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alexgriva [62]
3 years ago
14

A 2500 N force accelerates a car at a rate of 3.0 m/s^2. What is the car’s mass? 250 kg

Physics
1 answer:
Ronch [10]3 years ago
5 0

Apply Newton's second law to the car's motion:

F = ma

F = net force, m = mass, a = acceleration

Given values:

F = 2500N, a = 3.0m/s²

Plug in and solve for m:

2500 = m(3.0)

m = 830kg

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adell [148]

Answer:

q = 2.65 10⁻⁶ C

Explanation:

For this exercise we use Coulomb's law

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In this case they indicate that the load is of equal magnitude

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the force is attractive because the signs of the charges are opposite

       F = k \ \frac{q^2}{r^2}

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        q = \sqrt{\frac{0.7 \ 0.3^2 }{9 \ 10^9}  }

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Consider three force vectors Fi with magni- tude 53 N and direction 116º, F2 with mag- nitude 57 N and direction 217°, and F3 wi
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Answer:

a. Fnet =37.67N

b. The direction = 133.4 from the x axis counter clockwise.

c. Option 2

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Given that F1 is 53N at 116°, then it will be at a direction of 116-90=26° in the second quadrant.

Given that F2 is 57N at 116°, then it will be at a direction of 217-180=37° in the third quadrant..

Given that F1 is 71N at 20°, then it is in the first quadrant.

a. Fnet= F1+F2+F3

Fnet= -F1sin26i+F2cos26j-F2cos37i-F2sin37j+F3cos20i+F3sin20j

Fnet= 53sin26i+53cos26j-57cos37i-57sin37j+71cos20i+71sin20j

Resolving the vectors into x and y components.

Fnet= -2.04i+37.62j

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Fnet= √((-2.04)^2+(37.62)^2)

Fnet= 37.67N

Fnet is approximately 38N.

b. Direction of the Fnet.

Angle=arctan(y/x)

Angle=arctan(-37.61/2.04)

Angle= -43.37°

The angle is in the negative x axis and positive y axis.

Then the direction becomes 180-43.37

Therefore, the direction of the net force is 133.37°.

c. The instantaneous velocity of a body is always in the direction of the net force at that instant. Option 2 is correct.

Fnet=ma

Fnet= mv/t

So the velocity is in the direction of the Fnet.

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