-to-1 s
melisa1 [442]

Answer:

a) 12.74 V

b) Two pairs of diode will work only half of the cycle

c) 8.11 V

d) 8.11 mA

Explanation:

The voltage after the transformer is relationated with the transformer relationshinp:

V_o=Vrms*\frac{1}{12}\\V_o=10Vrms

the peak voltage before the bridge rectifier is given by:

V_{op}=Vo*\sqrt{2}\\V_{op}=14.14V

The diodes drop 0.7v, when we use a bridge rectifier only two diodes are working when the signal is positive and the other two when it's negative, so the peak voltage of the load is:

V_l=V_{op}-2(0.7)\\V_l=12.74V

As we said before only two diodes will work at a time, because the signal is half positive and half negative,so two of them will work only half of the cycle.

The averague voltage on a full wave rectifier is given by:

V_{avg}=2*\frac{V_l}{\pi}\\V_{avg}=8.11V

Using Ohm's law:

I_{avg}=\frac{V_{avg}}{R}\\\\I_{avg}=8.11mA

7 0
3 years ago
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