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2 years ago

12

The point on the graph that lies on the y-axis (vertical axis) is called the y-intercept. What does the y-intercept tell you abo

ut the runner? ( look at photo)
the ending position of the runner

Choices:

the runners distance from the state

the starting position of the runner

the runners speed

1 answer:

jekas [21]2 years ago

7 0

Answer:

The starting position of the runner.

Explanation:

When you look at the graph, you can see that the first point on the graph is twenty on the y-axis.

The runner starts at twenty, and ends at thirty.

Therefore, the runner starts at twenty on the y-axis, so it's the starting position of the runner.

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

xxTIMURxx [149]

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.

3 0

2 years ago

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A lens of focal length 15.0 cm is held 10.0 cm from a page (the object ). Find the magnification .

nevsk [136]

Answer:

Magnification, m = 3

Explanation:

It is given that,

Focal length of the lens, f = 15 cm

Object distance, u = -10 cm

Lens formula :

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

v is image distance

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-10)}\\\\v=-30\ cm$

Magnification,

$m=\dfrac{v}{u}\\\\m=\dfrac{-30}{10}\\\\m=3$

So, the magnification of the lens is 3.

3 0

2 years ago

|| Climbing ropes stretch when they catch a falling climber, thus increasing the time it takes the climber to come to rest and r

Otrada [13]

To solve this problem it is necessary to apply the concepts related to Newton's second law and the kinematic equations of movement description.

Newton's second law is defined as

$F = ma$

Where,

m = mass

a = acceleration

From this equation we can figure the acceleration out, then

$a = \frac{F}{m}$

$a = \frac{11*10^3}{80}$

$a = 137.5m/s$

From the cinematic equations of motion we know that

$v_f^2-v_i^2 = 2ax$

Where,

$v_f =$ Final velocity

$v_i =$ Initial velocity

a = acceleration

x = displacement

There is not Final velocity and the acceleration is equal to the gravity, then

$v_f^2-v_i^2 = 2ax$

$0-v_i^2 = 2(-g)x$

$v_i =\sqrt{2gx}$

$v_i = \sqrt{2*9.8*4.8}$

$v_i = 9.69m/s$

From the equation of motion where acceleration is equal to the velocity in function of time we have

$a = \frac{v_i}{t}$

$t = \frac{v_i}{a}$

$t =\frac{9.69}{137.5}$

$t = 0.0705s$

Therefore the time required is 0.0705s

4 0

2 years ago

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A physics student shoves a 0.50-kg block from the bottom of a frictionless 30.0° inclined plane. The student performs 4.0 j of w

Musya8 [376]

Answer:1.63 m

Explanation:

Given

mass of block $m=0.5 kg$

inclination $\theta =30^{\circ}$

Amount of work done $W=4 J$

block slides a distance s along the Plane

Work done =change in Potential Energy

Increase in height of block is $s\sin \theta$

Change in Potential Energy $=mg(\sin \theta -0)$

$\Delta P.E.=0.5\times 9.8\times s\sin 30$

$4=0.5\times 9.8\times s\sin 30$

$s=\frac{4}{2.45}$

$s=1.63 m$

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2 years ago

a small bar magnet is suspended horizontally by a string. When placed in a uniform horizontal magnetic field, it will

Reil [10]

Answer:

It will neither translate in the opposite direction nor .rotate so as to be at right angles, it will also neither rotate so as to be vertical direction

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2 years ago