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4 years ago

A cheetah can run from 0 to 70 mph in 2.2 seconds. what is the cheetah's acceleration in m/s?

Physics

2 answers:

Varvara68 [4.7K]4 years ago

6 0

14.2 m/s2

hope this helps...................

tankabanditka [31]4 years ago

5 0

(70mph-0mph)/(2.2s-0s) = 31.8181 mph/s

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1.) If you make a false statement or commit a forgery about your motor vehicle insurance you can be guilty of a _____ degree mis

krok68 [10]

A person is guilty of second degree misdemeanor if he or she makes a false statement or commits forgery about their motor vehicle’s insurance.

You would be required to have bodily injury liability insurance in Florida if you are involved in a crash where your vehicle has caused damage to the property of others.

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6 0

3 years ago

Read 2 more answers

What does a mechanical advantage less then one mean? What is gained?

GrogVix [38]

A mechanical advantage less than ' 1 ' means the output force is less than the input
force. But distance is gained ... the load moves farther than the driving force moves.

An example of all this is a big father and his little 4-year-old daughter playing
on the see-saw in the park. He adjusts the board so that he's much closer
to the pivot than she is, and that way, their weights can balance each other.

It's set up so that the mechanical advantage from him to her is less than ' 1 '.
Now, look at what happens when Dad sits down on his end of the see-saw,
and pushes it down with his 180 pounds of force. At the other end, the board
is barely lifting her with a force of just 30 pounds. BUT ... as he sinks down
only 1 foot against his end of the board, her end rises 6 feet off the ground.

7 0

3 years ago

How long does it take a rocket traveling at a constant velocity of 100 m/s to climb from 200m to 1200m

neonofarm [45]

Answer:

10 seconds

Explanation:

you multiply 100 by 10 in order to get 1000m

you add 200m to 1000m to get to the maximum of 1200m

since you travel at 100 per second, it took 10 seconds

7 0

3 years ago

Un the way to the moon, the Apollo astro-

kherson [118]

Answer:

Distance = 345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$