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Kitty [74]
3 years ago
6

16. How much force would it take to accelerate a 75kg object by 5 m/szą

Physics
1 answer:
patriot [66]3 years ago
7 0

Answer:

The force experienced is 375 N.

Explanation:

Here,

Mass of the object( M)= 75 kg

Acceleration of the body(a) =5 m/s²

Force required to accelerate the body( F) =?

Then, we know ,

From newton's second law of motion,

F=M*a

     =75* 5

    =375 kgm/s²

   = 375 Newton

So, The force experienced by the 75 kg body that undergoes an acceleration of 5 m/s² is 375 N.

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Jill is the input, as she creates the force. The wrench is the output because it gives the force to the finish peace of the chain.
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3 years ago
You are a pirate working for dread pirate roberts. you are in charge of a cannon that exerts a force 10000 n on a cannon ball wh
KATRIN_1 [288]
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7 0
3 years ago
Cheetahs can accelerate to a speed of 20.0 m/s in 2.50 s and can continue to accelerate to reach a top speed of 29.5 m/s. Assume
insens350 [35]

Answer:

boy if you dont get t f outa here

4 0
3 years ago
If a steady-state heat transfer rate of 3 kW is conducted through a section of insulating material 1.0 m2 in cross section and 2
kaheart [24]

Answer:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

So then the difference of temperature across the material would be \Delta T = 375 K

Explanation:

For this case we can use the Fourier Law of heat conduction given by the following equation:

Q = -kA \frac{\Delta T}{\Delta x}   (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 represent the cross sectional area

Q= 3KW represent the rate of heat transfer

\Delta T is the temperature of difference that we want to find

\Delta x=2.5 cm =0.025 m represent the thickness of the material

If we solve \Delta T in absolute value from the equation (1) we got:

\Delta T =\frac{Q \Delta x}{Ak}

First we convert 3KW to W and we got:

Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W

And we have everything to replace and we got:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

So then the difference of temperature across the material would be \Delta T = 375 K

5 0
3 years ago
A rotating wheel requires 5.00 s to rotate 28.0 revolutions. Its angular velocity at the end of the 5.00-s interval is 96.0 rad/
torisob [31]

Answer:

The constant angular acceleration of the wheel is 12.16 rad/s²

Explanation:

Given;

initial angular distance, θ = 28

time of the motion, t = 5 s

initial angular velocity is calculated as;

\omega _i = \frac{\theta}{t} = \frac{28}{5}.\frac{rev}{s}  \ \times \ \ \frac{2 \pi \ rad}{1 \ \ rev} = 35.19 \ rad/s

final angular velocity is given as, \omega _f = 96.0 \ rad/s

The constant angular acceleration is calculated as;

\alpha = \frac{\omega _f - \omega _i}{t} \\\\\alpha = \frac{96 - 35.19}{5} \\\\\alpha =  12.16 \ rad/s^2

Therefore, the constant angular acceleration of the wheel is 12.16 rad/s²

6 0
2 years ago
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