All categories

2 years ago

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

Physics

1 answer:

valina [46]2 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

You might be interested in

Derive the equation of motion of the block of mass m1 in terms of its displacement x. The friction between the block and the sur

Alenkasestr [34]

Answer:

the equivalent mass : $m_e = m_1+m_2+\frac{I}{R^2}$

the equation of the motion of the block of mass $m_1$ in terms of its displacement is = $(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

Explanation:

Let use m₁ to represent the mass of the block and m₂ to represent the mass of the cylinder

The radius of the cylinder be = R

The distance between the center of the pulley to center of the block to be = x

Also, the angles of inclinations of the cylinder and the block with respect to the ground to be $\phi$ and $\beta$ respectively.

The velocity of the block to be = v

The equivalent mass of the system = $m_e$

In the terms of the equivalent mass, the kinetic energy of the system can be written as:

$K.E = \frac{1}{2} m_ev^2$ --------------- equation (1)

The angular velocity of the cylinder = $\omega$ : &

The inertia of the cylinder about its center to be = I

The angular velocity of the cylinder can be written as:

$v = \omega R$

$\omega =\frac{v}{R}$

The kinetic energy of the system in terms of individual mass can be written as:

$K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I\omega^2$

By replacing $\omega$ with $\frac{v}{R}$ ; we have:

$K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I(\frac{v}{R})^2$

$K.E = \frac{1}{2}(m_1+ m_2+ \frac{I}{R} )v^2$ ------------------ equation (2)

Equating both equation (1) and (2); we have:

$m_e = m_1+m_2+\frac{I}{R^2}$

Therefore, the equivalent mass : $m_e = m_1+m_2+\frac{I}{R^2}$ which is read as;

The equivalent mass is equal to the mass of the block plus the mass of the cylinder plus the inertia by the square of the radius.

The expression for the force acting on equivalent mass due to the block is as follows:

$f_{block }=m_1gsin \beta$

Also; The expression for the force acting on equivalent mass due to the cylinder is as follows:

$f_{cylinder} = m_2gsin \phi$

Equating the above both equations; we have the equation of motion of the equivalent system to be

$m_e \bar x = f_{cylinder}-f_{block}$

which can be written as follows from the previous derivations

$(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

Finally; the equation of the motion of the block of mass $m_1$ in terms of its displacement is = $(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

8 0

3 years ago

How long does it take (in minutes) for light to reach venus from the sun, a distance of 1.152 × 108 km?

7nadin3 [17]

Using the precise speed of light in a vacuum ( $299,792,458 \ \frac{m}{s}$ ), and your given distance of $1.152 * 10^{8} km$ , we can convert and cancel units to find the answer. The distance in m, using $\frac{1000 \ m}{1 \ km}$ , is $1.152 * 10^{11} m$ . Next, for the speed of light, we convert from s to min, using $\frac{1 \ min}{60 \ s}$ , so we divide the speed of light by 60. Finally, dividing the distance between the Sun and Venus by the speed of light in km per min, we find that it is 6.405 min.

7 0

3 years ago

Three children are riding on the edge of a merry-go-round that is 100 kg, has a 1.60-m radius, and is spinning at 20.0 rpm. The

Kay [80]

Answer:

25.33 rpm

Explanation:

M = 100 kg

m1 = 22 kg

m2 = 28 kg

m3 = 33 kg

r = 1.60 m

f = 20 rpm

Let the new angular speed in rpm is f'.

According to the law of conservation of angular momentum, when no external torque is applied, then the angular momentum of the system remains constant.

Initial angular momentum = final angular momentum

(1/2 x M x r^2 + m1 x r^2 + m2 x r^2 + m3 x r^2) x ω =

(1/2 x M x r^2 + m1 x r^2 + m3 x r^2 ) x ω'

(1/2 M + m1 + m2 + m3) x 2 x π x f = (1/2 M + m1 + m3) x 2 x π x f'

( 1/2 x 100 + 22 + 28 + 33) x 20 = (1/2 x 100 + 22 + 33) x f'

2660 = 105 x f'

f' = 25.33 rpm

8 0

2 years ago

Examine the cross-sectional slice of a stem of a mint plant. Which statement most specifically describes mint?

Maslowich

The available options are:

Mint is a dicot.

Mint is a monocot.

Mint is an angiosperm.

Mint is a bulb plant.

Answer:

Mint is a dicot.

Explanation:

Given the fact that Mint is considered to be a member of Lamiaceae, an angiosperm plant which is characterized by typically having leaves that consist of reticulate vacation and appears like veins in structure. It also has a seed that contains two cotyledons.

Hence, it is considered a DICOT PLANT due to these characteristics. The botanical name of Mint is referred to as Mentha arvensis.

7 0

2 years ago

Frequency of the wave below?

agasfer [191]

It's hard to tell exactly what's happening in that 110 cm that you marked over the wave. What is under the ends of the long arrow ? How many complete waves ? I counted 4.5 complete waves ... maybe ?

If there are 4.5 complete waves in 110cm, then the length of 1 wave is (110/4.5)=24.44cm.

Frequency = speed/wavelength

Frequency = 2m/s /0.2444m

Frequency = 8.18 Hz

6 0

3 years ago