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2 years ago

A cheetah goes from 0 m/s to 25 m/s in 2.5s. what is the cheetahs rate of acceleration? A. 10m/s2

Physics

1 answer:

OleMash [197]2 years ago

5 0

The cheetah's rate of acceleration is 10 m/s².Option A is correct.

<h3>What is acceleration?</h3>

The rate of velocity change concerning time is known as acceleration.

Given data;

Initial velocity, u= 0 m/s

Final velocity, v= 25 m/s

Time elapsed, t = 2.5 seconds.

To find ;

Acceleration, a

The acceleration when the change in velocity is observed by the formula as:

$\rm a= \frac{v-u}{t}$

Substitute the given values:

$\rm a= \frac{25-0}{2.5} \\\\ a= 10 \ m/s^2$

The cheetah's rate of acceleration is 10 m/s²

Hence option A is correct.

To learn more about acceleration, refer to the link brainly.com/question/2437624

#SPJ1.

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distant galaxy emits light that has a wavelength of 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 nm.

a_sh-v [17]

Answer:

The speed of the galaxy relative to the Earth is $3.09\times 10^6\ m/s$ .

Explanation:

We have,

(a) Wavelength emitted by light at distant galaxy is 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 nm. It can be seen that the wavelength of light reduces as it reaches Earth. It is called Red shift. As per Doppler's effect, we can say that the galaxy is receding from the Earth.

(b) Let v is the speed of the galaxy relative to the Earth. It can be given by :

$v=c(\dfrac{\lambda'}{\lambda}-1)\\\\v=3\times 10^8\times (\dfrac{438.6 }{434.1 }-1)\\\\v=3\times 10^8\times (\dfrac{438.6}{434.1}-1)\\\\v=0.0103\cdot3\cdot10^{8}\\\\v=3.09\times 10^6\ m/s$

So, the speed of the galaxy relative to the Earth is $3.09\times 10^6\ m/s$ .

4 0

3 years ago

Describe the importance of the neutron in a atomic nuclei

RUDIKE [14]

Neutrons are required for the stability of nuclei, with the exception of the single-proton hydrogen nucleus. Neutrons are produced copiously in nuclear fission and fusion. They are a primary contributor to the nucleosynthesis of chemical elements within stars through fission, fusion, and neutron capture processes.

Hope it helps!

5 0

3 years ago

Read 2 more answers

Two runners start at the same point on a straight track. The first runs with constant acceleration so that he covers 98 yards in

charle [14.2K]

Answer:

94.13 ft/s

Explanation:

<u>Given:</u>

$t$ = time interval in which the rock hits the opponent = 10 s - 5 s = 5 s

$s$ = distance to be moved by the rock long the horizontal = 98 yards

$y$ = displacement to be moved by the rock during the time of flight along the vertical = 0 yard

<u>Assume:</u>

$u$ = magnitude of initial velocity of the rock

$\theta$ = angle of the initial velocity with the horizontal.

For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

$\therefore y = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow 0 = u\sin \theta 5 +\dfrac{1}{2}(-9.8)\times 5^2\\\Rightarrow u\sin \theta 5 =\dfrac{1}{2}(9.8)\times 5^2......(1)\\$

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

$\therefore u\cos \theta t = s\\\Rightarrow u\cos \theta 5 = 98.....(2)\\$

On dividing equation (1) by (2), we have

$\tan \theta = \dfrac{25}{20}\\\Rightarrow \tan \theta = 1.25\\\Rightarrow \theta = \tan^{-1}1.25\\\Rightarrow \theta = 51.34^\circ$

Now, putting this value in equation (2), we have

$u\cos 51.34^\circ\times 5 = 98\\\Rightarrow u = \dfrac{98}{5\cos 51.34^\circ}\\\Rightarrow u =31.38\ yard/s\\\Rightarrow u =31.38\times 3\ ft/s\\\Rightarrow u =94.13\ ft/s$

Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.

3 0

3 years ago

A child kicks a ball horizontally with a speed of 4.8 m/s off a deck 3.5 m off the ground. How far, in meters, from the deck doe

Vesna [10]

Answer:

2.605m

Explanation:

Using the formula for calculating Range (distance travelled in horizontal direction)

Range R = U√2H/g

U is the speed = 4.8m/s

H is the maximum height = ?

g is the acc due to gravity = 9.8m/s²

R = 3.5m

Substitute into the formula and get H

3.5 = 4.8√2H/9.8

3.5/4.8 = √2H/9.8

0.7292 = √2H/9.8

square both sides

0.7292² = 2H/9.8

2H = 0.7292² * 9.8

2H = 5.21

H = 5.21/2

H = 2.605m

Hence the height of the ball from the ground is 2.605m

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3 years ago

Identify the factors that cause melting and thus the formation of magma. Choose one or more: A. decompression B. compression C.

nordsb [41]

Answer: C Heat Transfer, E Addition of Volatiles and A Decompression

Explanation:

- Decompression melting takes place within Earth when a body of rock is held at approximately the same temperature but the pressure is reduced.

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3 years ago