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3 years ago

At which location could you place the south pole of a bar magnet so that it would be pulled toward the magnet shown?

Physics

1 answer:

stepan [7]3 years ago

3 0

Answer:

The north pole.

Explanation:

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ILL GIVE BRAINLY THING

Bogdan [553]

Answer:

About 3 trips

Explanation: if we do 2.5m*1.6m*0.75 it equals to 11000 then we divide that to 11m3 and it gives you 3.6 so it will be about 3 times

Thx

4 0

3 years ago

How long would it take 2.0x10^20 electrons to pass through a point in a conductor if the current was 10.0A?

inysia [295]

1 coulomb of electric charge is carried by 6.25 x 10^18 electrons

1 Ampere = 1 coulomb per second
10 A = 10 coulombs per second

(2.0 x 10^20 electrons) x (coul / 6.25 x 10^18 electrons) / (10 coul/sec) =

(2.0 x 10^20) / (6.25 x 10^18 x 10) sec = <em>3.2 seconds</em>

6 0

3 years ago

Light in vacuum is incident on the surface of a slab of transparent material. In the vacuum the beam makes an angle of 39.9° wit

wariber [46]

Answer:

n=2.053

Explanation:

We will use Snell's Law defined as:

$n_{1}*Sin\theta_{1}=n_{2}*Sin\theta_{2}$

Where n values are indexes of refraction and $\theta$ values are the angles in each medium. For vacuum, the index of refraction in n=1. With this we have enough information to state:

$1*Sin(39.9)=n_{2}*Sin(18.2)$

Solving for $n_{2}$ yields:

$n_{2}=\frac{Sin(39.9)}{Sin(18.2)}=2.053$

Remember to use degrees for trigonometric functions instead of radians!

6 0

3 years ago

The maximum wavelength that an electromagnetic wave can have and still eject electrons from a metal surface is 542 nm. What is t

Veseljchak [2.6K]

Answer:

2.295 eV

Explanation:

maximum wavelength, λ = 542 nm = 542 x 10^-9 m

The work function of the metal is defined as the minimum amount of energy falling on the metal so that the photo electrons just ejects the surface of metal.

$W_{o}=\frac{hc}{\lambda }$

where, h is the Plank's constant and c be the speed of light

h = 6.634 x 10^-34 Js

c = 3 x 10^8 m/s

$W_{o}=\frac{6.634\times 10^{-34}\times 3\times 10^{8}}{542\times10^{-9} }$

$W_{o}=3.67\times 10^{-19} J=\frac{3.67\times 10^{-19}}{1.6\times 10^{-19}} eV$

Wo = 2.295 eV

Thus, the work function of this metal is 2.295 eV.

5 0

2 years ago

A car wieghing 10,000 newtons is parked in a garage. what is true about the forces acting upon it

choli [55]

Since the car is at rest, the force experienced by car will be normal face(exerted by surface to which car is in contact) and weight of the car.
As the car is at rest, net force on the car should be zero.

4 0

3 years ago