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3 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]3 years ago

7 0

B. 18 to 26%

Hope this helps

xxTIMURxx [149]3 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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Martha was leaning out of the window on the second floor of her house and speaking to Steve. Suddenly, her glasses slipped from

DiKsa [7]

Answer:

s = 23.72 m

v = 21.56 m/s²

Explanation:

given

time to reach the ground (t) = 2.2 second

we know that

a) s = u t + 0.5 g t²

u = 0 m/s

g = 9.8 m/s²

s = 0 + 0.5 × 9.8 × 2.2²

s = 23.72 m

b) impact velocity

v = √(2gh)

v = √(2× 9.8 × 23.72)

v = √464.912

v = 21.56 m/s²

6 0

3 years ago

How many seconds of space should you keep between you if you're driving a 100 foot truck 30 mph?

Marrrta [24]

Answer:

 The space in seconds that will be kept = 2.27 seconds

Explanation:

S = d/t..................... Equation 1

making t the subject of formula in the equation above,

t = d/S.................... Equation 2

Where S = speed, d = distance, t = time.

Conversion: (i)if 1 mph = 0.44704 m/s,

then, 30 mph = 30×0.44704

= 13.41 m/s

(ii) If 1 foot = 0.3048 m

then, 100 foot = 30.48 m.

Given: S = 30 mph = 13.41 m/s, d = 100 foot = 30.48 m

Substituting these values into equation 2

t = 30.48/13.41

t = 2.27 seconds.

Therefore the space in seconds that will be kept = 2.27 seconds

7 0

3 years ago

A skier moving at 4.75 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220

disa [49]

First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force, $\mu m g$ . For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:
$ma=-\mu m g$
Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:
$a=-(0.220)(9.81 m/s^2)=-2.16 m/s^2$

Now we can use the following relationship to find the distance covered by the skier before stopping, S:
$2aS=v_f^2-v_i^2$
where $v_f=0$ is the final speed of the skier and $v_i=4.75 m/s$ is the initial speed. Substituting numbers, we find:
$S=- \frac{v_i^2}{2a}=- \frac{(4.75 m/s)^2}{2(-2.16 m/s^2)}=5.23 m$

5 0

3 years ago

Need help with exercises 3,4,5 Will give brainliest!!!

Harman [31]

I pretty sure that 3 is b and 4 is A and 5 I need a full picture

4 0

3 years ago

A bullet with a mass of 5 gramsand speed of 560 m/sis fired horizontally atablock of wood with a mass of 2 kg. The block rests o

Anni [7]

Momentum is conserved if and only if sum of all forces which are exserted on system equals zero. In our situation there are only internal forces, so by Newton's third law their vector sum is 0.

So $mv=(m+M)v' \Leftrightarrow v'=\frac{mv}{m+M}$ .

Kinetic energy of system at first: $\frac{mv^2}{2}=784\;\textbf{J}$ . After: $\frac{(m+M)\frac{m^2v^2}{(m+M)^2} }{2}=\frac{m^2v^2}{2(m+M)}\approx 1,96\; \textbf{J}$ . The secret is that other energy is in work of deformation forces (they in turn heat a bullet and a block).

Answer is A)

5 0

3 years ago