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2 years ago

Diamonds are a very dense material. Predict what would happen to the light ray if you projected it from air through a diamond

Physics

2 answers:

Anni [7]2 years ago

5 0

Answer:

a cat could eat it

Explanation:

Fluffy ate my diamond

Kipish [7]2 years ago

4 0

Answer:

It [the diamond] would act like a prism, and make a rainbow, or, the light would break up and disappear

Explanation:

that's what I would think at least

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an object with a mass of 6 kilograms accelerates 4.0 MS to the second when an unknown force is applied to it what is the amount

SVEN [57.7K]

Using the Newton's Secound Law, we have:

$F=ma \\ F=6*4 \\ \boxed {F=24N}$

7 0

3 years ago

What would you say to a friend who made this statement, “The visible-light spectrum of the Sun shows weak hydrogen lines and str

Oliga [24]

Explanation:

spectral lines or signatures of elements depend on temperature, the temperature of the sun is about 5800 K.

at this temperature most calcium atoms are excited to higher energy states than hydrogen atoms and this means that calcium atoms are gonna have more signatures than the atoms of hydrogen.

the statement that the sun shows weak hyrogen lines and strong calcium line is wrong because at the sun's temperature most of the hydrogen atoms are in lower energy states while calcium atoms are in higher energy states hence calcium has more or ''strong'' lines than hydrogen.

8 0

3 years ago

Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The

Greeley [361]

Answer:

I = 215.76 A

Explanation:

The direction of magnetic field produced by conductor 1 on the location of conductor 2 is towards left. Based on Right Hand Rule -1 and taking figure 21.3 as reference, the direction of force Fm due to magnetic field produced at C_2 is shown above. The force Fm balances the weight of conductor 2.

Fm = u_o*I^2*L/2*π*d

where I is the current in each rod, d = 0.0082 m is the distance 27rId

between each, L = 0.85 m is the length of each rod.

Fm = 4π*10^-7*I^2*1.1/2*π*0.0083

The mass of each rod is m = 0.0276 kg

F_m = mg

4π*10^-7*I^2*1.1/2*π*0.0083=0.0276*9.8

I = 215.76 A

note:

mathematical calculation maybe wrong or having little bit error but the method is perfectly fine

5 0

3 years ago

A current I flows down a wire of radius a.

Helga [31]

Answer:

(a) $K = \frac{I}{2\pi a}$

(b) $J = \frac{I}{2\pi as}$

Explanation:

(a) The surface current density of a conductor is the current flowing per unit length of the conductor.

$K = \frac{dI}{dL}$

Considering a wire, the current is uniformly distributed over the circumferenece of the wire.

$dL = 2\pi r$

The radius of the wire = a

$dL = 2\pi a$

The surface current density $K = \frac{I}{2\pi a}$

(b) The current density is inversely proportional

$J \alpha s^{-1}$

$J = \frac{k}{s}$ ......(1)

k is the constant of proportionality

$I = \int\limits {J} \, dS$

$I = J \int\limits \, dS$ ........(2)

substituting (1) into (2)

$I = \frac{k}{s} \int\limits\, dS$

$I = k \int\limits^a_0 \frac{1}{s} {s} \, dS$

$I = 2\pi k\int\limits\, dS$

$I = 2\pi ka$

$k = \frac{I}{2\pi a}$

substitute $J = \frac{k}{s}$

$J = \frac{I}{2\pi as}$

7 0

3 years ago

"The drawing shows three layers of different materials, with air above and below the layers. The interfaces between the layers a

My name is Ann [436]

Answer:

Angle of incidence that entered material b= 63.1°

Angle of incidence between a and b = 55.9°

Explanation: Using the formular:

n1sintheta1= n2sintheta2

The light ray which enters material B will be

1.4Sin72.8° = 1.5Sin theta

1.3373= 1.5Sintheta

sintheta = 1.3373/1.5

Sin^-1 0.8916 = Theta

63.1 = theta

When the ray hits interface with material a

1.5Sin63.1 = 1.3 Sin theta

1.3374 = 1.3Sin theta

Sintheta= 1.3374/1.3

Sin theta = 1.0877

There will be total reflection off the boundary b c because sin theta exceeded 1 in value.

The equation should be

1.4sin63.1 = 1.4 sin theta

Sin theta=72.8°

When the ray hits air-c boundary:

1.4sin72.8=1.00sin theta

Sin theta=1.3374/1 =1.3374

There is total reflection.

In material a,the ray will:

1.3sin72.8° = 1.00sin theta

There will be total reflection when the ray hits a-b boundary.

1.3sin72.8= 1.5sintheta

Sin theta= 1.2419/ 1.5

Sin theta =0.8279

Theta= Sin^-10.8279= 55.88°

When ray hits c-air boundary

1.4sin63.1= 1.00sintheta

1.2485= sin theta = Toal reflection.

Therefore when the ray of light pass through the layers of material a, b and c the boundary with air on top and bottom will be total reflection.

7 0

3 years ago