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guajiro [1.7K]
3 years ago
12

Suppose you have two magnets. Magnet A doesn't have its poles labeled, but Magnet B does have a clearly labeled north and south

pole. If these two are brought into contact with one another, which of the following could you expect? 
A. The side of Magnet A that's repelled by Magnet B's south pole must be Magnet A's north pole.

B. The side of Magnet A that's repelled by Magnet B's north pole must be Magnet A's south pole.

C. The side of Magnet A that's attracted to Magnet B's south pole must be Magnet A's north pole.

D. The side of Magnet A that's attracted to Magnet B's north pole must be Magnet A's north pole.
Physics
2 answers:
Serggg [28]3 years ago
7 0
The opposite pole attracts each other and the similar pole repel each other. Which means only C is true.
Elina [12.6K]3 years ago
3 0

Before going to answer this question first we have to know the fundamental principle of magnetism.

A magnet have two poles .The important characteristic of a magnet is that like poles will repel each other while unlike poles will attract each other.

Through this concept the question can be answered  as explained below-

A-As per first option the side of  magnet A is repelled by the south pole  of magnet B. Hence the pole of a must be south .It can't be north as it will lead to attraction.

B-The side of magnet A is repelled by the  north pole of magnet B. Hence the side of A must be  north pole.It can't be a south pole.

C-The side of magnet A is attracted by the south pole  of magnet  B .Hence the side of magnet A must be north.Hence this is right

D-The side of magnet A is attracted by the north pole of magnet B. Hence the side of A must south.It can't be north as it will lead to repulsion.

Hence the option C is right.

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what is the wavelength λ2λ2lambda 2 of the second laser that would place its second maximum at the same location as the fourth m
nordsb [41]

The distance separating two wave crests is known as the wavelength. The wavelength of the second laser, where its second maximum would be located, is 0.109375 mm.

Double slit interference (constructive) can be described by the equation dsinθ=m λ

Here, the order is (m), the wavelength is (), and the slit distance is (d).

Information provided-

The laser's wavelength is d/8.

The slit separation is 0.500 mm in length.

Put the distance into consideration to determine the wavelength values as,

λ1=0.5 x 10⁻³/8

λ1=0.0625 x 10⁻³

Due to the second laser's wavelength, its second maximum would be at the same spot as the first laser's fourth minimum.

Therefore, we must first determine the fourth order minima as,

dsinθ= (4-1/2) λ1

dsinθ= (7/2) λ1

Using the second maxima, dsinθ= 2λ₂

Put the following value for in the equation above:

7/2λ1 = 2λ₂

7/2 x (0.0625 x 10⁻³) = 2λ₂

λ₂=0.000109375m.

Therefore, the wavelength of the second laser, where its second maximum would be located, is 0.109375 mm.

Learn more about wavelength here-

brainly.com/question/13533093

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4 0
1 year ago
The circuit has a 3 volt EMF and two ohm resistors. How much power in watts does this circuit draw? A) 4.5 , B) 24, C) 1.13 D) 2
dlinn [17]

Answer:

P = 4.5 watts

Explanation:

Given that,

EMF of the circuit, E = 3 volt

The resistance  of the resistors, R = 2 ohms

We need to find the power of this circuit. The relation between power, emf and resistance is given by the formula as follows :

P=\dfrac{V^2}{R}

Substitute all the values,

P=\dfrac{3^2}{2}\\\\P=4.5\ W

So, the power of this circuit is equal to 4.5 watts.

5 0
3 years ago
Which statement explains how planets move in orbit as supported by Newton’s first law of motion?
GuDViN [60]

Answer:

C

Explanation:

got a one hundred on the test

5 0
3 years ago
Read 2 more answers
Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, coun
iris [78.8K]

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})     (1)

n1: final state = 5

n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV

B. The energy of the emitted photon is given by the following formula:

E=h\frac{c}{\lambda}   (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J

Next, you use the equation (2) and solve for λ:

\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm

C. The radius of the orbit is given by:

r_n=n^2a_o   (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

r_5=5^2(2.380)=59.5

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

8 0
3 years ago
8. An effort force of 15 Newtons is applied to an ideal pulley system to lift up a 16 Newton object. If the effort force is exer
Sonbull [250]

Answer:

the distance that the object is raised above its initial position is 5.625 m.​

Explanation:

Given;

applied effort, E = 15 N

load lifted by the ideal pulley system, L = 16 N

distance moved by the effort, d₁ = 6 m

let the distance moved by the object = d₂

For an ideal machine, the mechanical advantage is equal to the velocity ratio of the machine.

M.A = V.R

M.A = \frac{Load}{Effort} = \frac{L}{E} \\\\V.R = \frac{disatnce \ moved \  by \ the \ effort}{disatnce \ moved \  by \ the \ load} = \frac{d_1}{d_2} \\\\For \ ideal \ machine; \ M.A = V.R\\\\\frac{L}{E} = \frac{d_1}{d_2} \\\\d_2 = \frac{E \times d_1}{L} \\\\d_2 = \frac{15 \times 6}{16} \\\\d_2 = 5.625 \ m

Therefore, the distance that the object is raised above its initial position is 5.625 m.​

3 0
2 years ago
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