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Anit [1.1K]
2 years ago
13

Assume that the length of the magnet is much smaller than the separation between it and the charge. As a result of magnetic inte

raction (i.e., ignore pure Coulomb forces) between the charge and the bar magnet, the magnet will experience which of the following?
a. No torque at all
b. A torque only if one magnetic pole is slightly closer to the charge than the other
c. A torque due to the charge attracting the south pole of the magnet
d. A torque due to the charge attracting the north pole of the manget
Physics
1 answer:
faltersainse [42]2 years ago
5 0

Answer:

Assuming that the length of the magnet is much smaller than the separation between it and the charge. As a result of magnetic interaction (i.e., ignore pure Coulomb forces) between the charge and the bar magnet, the magnet will not experience any torque at all - option A

Explanation:

Assuming that the length of the magnet is much smaller than the separation between it and the charge. As a result of magnetic interaction (i.e., ignore pure Coulomb forces) between the charge and the bar magnet, the magnet will not experience any torque at all; the reason being that: no magnetic field is being produced by a charge that is static. Only a moving charge can produce a magnetic effect. And the magnet can not have any torque due to its own magnetic lines of force.

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Alex is standing still and throws a football with a speed of 10 m/s to his friend, who is also standing still. The two friends a
Phantasy [73]

The question is incomplete. It comes with a set of answer choices.


These are the answer choices:


Alex observes it as 10 m/s, and his friend observes it as less than 10 m/s.


Alex observes it as less than 10 m/s, and his friend observes it as 10 m/s.


Both Alex and his friend observe it as 10 m/s.


Both Alex and his friend observe it as less than 10 m/s.



Answer: Both Alex and his friend observe it as 10 m/s.


Justification:


1) The speed is relative to the frame of reference.


2) It is said that the both Alex and his friend are standing still.


3) Then, the speed they both see is the same, 10 m/s, respect the Earth (where they are standing still).


Of course, Alex is watching the ball moving away and his friend is seing it approaching, but it is not relevant for the question, as it deals with the speed which is only about magnitude, not direction.

7 0
2 years ago
Read 2 more answers
You have negotiated with the Omicronians for a base on the planet Omicron Persei 7. The architects working with you to plan the
steposvetlana [31]

Answer:

5.724 meters / second^2

Explanation:

We are given two pieces of information, 5.24 flurg = 1 meter, 1 grom = 0.493 second. If that is so, we can say that there are two possible conversion units,  5.25 flurg / meter, and 0.493 second / grom.

_____

We want to convert 7.29 flurg / grom^2 ( I believe? ) to the units meters / second^2. But, let's break this down into bits. It would be convenient to first convert 7.29 flurg / grom^2 to the units meters / grom^2, by dividing the conversion factors as to cancel out the appropriate things, which we will go into detail on a bit later ( using the first conversion factor ). Respectively we can convert meters / grom^2 to meters / grom * s, canceling out the flurg ( through the second conversion factor ). And now we would need to get rid of the grom, dividing similarly.

_____

( 1 ) ( flurg / grom^2 ) / ( flurg / meters  ) - first conversion unit

= flurg / grom^2 * meters /flurg

= ( meters * flurg ) / ( grom^2 * flurg )  

= meters /grom^2,

7.29 flurg / grom^2 / 5.24 flurg / meter = ( About ) 1.39 meter / grom^2

( 2 ) ( meter / grom^2 ) / ( second / grom  ) - second conversion unit

= meter / grom^2 * grom / second

= ( meter * grom ) / ( grom^2 * second )

= meter / ( grom * second ),

( 1.39 meter / grom^2 ) / 0.493 second / grom = ( About ) 2.82195 meter /  grom * second

( 3 ) ( 2.82195 meter / ( grom * second ) ) / 0.493 second / grom = 5.724 meter / second^2

( And thus, the value of gOP7 in the units the architects will use should be about 5.724 meters / second^2 )

8 0
2 years ago
A flat surface is in a uniform magnetic field. Given only the area of the surface and the magnetic flux through the surface, it
Tasya [4]

Answer:

Given the area A of a flat surface and the magnetic flux through the surface \Phi it is possible to calculate the magnitude \frac{\Phi}{A}=B\ cos \theta.

Explanation:

The magnetic flux gives an idea of how many magnetic field lines are passing through a surface. The SI unit of the magnetic flux \Phi is the weber (Wb), of the magnetic field B is the tesla (T) and of the area A is (m^{2}). So 1 Wb=1 T.m².

For a flat surface S of area A in a uniform magnetic field B, with \theta being the angle between the vector normal to the surface S and the direction of the magnetic field B, we define the magnetic flux through the surface as:

                                                     \Phi=B\ A\ cos\theta

We are told the values of \Phi and B, then we can calculate the magnitude

                                                      \frac{\Phi}{A}=B\ cos\theta

3 0
1 year ago
We can learn a lot about the properties of a star by studying its spectrum. All of the followingstatements are true except one.
Kisachek [45]

Answer:

B. The total amount of light in the spectrum tells us the star’s radius.

Explanation:

A.

The effective temperature of a star can be determined by means of its spectrum¹ and Wien's displacement law.                    

Since stars behave in a local way as a blackbody, it will take the wavelength at which is the peak of emission greater in the continuum (see the image below).

Then, the maximum peak of emission (\lambda_{max}) will be replaced in the next equation of the Wien's displacement law:

T = \frac{2.898x10^{-3} m. K}{\lambda max}  (1)

Where T is the effective temperature of the star.

Bodies that are hot enough emits light as consequence of its temperature. For example, a iron bar in contact with fire will start to change colors as the temperature increase, until it gets to a blue color, which its know as Wien's displacement law. Which establishes that the peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase.

The same scenario described above can be found in the stars, a star whit higher temperature will have a blue color and one with lower temperature, a red color.

B.

Since star does not have the same size, they have different brightness, That is because the photons have a free mean path greater in a bigger radius.

So a star brightness is a consequence of its radius.

               

C.  

Spectral lines will be shifted to the blue part of the spectrum1 if the source of the observed light is moving toward the observer, or to the red part of the spectrum when it is moving away from the observer (that is known as the Doppler effect).        

By using that shift in the spectral lines, the Doppler velocity can be determined.

v = c\frac{\Delta \lambda}{\lambda_{0}}  (2)

Where \Delta \lambda is the wavelength shift, \lambda_{0} is the wavelength at rest, v is the velocity of the source and c is the speed of light.

   

D.

When a photon is absorbed by an electron in an atom of a particular element in the star photosphere, the electron will be pass to a higher state, when it comes back to the ground state, a photon will be emitted again. If the emitted photon does not go in the same direction of the incident photon an absorption line will be created in the spectrum of the star.          

This patterns of spectral lines in the spectrum of the star are compared with the patterns that are got by lamps of that element in a laboratory.

Key term:

¹Spectrum: decomposition of light in its characteristic colors (wavelengths).

3 0
1 year ago
A 15-cm-tall closed container holds a sample of polluted air containing many spherical particles with a diameter of 2.5 μm and a
TEA [102]

Answer:

t = 224 s

Explanation:

given,

length of container = 15 cm = 0.15 m

diameter of spherical particle = 2.5 μm

mass of particle = 1.9 x 10⁻¹⁴ kg

viscosity of air = μ = 1.18 x 10⁻⁵ kg/m.s

time taken by the particle to stop = ?

radius of particle = 2.5/2 = 1.25 μm

volume of particle = \dfrac{4}{3}\pi r^3

                              =\dfrac{4}{3}\pi (1.25 \times 10^{-6})^3

Density =\dfrac{mass}{volume}

Density =\dfrac{1.9 \times 10^{-14}}{\dfrac{4}{3}\pi (1.25 \times 10^{-6})^3}

ρ = 2322 kg/m³

terminal velocity

v_t = \dfrac{2}{9}\ \dfrac{gR^2(\rho - \rho_{air})}{\mu}

v_t = \dfrac{2}{9}\ \dfrac{9.8 \times (1.25 \times 10^{-6})^2(2322 - 1)}{1.18 \times 10^{-5}}

v_t = 6693 x 10⁻⁷ m/s

t = \dfrac{d}{v_t}

t = \dfrac{0.15}{6693 \times 10^{-7}}

t = 224 s

7 0
2 years ago
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