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Marina86 [1]
3 years ago
10

A crane exerts a net force of 900 N upward on a 750-kilogram car as the crane starts to lift the

Physics
1 answer:
Darya [45]3 years ago
8 0

The acceleration of the car is 1.2 m/s^2 upward

Explanation:

We can solve the question by using Newton's Second Law of motion, which states that:

F=ma

where

F is the net force exerted on an object

m is the mass of the object

a is its acceleration

In this problem, we know that:

F = 900 N is the net force exerted on the car

m = 750 kg is the mass of the car

Solving the equation for a, we find the acceleration of the car:

a=\frac{F}{m}=\frac{900}{750}=1.2 m/s^2

and it is in the same direction as the force (upward).

Learn more about Newton's second law of motion here:

brainly.com/question/3820012

#LearnwithBrainly

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The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m
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Answer:

  • The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

F = k \frac{q_1 q_2}{d^2}

where k is Coulomb's constant, q_1 and q_2 are the charges and d is the distance between the charges.

Working a little the equation, we can take:

d^2 = k \frac{q_1 q_2}{F}

d = \sqrt{ k \frac{q_1 q_2}{F}}

And this equation will give us the distance between the charges. Taking the values of the problem

k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00

(the force has a minus sign, as its attractive)

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 28,462,500 \ m^2}}

d = 5,335.026 m

And this is the distance between the charges.

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