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If a cheetah could maintain its top speed of 120 km/h for 20 minutes, how far would it run?

GuDViN [60]

Alright. If it’s running at a constant speed of 120 km/h, in one hour it will have traveled 120 kilometers. hence “kilometers per hour”

20 minutes is 1/3 of an hour so 120/3 will be 40

therefore, in 20 minutes the cheetah would’ve ran 40km if it maintained 120km/h for 20 minutes.

3 0

3 years ago

What is the speed of sound for a noise that travels 2km in 5.8s?

Gnesinka [82]

Answer:

Explanation:

Speed is defined as the rate at which an object covers a particular distance. So the formula for determining speed is given as the ratio of distance to time taken for covering that distance.

Speed = Distance/Time

As here the distance is given in km units and time in s units, so the units of any one parameter should be changed. Since we know that speed of sound is always about 300 m/s. So it is better to convert the unit of distance from km to m.

Hence, now the distance traveled by the noise is 2000 m and time taken is 5.8 s.

So the speed of noise = Distance/Time = 2000/5.8=345 m/s.

Thus, the speed of noise is slightly greater than the speed of sound and it is found to be 345 m/s.

5 0

4 years ago

Hewo fwiends !! What is the different between Kinetic Energy and Kinetic Potential Energy???

Goshia [24]

Answer:

Potential refers to stored energy while kinetic is energy in motion. All energy, whether potential or kinetic, is measured in Joules (J).

Explanation:

Tell me if it is wrong

A brainliest would be appreciated!

Have a good day!

8 0

3 years ago

Read 2 more answers

The burning rates of two different solid-fuel propellants used in aircrew escape systems are being studied. It is known that bot

Katena32 [7]

Answer: $z_{\text{calc}}=-0.32$

Explanation:

The test statistic for difference of two population means :-

$z=\dfrac{\mu_1-\mu_2}{\sigma\sqrt{\dfrac{1}{n_1}+\dfrac{1}{n_2}}}$

Given : $\sigma=\sigma_1=\sigma_2=3$

$n_1 = n_2 = 20$

$\mu_1=23.7$

$\mu_2=24$

Then , $z=\dfrac{23.7-24}{(3)\sqrt{\dfrac{1}{20}+\dfrac{1}{20}}}$

$\Rightarrow\ z=-0.316227766017\approx-0.32$

Hence, the value of $z_{\text{calc}}=-0.32$

5 0

4 years ago

A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of

frosja888 [35]

1. 27.3 m/s

The velocity of the gazelle at any time is given by:

$v=u+at$

where

u is the initial velocity

a is the acceleration

t is the time

Here we have:

u = 0 (the gazelle starts from rest)

$a=4.2 m/s^2$

t = 6.5 s

Substituting the data, we find the gazelle's top speed:

$v=0+(4.2)(6.5)=27.3 m/s$

2. 3.8 s

The distance covered by the gazelle is

d = 30 m

We know that the gazelle accelerates during the first part of the motion and then it continues at constant speed. We need to find first if the gazelle completes the race during the first part of its motion (accelerated motion); to do this, we can calculate what would be the distance covered by the gazelle before reaching the top speed, after t = 6.5 s:

$d'=\frac{1}{2}at^2 = \frac{1}{2}(4.2)(6.5)^2=88.7 m$