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2 years ago

A 15 N force and a 45 N force act on in object in opposite directions. What is the net force on the object?

Physics

1 answer:

vovikov84 [41]2 years ago

7 0

Answer:

30N in the direction the 45N acts.

Explanation:

Fnet = F1 + F2 (the vector sum of the forces)

Assigning a positive direction to the 45N force and a negative direction to the 15N force gives:

Fnet = 45 - 15

Fnet = 30N

Since the answer is positive, it is in the direction the 45N force acts.

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2 years ago

What amount of force is needed to propel and object of 27 kg to an acceleration of 11,550 m/s^2? (1 point)

aliya0001 [1]

Answer:

311,850 N

Explanation:

We can solve the problem by using Newton's second law:

$F=ma$

where

F is the net force applied on an object

m is the mass of the object

a is its acceleration

For the object in this problem,

m = 27 kg

$a=11550 m/s^2$

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2 years ago

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3 years ago

In the diagram, q1= -2.60*10^-9 C and

Alekssandra [29.7K]

Answer:

The magnitude of the net electric field is:

$E_{net}=90.37\: N/c$

Explanation:

The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).

On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).

Knowing this, the magnitude of the net electric field will be the E1 + E2.

Let's find first E1 and E2.

The electric field equation is given by:

$|E_{1}|=k\frac{|q_{1}|}{d_{1}^{2}}$

Where:

k is the Coulomb constant (k = 9*10^{9} Nm²/C²)
q1 is the first charge
d1 is the distance from q1 to P

$|E_{1}|=(9*10^{9})\frac{|-2.60*10^{-9}|}{0.538^{2}}$

$|E_{1}|=80.84\: N/C$

And E2 will be:

$|E_{2}|=k\frac{|q_{2}|}{d_{2}{2}}$

$|E_{2}|=(9*10^{9})\frac{|-8.30*10^{-9}|}{1.36^{2}}$

$|E_{2}|=40.39\: N/C$

Finally, we need to use the Pythagoras theorem to find the magnitude of the net electric field.

$E_{net}=\sqrt{E_{1}^{2}+E_{2}^{2}}$

$E_{net}=\sqrt{80.84^{2}+40.39^{2}}$

$E_{net}=90.37\: N/c$

I hope it helps you!

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Answer:

Explanation:

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