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LekaFEV [45]
2 years ago
6

Two bar magnets are labeled A and B. The ends of each magnet are numbered 1 or 2, but the poles are not labeled. When A1 is brou

ght near B1, the bars repel. Which conclusion is best supported by the data?
A. A1 is the north pole of a magnet, and B1 is the south pole of a magnet.
B. A1 is the north pole of a magnet, and B1 is the north pole of a magnet.
C. A1 and B1 are opposite poles, but there is not enough information to tell which ones.
D. A1 and B1 are like poles, but there is not enough information to tell whether they are north poles or south poles.
Physics
2 answers:
IceJOKER [234]2 years ago
6 0

Answer: D

Explanation:

vichka [17]2 years ago
3 0
It is definitely letter D. <span>A1 and B1 are like poles, but there is not enough information to tell whether they are north poles or south poles.

A1 and B1 is either both north poles or both south poles. Repulsion of both magnets says it all--like poles always repel while opposite poles always attract. Thus, the best conclusion to this would be choice D.</span>
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−5.0  

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An owl has a mass of 4.00 kg. It dives to catch a mouse, losing 800.00 J of its GPE. What was the starting height of the owl, in
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Answer:

height =20m

Explanation:

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2 years ago
The Slowing Earth The Earth's rate of rotation is constantly decreasing, causing the day to increase in duration. In the year 20
NNADVOKAT [17]

Answer:

The average angular acceleration of the Earth, α  = 6.152 X 10⁻²⁰ rad/s²

Explanation:

Given data,

The period of 365 rotation of Earth in 2006, T₁ = 365 days, 0.840 sec

                                                                                  = 3.1536 x 10⁷ +0.840

                                                                                 = 31536000.84 s

The period of 365 rotation of Earth in 2006, T₀ = 365 days

                                                                               = 31536000 s

Therefore, time period of one rotation on 2006, Tₐ = 31536000.84/365

                                                                                   = 86400.0023 s

The time period of rotation is given by the formula,

                                <em>Tₐ = 2π /ωₐ</em>

                                 ωₐ = 2π / Tₐ

Substituting the values,

                                  ωₐ = 2π /  365.046306        

                                      = 7.272205023 x 10⁻⁵ rad/s

Therefore, the time period of one rotation on 1906, Tₓ = 31536000/365

                                                                                    = 86400 s

Time period of rotation,

                                   Tₓ = 2π /ωₓ

                                    ωₓ = 2π / T

                                           =  2π /86400

                                          = 7.272205217  x 10⁻⁵ rad/s

The average angular acceleration

                                   α  = (ωₓ -   ωₐ) /  T₁

             = (7.272205217  x 10⁻⁵ - 7.272205023 x 10⁻⁵) / 31536000.84

                                    α  = 6.152 X 10⁻²⁰ rad/s²

Hence the average angular acceleration of the Earth, α  = 6.152 X 10⁻²⁰ rad/s²

3 0
2 years ago
A book is pushed with an initial horizontal velocity of 5.0 meters per second off the top of a 1.19 meter high desk. How far awa
kipiarov [429]

Answer:

2.45 m

Explanation:

First of all, we have to calculate the time of flight of the book, by using the equation for the vertical motion:

h=\frac{1}{2}gt^2

where

h = 1.19 m is the vertical distance covered by the book

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

t is the time of flight

Solving for t,

t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(1.19)}{9.8}}=0.49 s

Now we can find the horizontal distance covered by the book, which is given by

d=v_x t

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v_x = 5.0 m/s is the horizontal velocity

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Substituting,

d=(5.0)(0.49)=2.45 m

So the book lands 2.45 m away.

8 0
2 years ago
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