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aalyn [17]
3 years ago
10

calculate the acceleration of a 520 kg car that had 5050N of net force exerted on it by the engine. explain how you solve the pr

oblem please
Physics
1 answer:
elena55 [62]3 years ago
3 0

I'll explain first:

You use the formula for Newton's second law of motion.
The formula is so simple, and it solves so many problems,
that it's actually a good idea to memorize it.
(I learned it 58 years ago, and I still remember it.)

The formula is                 F  =  M A

It means                    Force = (mass) x (acceleration)

In this question, you know the force, and you know the mass,
so calculating the acceleration is a piece-o-cake.

                                                     Force = (mass) x (acceleration)

                                                (5,050 N) = (520 kg) x (acceleration)

Divide each side by  520 kg :  (5,050 N / 520 kg) = acceleration

                                                 Acceleration = 9.712 m/s² (rounded)
 
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7 0
2 years ago
If a force of 32000N exerted pressure of 160N/m² , find the area on which the force acts.​
Alex17521 [72]

Answer: 200m^2

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160N=32000N/x

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x=200m^2

4 0
1 year ago
Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e
Nataliya [291]

Answer:

q=1.95*10^{-7}C

Explanation:

According to the free-body diagram of the system, we have:

\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)

So, we can solve for T from (1):

T=\frac{mg}{cos(15^\circ)}(3)

Replacing (3) in (2):

(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e

The electric force (F_e) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)

Finally, we replace (5) in (4) and solving for q:

mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C

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3 years ago
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Read 2 more answers
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