Respuesta :
Answer:
[tex]P_1 = 1166.7 Watt[/tex]
[tex]P_2 = 2000 Watt[/tex]
Explanation:
Average power for the human sprinter is given as
[tex]Power = \frac{\Delta E}{\Delta t}[/tex]
so we have
[tex]P = \frac{\frac{1}{2}mv^2 - 0}{\Delta t}[/tex]
[tex]P = \frac{\frac{1}{2}(70)(10^2) - 0}{3}[/tex]
[tex]P_1 = 1166.7 Watt[/tex]
Average power for greyhound is given as
[tex]P = \frac{\frac{1}{2}mv^2 - 0}{\Delta t}[/tex]
[tex]P = \frac{\frac{1}{2}(30)(20^2) - 0}{3}[/tex]
[tex]P_2 = 2000 Watt[/tex]
Power is said to be the rate of doing the work. Its' si unit is the watt. The average power of the first and second cases will be 1166.7 watts and 2000 watts respectively.
What is power?
Power is said to be the rate of doing the work. Its' si unit is the watt. Horsepower is another unit of power. Mathematically it is given by the total work done to the total time taken by the body.
The given data in the problem will be ;
m is the mass of a human = 70 kg
u is the initial velocity=0 m/sec
v₁ is the final velocity for case 1= 10 m/sec.
v₂ is the final velocity for case 2=20 m/sec.
t is the time elapsed=3.0 sec.
p₁ is the average power output for case 1=?
p₂ is the average power output for case 2=?
The average power is defined as the ratio of change of work done or energy to the time elapsed.
The average power for case 1,
[tex]\rm p_1=\frac{\frac{1}{2} m(v_1^2-u_1^2)}{t} \\\\\rm p_1=\frac{\frac{1}{2} mv_1^2}{t} \\\\p_1=\frac{\frac{1}{2} \times70\times (10)^2}{3} \\\\\rm p_1=1166.7 \;watt.[/tex]
Hence the average power of the first case will be 1166.7 watts.
The average power for case 2,
[tex]\rm p_2=\frac{\frac{1}{2} m(v_2^2-u_2^2)}{t} \\\\\rm p_2=\frac{\frac{1}{2} mv_2^2}{t} \\\\p_2=\frac{\frac{1}{2} \times30\times (20)^2}{3} \\\\\rm p_2=2000 \;watt.[/tex]
Hence the average power of the second case will be 2000 watts.
To learn more about the power refer to the link;
https://brainly.com/question/1334166
