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2 years ago

Human muscles have an efficiency of about __________. A. 9 to 17% B. 18 to 26% C. 27 to 35% D. 36 to 44%

Physics

2 answers:

mafiozo [28]2 years ago

7 0

B. 18 to 26%

Hope this helps

xxTIMURxx [149]2 years ago

3 0

Answer:

The correct answer would be B. 18 to 26%.

The muscle efficiency is calculated by dividing mechanical work output by total metabolic cost.

It is estimated that human muscles have an efficiency of about 18% to 26%.

The efficiency is low because most of the energy is lost when food energy is converted into ATP (adenosine triphophate).

In addition, there is second energy loss when energy in the form of ATP is converted into the mechanical energy such as rowing, cycling et cetera.

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If earth's axis was tilted 45 degrees at what latitudes

givi [52]

23.5 DEGREES is the answer.

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2 years ago

Biotic factors are the blank and blank part of the ecosystem

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3 years ago

A 2.0-kg object moving with a velocity of 5.0 m/s in the positive x direction strikes and sticks to a 3.0-kg object moving with

Andrej [43]

Answer:

5.4 J.

Explanation:

Given,

mass of the object, m = 2 Kg

initial speed, u = 5 m/s

mass of another object,m' = 3 kg

initial speed of another orbit,u' = 2 m/s

KE lost after collusion = ?

Final velocity of the system

Using conservation of momentum

m u + m'u' = (m + m') V

2 x 5 + 3 x 2 = ( 2 + 3 )V

16 = 5 V

V = 3.2 m/s

Initial KE = $\dfrac{1}{2}mu^2 + \dfrac{1}{2}m'u'^2$

= $\dfrac{1}{2}\times 2\times 5^2 + \dfrac{1}{2}\times 3 \times 2^2$

= 31 J

Final KE = $\dfrac{1}{2} (m+m')V^2 = \dfrac{1}{2}\times 5 \times 3.2^2 = 25.6 J$

Loss in KE = 31 J - 25.6 J = 5.4 J.

4 0

3 years ago

An object is thrown upward with some velocity. If the object rises 77.5 m above the point of release, (a) how fast was the objec

jolli1 [7]

Answer:

$v_o=39\ m/s\\t_m=4\ s$

Explanation:

<u>Vertical Launch Upwards</u>

In a vertical launch upwards, an object is launched vertically up from a height H without taking into consideration any kind of friction with the air.

If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:

$\displaystyle h_m=H+\frac{v_o^2}{2g}$