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3 years ago

How much metabolic energy is required for a 68kg person to run at a speed of 15km/hr for 15min ?

1 answer:

8 0

What I have here is exactly the same problem, however, with the time changed to 19 mins:

metabolic energy = metabolic power*time = 1.150*19*60 = 1.311 kJ..corresponding to 1.311/4.186 = 313,2 Cal or kcal

If we reasonably assume a metabolic eff.cy of 20%, it means we need to assume food for 1500 Cal approx.

Just plug the value t=15min to the equation and you will surely get the correct answer.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

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Explain solar and lunar eclipse.with diagram.

Lunna [17]

Explanation:

an eclipse happens when one

astronomical body block light

from or to another, the moon moves into

the shadow of earth cast by sun... In a solar

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3 years ago

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What is the orbital period in years of a planet with a semi major axis of 35 au

mezya [45]

Answer:

Orbital period of the planet will be 207.06 year

Explanation:

We have given the planet have the semi major axis as 35 au

We have to find the orbital period of the planet

From Keplar's third law there is relation between the orbital period and semi major axis which is t $T^2=R^3$

So $T^2=35^3$

$T^2=42875$

$T=207.06year$

So orbital period of the planet will be 207.06 year

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3 years ago

A block is projected up a frictionless inclined plane with initial speed v0 = 7.14 m/s. The angle of incline is θ = 36.5°. (a) H

Wewaii [24]

Answer:

$(a)x=4.37m\\\\(b)t=1.225s\\\\(c) v_{f}=7.14m/s$

Explanation:

Given data

$v_{o}=7.14m/s\\\alpha =36.5^o$

For Part (a)

Starting with the -ve acceleration of the body (opposite to the gravitational force)

$a=-gSin\alpha \\a=-(9.8m/s^2)Sin(36.5)\\a=-5.83m/s^2$

Using equation of motion

$v_{f}^2=v_{o}^2+2ax\\(0m/s)^2=(7.14m/s)^2+2(-5.83m/s^2)x\\-(7.14m/s)^2=2(-5.83m/s^2)x\\x=\frac{-(7.14m/s)^2}{2(-5.83m/s^2}\\ x=4.37m$

For Part (b)

Using the result in Part (a) we can substitute in other equation of motion to get time t:

$x=\frac{1}{2}vt\\ 4.37m=\frac{1}{2}(7.14m/s)t\\ (7.14m/s)t=2*(4.37)\\t=8.744/7.14\\t=1.225s$

For Part (c)

At state 2 where vo=0m/s and the acceleration is positive (same direction as the gravitational force)

$a=gSin\alpha \\a=(9.8m/s^2)Sin(36.5)\\a=5.83m/s^2\\\\\\v_{f}^2=v_{o}^2+2ax\\v_{f}^2=(0m/s)^2+2(5.83m/s^2)(4.37m)\\v_{f}^2=50.95\\v_{f}=\sqrt{50.95}\\ v_{f}=7.14m/s$

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3 years ago

The kinetic energy of a ball with a mass of 0.5 kg and a velocity of 10 m/s is J.

Jlenok [28]

Hello!

We have the following data:

KE (Kinetic Energy) = ? (in Joule)
m (mass) = 0.5 Kg
v (speed) = 10 m/s

Formula to calculate kinetic energy:

 $KE = \frac{1}{2} *m*v^2$

Solving:

 $KE = \frac{1}{2} *m*v^2$

$KE = \frac{1}{2} *0.5*10^2$

$KE = \frac{1}{2} *0.5*100$

$KE = \frac{50}{2}$

$\boxed{\boxed{KE = 25\:Joule}}\Longleftarrow(kinetic\:energy\:of\:a\:ball)\end{array}}\qquad\quad\checkmark$

I hope this helps! =)

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Vikentia [17]

Answer:

mabye 8

Explanation:

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