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3 years ago

If a cheetah runs 352 meters in 20 seconds, what is the speed of the cheetah?

Physics

1 answer:

valina [46]3 years ago

5 0

Answer:

63360 mi/h

Explanation:

<u>How to find the speed of an object</u>

Calculate speed, distance, or time using the formula d = st, distance equals speed times time. The Speed Distance Time Calculator can solve for the unknown SDT value given two known values.

Time can be entered or solved for in units of second S (s), minutes (min), hours (hr), or hours and minutes and seconds (hh:mm: ss). See shortcuts for time formats below.

To solve for distance use the formula for distance D = st, or distance equals speed times time.

distance = speed x time

Rate and speed are similar since they both represent some distance per unit time like miles per hour or kilometers per hour. If rate r is the same as speed s, r = s = d/t. You can use the equivalent formula d = rt which means distance equals rate times time.

distance = rate x time

To solve for speed or rate use the formula for speed, s = d/t which means speed equals distance divided by time.

speed = distance/time

To solve for time use the formula for time, t = d/s which means time equals distance divided by speed.

time = distance/speed

Therefore, the speed = 63360 miles per hour

= 63360 mi/h

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An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the radius were changed to 1000 m, but the sp

laila [671]

Answer:

The centripetal acceleration changed by a factor of 0.5

Explanation:

Given;

first radius of the horizontal circle, r₁ = 500 m

speed of the airplane, v = 150 m/s

second radius of the airplane, r₂ = 1000 m

Centripetal acceleration is given as;

$a = \frac{v^2}{r}$

At constant speed, we will have;

$v^2 =ar\\\\v = \sqrt{ar}\\\\at \ constant\ v;\\\sqrt{a_1r_1} = \sqrt{a_2r_2}\\\\a_1r_1 = a_2r_2\\\\a_2 = \frac{a_1r_1}{r_2} \\\\a_2 = \frac{a_1*500}{1000}\\\\a_2 = \frac{a_1}{2} \\\\a_2 = \frac{1}{2} a_1$

a₂ = 0.5a₁

Therefore, the centripetal acceleration changed by a factor of 0.5

7 0

4 years ago

An object is at rest. There are several forces acting on the object, but the net force is zero. If all the

pantera1 [17]

It would not. Imagine four forces equal in magnitude but opposite in direction (e.g. north, east, south, and west). If these forces were to double in magnitude they would still have the same magnitude, meaning the net force is still equal to zero.

3 0

2 years ago

A 3.9 kg ball traveling towards a soccer player at a velocity of -3.5 m/s rebounds off the soccer player's foot at a velocity of

Tresset [83]

Answer: 2.92 s

Explanation:

Given

Mass of ball is $m=3.9\ kg$

The initial velocity of the ball is $u=-3.5\ m/s$

Velocity after the rebound is $v=15.9\ m/s$

Force during the contact is $F=25.9\ N$

We know, change in momentum is Impulse

$\Rightarrow F\cdot \Delta t=m(\Delta v)$

$\Rightarrow 25.9\cdot \Delta t=3.9(15.9-(-3.5))\\\\\Rightarrow \Delta t=\dfrac{3.9\times 19.4}{25.9}=2.92\ s$

Thus, the force is applied for 2.92 s

4 0

3 years ago

Recall that the spring constant is inversely proportional to the number of coils in the spring, or that shorter springs equate t

ruslelena [56]

Answer:

$x_1= 0.0425m$

Explanation:

Using the tension in the spring and the force of the tension can by describe by

T = kx

, T = mg

Therefore:

$m*g = k*x$

With two springs, let, T1 be the tension in each spring, x1 be the extension of each spring. The spring constant of each spring is 2k so:

$T_1 = 2k*x_1$

$2T_1 = m*g=4k x_1$

Solve to x1

$x_1=\frac{m*g}{4k}$

$x_1=\frac{k*x}{4*k}$

$x_1=\frac{x}{4}$

$x_1 = 0.170 / 4$

$x_1= 0.0425m$

7 0

3 years ago

If the distance between us and a star is doubled, with everything else remaining the same, the luminosity Group of answer choice

Savatey [412]

Answer:

remains the same, but the apparent brightness is decreased by a factor of four.

Explanation:

A star is a giant astronomical or celestial object that is comprised of a luminous sphere of plasma, binded together by its own gravitational force.

It is typically made up of two (2) main hot gas, Hydrogen (H) and Helium (He).

The luminosity of a star refers to the total amount of light radiated by the star per second and it is measured in watts (w).

The apparent brightness of a star is a measure of the rate at which radiated energy from a star reaches an observer on Earth per square meter per second.

The apparent brightness of a star is measured in watts per square meter.

If the distance between us (humans) and a star is doubled, with everything else remaining the same, the luminosity remains the same, but the apparent brightness is decreased by a factor of four (4).

Some of the examples of stars are;

- Canopus.

- Sun (closest to the Earth)

- Betelgeuse.

- Antares.

- Vega.

8 0

3 years ago