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andrey2020 [161]
3 years ago
6

You may need a piece of paper to write this down and understand. they have a box on a table with 3 arrows. one of them is pointi

ng both up and down and have a 4N above it and -4N on the bottom . they have a arrow on the left pointing to the left that has -2N to the side of it and the last arrow is on the rights pointing to the right that has 5N. and it says/ asks this
Analyzing the Net Force

Your job is to analyze the forces acting on this box to determine the net force acting on it. Look at the following diagram and read the information. Then answer the question.
A downward force of 4 N acts on the box. This is the box’s weight. Assume the table is strong enough to support the box, so the normal force acting on the box has a magnitude of 4 N. A left-pulling force of 2 N and a right-pulling force of 5 N act on the box too.

So, after considering the effects of all these forces, what is the net force acting on the box?

Remember that a force has both a magnitude (a number) and a direction. Therefore, be sure to include both of these in your answer and explain how you found your answer.
Physics
2 answers:
Jet001 [13]3 years ago
7 0
Is there 4 arrows? Cause from the question, it looks like there is four arrows, anyways, net force is defined as the sum of the forces. We can look at this from a component view point.

First lets take a look at vertical forces. There is a force pointing downwards by -4N, there is another force pointing upwards which is +4N. The sum of these forces is 0, so the net force is 0. This tells us that there is no force acting on the box in the vertical direction.

Next let's examine the horizontal forces. There is a force pointing back -2N and another force pointing in the positive direction 5N. The sum of these forces is 3N, meaning there is a force of 3N being applied in the positive x-direction, or to the right.

So the total net force is 3N to the right.
iris [78.8K]3 years ago
6 0

12N to the right on E2020.

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Two stationary point charges of 3.00 nC and 2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at
andrew-mc [135]

Answer:

1. the electric potential energy of the electron when it is  at the midpoint is - 2.9 x 10^{-17} J

2. the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge is - 5.04 x  10^{-17} J

Explanation:

given information:

q_{1} =  3 nC = 3 x 10^{-9} C

q_{2} =  2 nC = 2 x 10^{-9} C

r = 50 cm = 0.5 m

the electric potential energy of the electron when it is  at the midpoint

potential energy of the charge, F

F = k \frac{q_{e}q}{r}

where

k = constant (8.99 x 10^{9} Nm^{2} /C^{2})

electron charge, q_{e} = - 1.6 x 10^{-19} C

since it is measured at the midpoint,

r = \frac{0.5}{2}

  = 0.25 m

thus,

F = F_{1}+ F_{2}

  = k\frac{q_{e} q_{1} }{r} + k\frac{q_{e} q_{2} }{r}

  = \frac{kq_{e} }{r} (q_{1} +q_{2})

  = (8.99 x 10^{9})( - 1.6 x 10^{-19} )(3 x 10^{-9} +2 x 10^{-9})/0.25

  = - 2.9 x 10^{-17} J

the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge

r_{1} = 10 cm = 0.1 m

r_{2} = 0.5 - 0.1 = 0.4 m

F = k\frac{q_{e} q_{1} }{r} + k\frac{q_{e} q_{2} }{r}

  = kq_{e}(\frac{q_{1} }{r_{1} }+\frac{q_{2} }{r_{2} })

  = (8.99 x 10^{9})( - 1.6 x 10^{-19} )(3 x 10^{-9} /0.1+2 x 10^{-9}/0.4)

  = - 5.04 x  10^{-17} J

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Explanation:

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lara [203]

Answer:

c > √(2ab)

Explanation:

In this exercise we are asked to find the condition for c in such a way that the results have been real

The given equation is

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3 years ago
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