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pav-90 [236]
4 years ago
13

Barton conducts an experiment using three metallic bars that might be magnets. The bars are labeled A, B, and C. The ends of eac

h bar are numbered 1 or 2.
1A2 | 1B2 | 1C2

He places the end of one bar close to an end of a second bar and records his results in the table shown.

Based on the data, which prediction should he expect to occur?


A) A2 repels B1.
B) C2 attracts B2.
C) B1 repels C1.
D) A1 attracts C2.

Physics
2 answers:
djyliett [7]4 years ago
5 0

The answer is B.) C2 attracts B2.

sineoko [7]4 years ago
4 0

Answer:

A) A2 repels B1

C) B1 repels C1

Explanation:

First, from the rule of magnetization, like poles repel while unlike poles attract, therefore, from the data given;

Record 1: If A1 and B1 attracts

This means that A1 and B1 are opposite poles of a magnet

Therefore, we will assume that A1 is NORTH POLE while A2 is SOUTH  POLE of magnet A.

Then we will have B1 as SOUTH POLE while B2 is NORTH  POLE

Record 2: If A2 and C1 repel

This means A2 and C1 are like poles of a magnet.

Since A2 is a SOUTH POLE, C1 is also a SOUTH POLE

If C1 is SOUTH POLE, then  C2 is NORTH POLE.

Record 3: If B2 and A1 attracts

This means B2 and A1 are opposite poles.

Now we have all the poles identified.

Therefore, the predictions that will occur are as follows;

A) A2 repels B1.    because A2 and B1 are SOUTH POLES

C) B1 repels C1. because B1 and C1 are SOUTH POLES

However, the following predictions will not occur;

B) C2 attracts B2 will not occur because C2 and B2 are NORTH POLES

D) A1 attracts C2 will not occur because A1 and C2 are NORTH POLES

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A parallel beam of light of wavelength 4.5 x 10^-7 m is incident on a pair of slits that are 5.0 x 10^-4 m apart. The interferen
RSB [31]

Answer:

1.8x10⁻³m

Explanation:

From the question above, the following information was used to solve the problem.

wavelength λ = 4.5x10⁻⁷m

Length L = 2.0 meters

distance d = 5 x 10₋⁴m

ΔY = λL/d

= 4.5x10⁻⁷m (2) / 5 x 10₋⁴m

= 0.00000045 / 0.0005

= 0.0000009/0.0005

= 0.0018

= 1.8x10⁻³m

from the solution above The separation between two adjacent bright fringes is most nearly 1.8x10⁻³m

thank you!

3 0
3 years ago
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Alex73 [517]

Answer:

number 6 is wrong

the answer is pesticides

2.combines are used for harvesting or threshing.so number <em>2</em><em> </em><em>is</em><em> </em><em>w</em><em>rong</em>

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The mechanical energy of an object equals its
vova2212 [387]
B. Kinetic energy plus its potential energy.
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How is wind energy converted into electrical energy.
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3 years ago
5) Consider pushing a 50.0 kg box through a 5.00 m displacement on both a flat surface and up a
Svetach [21]

a) The work done by the gravitational force on the flat surface is zero

b) The work done by the gravitational force on the ramp is -634 J

c) The work done by the applied force on the flat surface is 500 J

d) The work done by the applied force on up along the ramp is 500 J

Explanation:

a)

The work done by a force is given by the equation

W=Fdcos \theta

where

F is the magnitude of the force

d is the dispalcement of the object

\theta is the angle between the direction of the force and of the displacement

In this problem, we want to calculate the work done by the gravitational force as the box is pushed across the flat ground.

We immediately notice that the gravitational force acts downward, while the displacement is horizontal: therefore, the angle between force and displacement is 90^{\circ}; this means that cos 90^{\circ}=0, and therefore, the work done is zero:

W=0

b)

In this case, the box is pushed along the ramp. We have:

F=mg=(50.0)(9.8)=490 N is the magnitude of the force of gravity, where

m = 50.0 kg is the mass of the box

g=9.8 m/s^2 is the acceleration of gravity

d = 5.00 m is the displacement of the box along the ramp

The ramp is inclined to the horizontal by 15.0^{\circ}, therefore the angle between the force of gravity and the displacement of the box (moving up along the ramp) is:

\theta=90^{\circ}+15^{\circ}=105^{\circ}

Therefore, the work done by gravity in this case is:

W=(490)(5.00)(cos 105^{\circ})=-634 J

c)

In this case, we want to calculate the work done by the force you apply as the box is pushed across the flat ground.

Here we have:

F = 100.0 N (force applied)

d = 5.00 m (displacement of the box)

\theta=0^{\circ} (the force is applied parallel to the flat surface, therefore force and displacement have same direction)

Therefore, the work done by the force you apply on the flat ground is:

W=(100.0)(5.00)(cos 0^{\circ})=500 J

d)

In this last case, we want to calculate the work done by the force you apply as the box is pushed up along the ramp.

This time we have:

F = 100.0 N (force applied is the same)

d = 5.00 m (displacement of the box is also the same)

\theta=0^{\circ} (the force is applied parallel to the ramp, therefore force and displacement have again same direction)

Therefore, the work done by the force you apply while pushing the box along the ramp is:

W=(100.0)(5.00)(cos 0^{\circ})=500 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

8 0
4 years ago
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