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3 years ago

What is the direction of magnetic field lines inside any magnet?

Physics

2 answers:

likoan [24]3 years ago

7 0

<h2>Answer:</h2>

The right option is (C)They travel in a loop.

<h2>Explanation:</h2>

The magnetic field lines point from the north pole to the south pole. Magnetic lines of force always forms closed loops. However, the magnetic field lines do not just end at the tip of the magnet. It is equally strong at the north pole when compared with the south pole.They go right through it, so that inside the magnet the magnetic field points from the south pole to the north pole.

Kruka [31]3 years ago

3 0

the answer is a because they start on the north side

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The indices of refraction for violet light (λ = 400 nm) and red light (λ = 700 nm) in diamond are 2.46 and 2.41, respectively. A

JulijaS [17]

Answer:

0.42°

Explanation:

Using Snell's law of refraction which states that the ratio of the angle of sin of incidence to angle of sine of refraction is equal to a constant for a given pair of media. Mathematically,

Sin(i)/sin(r) = n

n is the refractive index of the medium

FOR VIOLET LIGHT:

n = 2.46

i = 51°

r = ?

To get r, we will use the Snell's law formula.

2.46 = sin51°/sinr

Sinr = sin51°/2.46

Sinr = 0.316

r = sin^-1(0.316)

rv = 18.42°

FOR RED LIGHT:

n = 2.41

i = 51°

r = ?

To get r, we will use the Snell's law formula.

2.41 = sin51°/sinr

Sinr = sin51°/2.41

Sinr = 0.323

r = sin^-1(0.323)

rd = 18.84°

The angular separation between these two colors of light in the refracted ray will be the difference between there angle of refraction.

Angular separation = rd - rv

= 18.84° - 18.42°

= 0.42°

6 0

2 years ago

What is the energy of light that must be absorbed by a hydrogen atom to transition an electron from n = 3 to n = 5?

alina1380 [7]

Answer:

The energy absorbed by a hydrogen atom is 1.549 X10⁻¹⁹ J

Explanation:

Using Bohr's equation; the energy absorbed by the hydrogen atom can be calculated as follows:

$\delta E = (\frac{1}{n_2{^2}} -\frac{1}{n_1^{2}})13.6eV$

When an electron moves from a lower energy level to a higher energy level, energy is absorbed by the atom.

Lower energy level (n₂) = 3

Higher energy level (n₁) = 5

1 eV = 1.602X10⁻¹⁹ C

$\delta E = (\frac{1}{3{^2}} -\frac{1}{5^{2}})13.6X1.602X10^{-19}$

ΔE = 1.549 X10⁻¹⁹J

The energy absorbed by a hydrogen atom to transition an electron from n = 3 to n = 5 is 1.549 X10⁻¹⁹ J

4 0

3 years ago

What does decelerate mean

fenix001 [56]

To slow down or reduce speed.

5 0

2 years ago

Find an expression for the electric field e⃗ at the center of the semicircle. hint: a small piece of arc length δs spans a small

ahrayia [7]

Let l = Q/L = linear charge density. The semi-circle has a length L which is half the circumference of the circle. So w can relate the radius of the circle to L by

C = 2L = 2*pi*R ---> R = L/pi 

Now define the center of the semi-circle as the origin of coordinates and define a as the angle between R and the x-axis. 

we can define a small charge dq as 

dq = l*ds = l*R*da 

So the electric field can be written as: 

dE =kdq*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) 

dE = k*I*R*da*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) 

E = k*I*(sin(a)/R I_hat - cos(a)/R^2 j_hat) 

E = pi*k*Q/L(sin(a)/L I_hat - cos(a)/L j_hat)

8 0

3 years ago

Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor

fenix001 [56]

Answer:

a) $F_{r}= -583.72MN i + 183.47MN j + 6.05GN k$

b) $E=3.04 \frac{GN}{C}$

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

$r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}$

which yields:

$r_{AP}^{2}= 43 m^{2}$

(I will assume the positions are in meters)

Next, we can make use of the force formula:

$F=k_{e}\frac{q_{1}q_{2}}{r^{2}}$

so we substitute the values:

$F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}$

which yields:

$F_{AP}=418.14 MN$

Now we can find its components:

$F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i$

$F_{APx}=191.30 MNi$

$F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j$

$F_{APy}=191.30MN j$

$F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k$

$F_{APz}=318.83 MN k$

And we can now write them together for the first force, so we get:

$F_{AP}=(191.30i+191.30j+318.83k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}$

which yields:

$r_{BP}^{2}= 33 m^{2}$

Next, we can make use of the force formula:

$F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}$

which yields:

$F_{BP}=2.18 GN$

Now we can find its components:

$F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i$

$F_{BPx}=758.98 MNi$

$F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j$

$F_{BPy}=758.98MN j$

$F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k$

$F_{BPz}=1.897 GN k$

And we can now write them together for the second, so we get:

$F_{BP}=(758.98i + 758.98j + 1897k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}$

which yields:

$r_{CP}^{2}= 30 m^{2}$

Next, we can make use of the force formula:

$F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}$

which yields:

$F_{CP}=4.20 GN$

Now we can find its components:

$F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i$

$F_{CPx}=-1.534 GNi$

$F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j$

$F_{CPy}=-766.81 MN j$

$F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k$

$F_{CPz}=3.83 GN k$

And we can now write them together for the third force, so we get:

$F_{CP}=(-1.534i - 0.76681j +3.83k)GN$

So in order to find the resultant force, we need to add the forces together:

$F_{r}=F_{AP}+F_{BP}+F_{CP}$

so we get:

$F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN$

So when adding the problem together we get that:

$F_{r}=(-0.583.72i + 0.18347j +6.05k)GN$

which is the answer to part a), now let's take a look at part b).

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

$F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}$

which yields:

$F_{r}=6.08 GN$

and now we take the formula for the electric field which is:

$E=\frac{F_{r}}{q}$

so we go ahead and substitute:

$E=\frac{6.08GN}{2C}$

$E=3.04\frac{GN}{C}$

7 0

3 years ago