A cheetah ran 300 feet in 2.92 seconds. What was the cheetah’s average speed in miles per hour?

Answers

Answer 1

Answer: We will first convert feet to miles:
1 foot = 0.00018939 miles
300 foot = 0.056817 miles
Than seconds to hours:
2.92 seconds = 0.04867 minutes = 0.0008111 hours
Average speed :
v = d / t
v = 0.056817 miles / 0.008111 h
v = 70.05 miles per hour

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Dense substances sinking into Earth's core heated Earth's interior through _________.A. friction
B. volcanism
C. radioactive decay
D. mantle convection

Answers

Dense substances sink into the earth's core heated earth's interior through friction. Hope this helped!

A skier reaches the bottom of a slope with a velocity of 12 meters per second north. If the skier comes to a complete stop in 3 seconds, what is the acceleration

Answers

The skier's acceleration is 4 meter per square seconds in south direction.

What is acceleration?

Acceleration is the rate at which speed and direction of velocity vary over time. A point or object going straight ahead is accelerated when it accelerates or decelerates.

Even if the speed is constant, motion on a circle accelerates because the direction is always shifting. Both effects contribute to the acceleration for all other motions.

Acceleration is a vector quantity since it has both a magnitude and a direction.

The skier's acceleration is = change in velocity ÷ time interval

= ( final velocity - initial velocity) ÷ time interval

= (0 m/s - 12 m/s) ÷ 3 seconds

= - 4 meter per square seconds.

Hence, the skier's acceleration is 4 meter per square seconds in south direction.

Learn more about acceleration here:

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Answer:

A skier reaches the bottom of a slope with a velocity of 12 meters per second north. If the skier comes to a complete stop in 3 seconds. A. 36 m/s (squared) north

Explanation:

a state reached when particles continue to move but in equal amounts in and out of the cell is called

Answers

The state you are referring to is called dynamic equilibrium. In the context of a cell, this means that the concentration of molecules or ions is the same inside and outside the cell,

but particles continue to move across the membrane in both directions at equal rates.

This state is reached when the membrane is selectively permeable and allows for the free diffusion of the particles across the membrane until an equal concentration is reached on both sides.

At dynamic equilibrium, there is no net movement of particles across the membrane, although individual particles continue to move back and forth in both directions.

This is an important concept in cellular physiology, as it allows for the maintenance of homeostasis and proper functioning of the cell.

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How do you calculate the initial speed given time and displacement?The question is: The longest kick in CFL history was 83.2m. If the ball remained in the air for 4.12s, determine its initial speed.
I calculated it several times and got the same answer, 40.38m/s, but the textbook says it's 28.6m/s.

Answers

Let's first find the velocity in the x-direction which is constant.
$d = vt \n v = (d)/(t) \nv = (83.2m)/(4.12s) \n v= 20.2m/s$

Now, we need the velocity in the y-direction
$d = v_(o)t + (1)/(2)a t^(2) \n d - (1)/(2)at^(2) = v_(o)t\nv_(o) = (0 - (1)/(2)(-9.81)(4.12)^(2))/(4.12)\nv_(o) = 20.2m/s$

Finally, we need to put the two velocities together using pythagorean theorem
$v_(t)^(2) = v_(x)^(2) + v_(y)^(2)$
$v_(t) = \sqrt{20.2^(2) + 20.2^(2)}$
v = 28.6m/s

How is velocity of a body in uniform circular motion related to its time period

Answers

If the radius remains constant, then the speed of an object in uniform
circular motion and its period of revolution are inversely proportional.

stone thrown vertically upward with speed of 15.5m/s from edge of cliff 75.0m high. how much later does it reach bottom of cliff?

Answers

Answer:

Approximately $5.66\; {\rm s}$ , assuming that $g = 9.81\; {\rm m\cdot s^(-2)}$ and that air resistance is negligible.

Explanation:

Let upward be the positive direction. Under the assumptions, acceleration of the stone would be $a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}$ (negative since the stone is accelerating downward.)

The duration of the flight can be found in the following steps:

Find velocity right before landing given displacement, initial velocity, and acceleration.
Find duration of the flight from acceleration and the change in velocity.

In SUVAT equation $v^(2) - u^(2) = 2\, a\, x$ :

$v$ is the final velocity right before landing (needs to be found,)
$u = 15.5\; {\rm m\cdot s^(-1)}$ is the initial velocity,
$a = (-9.81)\; {\rm m\cdot s^(-2)}$ is acceleration, and
$x = (-75.0)\; {\rm m}$ is displacement (downward because the stone landed below where it was launched.)

Rearrange this equation to find $v$ :

$\begin{aligned}v &= \sqrt{u^(2) + 2\, a\, x} \n &= \sqrt{(15.5)^(2) + 2\, (-9.81)\, (-75.0)} \; {\rm m\cdot s^(-1)} \n &\approx -40.05\; {\rm m\cdot s^(-1)}\end{aligned}$ .

In other words, the velocity of this stone has changed from the initial value of $u = (15.5)\; {\rm m\cdot s^(-1)}$ to $v \approx (-40.05)\; {\rm m\cdot s^(-1)}$ during the flight. Divide the change in velocity by acceleration $a = (-9.81)\; {\rm m\cdot s^(-2)}$ (the rate of change in velocity) to find the duration of the flight:

$\begin{aligned}t &= (v - u)/(a) \n &\approx ((-40.05) - (15.5))/((-9.81))\; {\rm s} \n &\approx 5.66\; {\rm s}\end{aligned}$ .

In other words, the stone would be in the air for approximately $5.66\; {\rm s}$ .

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