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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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A 269-turn solenoid is 102 cm long and has a radius of 2.3 cm. It carries a current of 3.9 A. What is the magnetic field inside
RUDIKE [14]

Answer:

Magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T

Explanation:

Given;

number of turns of solenoid, N = 269 turn

length of the solenoid, L = 102 cm = 1.02 m

radius of the solenoid, r = 2.3 cm = 0.023 m

current in the solenoid, I = 3.9 A

Magnitude of the magnetic field inside the solenoid near its centre is calculated as;

B = \frac{\mu_o NI}{l} \\\\

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

B = \frac{4\pi*10^{-7} *269*3.9}{1.02} \\\\B = 1.293 *10^{-3} \ T

Therefore, magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T

8 0
4 years ago
A diffraction pattern forms when light passes through a single slit. The wavelength of the light is 691 nm. Determine the angle
expeople1 [14]

Explanation:

Given that,

Wavelength of the light, \lambda=691\ nm=691\times 10^{-9}\ m

(a) Slit width, a=3.8\times 10^{-4}\ m

The angle that locates the first dark fringe is given by :

sin\theta=\dfrac{\lambda}{a}

sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-4}}

\theta=0.104^{\circ}

(b) Slit width, a=3.8\times 10^{-6}\ m

The angle that locates the first dark fringe is given by :

sin\theta=\dfrac{\lambda}{a}

sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-6}}

\theta=10.47^{\circ}

Hence, this is the required solution.

7 0
3 years ago
If a suspended object A is attracted to a charged object B, can we conclude that A is charged?
spayn [35]
Not necessarily, object A could also be neutral, and becoming a dipole due to object B's charge. A charged object can induce a dipole in a neutral object, and that object would then become attracted without being charged.
8 0
3 years ago
Is there like an actual definition for momentum?
krok68 [10]

Yes there is.

Momentum of an object is  (its mass) times (its speed) .

6 0
3 years ago
How far will it go, given that the coefficient of kinetic friction is 0.10 and the push imparts an initial speed of 3.9 m/s ?
Akimi4 [234]

Answer:19.5 m

Explanation:

Given

coefficient of kinetic Friction \mu _k=0.10

Initial speed u=3.9\ m/s

Friction is present so it tries to stop to the object and stops it completely after moving certain distance let say s

maximum deceleration provided by friction is

a_{max}=\mu _k\cdot g

a_{max}=0.1\times 3.9=0.39\ m/s^2

using equation of motion

v^2-u^2=2 as

where v=final\ velocity

u=initial velocity

a=acceleration

s=displacement

(0)-(3.9)^2=2\times (-0.39)\times s

s=19.5\ m

6 0
3 years ago
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