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3 years ago

A 10 kg computer accelerates at a rate of 5 m/s2. How much force was applied to the computer?

Physics

2 answers:

worty [1.4K]3 years ago

8 0

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 10 × 5

We have the final answer as

Hope this helps you

AURORKA [14]3 years ago

6 0

Answer:

<h2><u>Given </u><u>:</u><u>-</u><u> </u></h2>

Mass = 10 kg
Acceleration = 5 m/s²

Force applied

<h2>SoluTion</h2>

As we know that Force applied is the product of mass and acceleration

$\sf \: force \: = 5 \times 10$

$\sf \: force \: = 50 \: n$

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3 years ago

You are skiing down a mountain with a vertical height of 1000 feet. the distance from the top of the mountain to the base is 200

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3 years ago

A rocket is launched from atop a 101 foot cliff with an initial velocity of 116 ft/s.

Katen [24]

Answer:

A. 16t2 -116t -101=0 B. 8.0s

Explanation:

Known parameters from the question:

Height of Cliff,h0 = 101ft

Velocity of rocket ,v= 116ft/s.

1.Substituting the above to the above formula we have;

h(t) = -16t2 + vt + h0

Since h(t) =0, it means the rocket is falling towards the ground, when it gets to the ground when it's at rest the height h(t) = 0m

{Note the rocket is launched from the height of the Cliff so that would be the initial height of the rocket,h0}

2.Substituting into h(t) = -16t2 + vt + h0

We have;

0 = -16t2 + 116t + 101=> 16t2 -116t -101=0

Using formula method for solving quadratic equation we have;

t = -(-116)+_√[(-116)^2 -( 4× 16 ×-101]/ (2× 16)

t = [116 +_(141.1382)]/32

t = (116 -141.1382)/32 or (116 +141.1382)/32

-0.786s or 8.036s

-0.8s or 8.0s to the nearest tenth.

Now time cannot be negative in real life situation hence the time is 8.0s

Note : the general equation of a quadratic equation with variable t is given below;

at2 + bt + c=0

Formula method for quadratic equation is :

t =( -b+_√[(b^2 -( 4× a×c)] ) / (2× a)

6 0

3 years ago

To start a lawn mower, you must pull on a rope wound around theperimeter of a flywheel. After you pull the rope for 0.95 s, thef

Molodets [167]

Answer:

29.76245 rad/s², -117.80972 rad/s²

28.2743 rad/s

3.95833

Explanation:

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation

t = Time taken

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{4.5\times 2\pi-0}{0.95}\\\Rightarrow \alpha=29.76245\ rad/s^2$

Angular acceleration during speed up is 29.76245 rad/s²

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-4.5\times 2\pi}{0.24}\\\Rightarrow \alpha=-117.80972\ rad/s^2$

Angular acceleration during spin down is -117.80972 rad/s²

Angular speed is given by

$\omega=2\pi 4.5=28.2743\ rad/s$

Maximum angular speed reached by the flywheel is 28.2743 rad/s

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=0\times t+\frac{1}{2}\times 29.76245\times 0.95^2\\\Rightarrow \theta=13.4303\ rad$

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=2\pi 4.5\times 0.24+\frac{1}{2}\times -117.80972\times 0.24^2\\\Rightarrow \theta=3.39292\ rad$

The ratio would be $\dfrac{13.4303}{3.39292}=3.95833$

3 0

3 years ago

A 10 kg computer accelerates at a rate of 5 m/s2. How much force was applied to the computer?​

A 10 kg computer accelerates at a rate of 5 m/s2. How much force was applied to the computer?