All categories

2 years ago

A light beam travels through a diamond (n=2.42) to pure water (n=1.33). The refracted ray will:

Physics

1 answer:

Marrrta [24]2 years ago

3 0

Answer:

b) bend away from the normal

Explanation:

According to snell's law , if i be the angle of incidence and r be the angle of refraction

sin i / sin r = 1.33 / 2.42

sin i / sin r = .55

Hence sin r > sin i

r > i

In other words angle of refraction will be more than angle of incidence . So, the ray will bend away from the normal .

You might be interested in

an engine has been design to work between source and the sink at temperature 177 degree Celsius and 27 degree Celsius respective

irina [24]

Given data

Source temperature (T₁) = 177°C = 177+273 = 450 K

Sink temperature (T₂) = 27°C = 27+273 = 300 K

Energy input (Q₁) = 3600 J ,

Work done = ?

We know that, efficiency (η) = Net work done ÷ Heat supplied

η = W ÷ Q₁

W = η × Q₁

First determine the efficiency ( η ) = ?

Also, we know that ( η ) = (T₁ - T₂) ÷ (T₁)

= 33.3% = 0.333

Now, Work done is W = η × Q₁

= 0.33 × 3600

W = 1188 J

Work done by the engine is 1188 J

4 0

3 years ago

An electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric

8090 [49]

Answer:

1.) 11 km/s

2.) 9.03 × 10^-5 metres

Explanation:

Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.

Electron q = 1.6×10^-19 C

Electron mass = 9.11×10^-31 Kg

(a) What is the speed of the electron 1.3 ns after entering this region?

E = F/q

F = Eq

Ma = Eq

M × V/t = Eq

Substitute all the parameters into the formula

9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19

V = 7.68×10^-18 /7.0×10^-22

V = 10971.43 m/s

V = 11 Km/s approximately

(b) How far does the electron travel during the 1.3 ns interval?

The initial velocity U = 64 km/s

S = ut + 1/2at^2

S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2

S =8.32×10^-5 + 7.13×10^-6

S = 9.03 × 10^-5 metres

3 0

2 years ago

A proton moving in the positive x direction with a speed of 9.9 105 m/s experiences zero magnetic force. When it moves in the po

Alex

Answer:

The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction

Explanation:

In order to calculate the magnitude and direction of the magnetic field, you take into account the following equation for the magnetic force on the proton:

$\vec{F_B}=q\vec{v}\ X\ \vec{B}$ (1)

v: speed of the proton = 9.9*10^5 m/s

q: charge of the proton = 1.6*10^-19C

B: magnetic field = ?

FB: magnetic force on the proton = 1.6*10^-13N

When the proton travels in the positive y direction (^j), you have that the proton experiences a force in the positive z direction (+^k). To obtain this direction of the magnetic force on the proton, it is necessary that the magnetic field points in the negative x direction, in fact, you have:

^j X (-^i) = -(-^k)=^k

To obtain the magnitude of the magnetic field you use:

$F_B=qvBsin90\°=qvB\\\\B=\frac{F_B}{qv}=\frac{1.6*10^{-13}N}{(1.6*10^{-19}C)(9.9*10^5m/s)}\\\\B=1.01T$

The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction

8 0

2 years ago

Use differentials to estimate the amount of metal in a closed cylindrical can that is 10 cm high and 4 cm in diameter if the met

ira [324]

The amount of metal in a closed cylindrical can that is 10 cm high and 4 cm in diameter if the metal on the top and the bottom is 0.1 cm thick and the metal on the sides is 0.05 cm thick is 8.8 cm.

The formula for calculating the volume of a cylinder is given below.

V = πr^2 h

Get the differential of the volume as shown:

dV = V/ h dh + V / r dr

V/ h = πr^2

V/ h = 2 πr h

Now, the differential becomes

dV = πr^2dh + 2πrh dr

Given the following parameters i.e. diameter and height

dh = 0.1 + 0.1 = 0.2 cm

dr = 0.05 cm

h = 10 cm