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3 years ago

A cheetah can go from the state of rest to running at 20m/s in just two seconds. What is the Cheetahs average acceleration

Physics

1 answer:

babymother [125]3 years ago

7 0

acceleration = change in velocity/change in time

so...

a = 20 m/s / 2 seconds

a = 10

hope that helps :)

P.S. found this from Brainly User, sometimes all you have to do is search to find the answer.

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The speed of light in vacuum is exactly 299,792,458 m/s. A beam of light has a wavelength of 651 nm in vacuum. This light propag

sukhopar [10]

Answer:

$v=2.58\times10^8m/s$

Explanation:

The index of refraction is equal to the speed of light c in vacuum divided by its speed v in a substance, or $n=\frac{c}{v}$ . For our case we want to use $v=\frac{c}{n}$ , which for our values is equal to:

$v=\frac{c}{n}=\frac{299792458m/s}{1.16}=258441774.138m/s$

Which we will express with 3 significant figures (since a product or quotient must contain the same number of significant figures as the measurement with the <em>least</em> number of significant figures):

$v=2.58\times10^8m/s$

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3 years ago

The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This

erica [24]

To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

$g = \frac{GM}{(d-R_{CM})^2}$

Where,

G = Gravitational Universal Constant

d = Distance

M = Mass

$R_{CM} =$ Radius earth center of mass

PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,

$g = \frac{GM}{(d-R_{CM})^2}$

$g = \frac{(6.67*10^{-11})(7.35*10^{22})}{(3.84*10^8-4700*10^3)^2}$

$g = 3.4*10^{-5}m/s^2$

PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

$\omega = \frac{2\pi}{T}$

At the same time we have that centripetal acceleration is given as

$a_c = \omega^2 r$

Replacing

$a_c = (\frac{2\pi}{T})^2 r$

$a_c = (\frac{2\pi}{26.3d(\frac{86400s}{1days})})^2 (4700*10^3m)$

$a_c = 3.34*10^{-5}m/s^2$

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3 years ago

What temperature can a solar flare reach

balandron [24]

It reaches 10 or 20 million degrees kelvin but it can get as high as 10 million degrees kelvin

5 0

3 years ago

Describe how newtons third law is related to fluid pressure.

Hatshy [7]

Fish swimming forward in the water, the water gets pushed backward because the fish moving forward is forcing the water to move backward, the motion forward and backward are the same, they are opposite and equal.

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3 years ago

A soccer player carries the ball for a distance of 40.0 m in the direction 42.0° west of south. find the westward component of t

Maksim231197 [3]

Actually what the problem meant about the westward component of the ball’s displacement is the horizontal component of the displacement. To help us better understand the problem, I attached a figure of the situation.

We can see from the figure that to solve for the value of the horizontal component, we have to make use of the sin function. That is:

sin θ = side opposite to the angle / hypotenuse of the triangle

sin 42 = x / 40 m

x = (40 m) sin 42

x = 26.77 m

Therefore the ball has a westward displacement of about 26.77 m

3 0

3 years ago