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3 years ago

Define rectilinear propagation of light.??

Physics

2 answers:

Tju [1.3M]3 years ago

5 0

Answer:

it relates to the light propensity to travel over one straight line without having any interference in its trajectory

Explanation:

olya-2409 [2.1K]3 years ago

4 0

Answer:

the light always moves in a straight line I guess that's it

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What is the mass of the object if it has a density of 657 g/mL and a volume of 32 mL? Show work!

Luden [163]

Answer:

The answer is 21024g/mL

Explanation:

Multiply 657 by 32:

657

× 32

⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻

21024

↳ 21024 g/mL

3 0

3 years ago

g A thin-walled hollow cylinder and a solid cylinder, both have same mass 2.0 kg and radius 20 cm, start rolling down from rest

ArbitrLikvidat [17]

Answer:

a. i. 3.43 m/s ii. 2.8 m/s

b. The thin-walled cylinder

Explanation:

a. Find translational speed of each cylinder upon reaching the bottom

The potential energy change of each mass = total kinetic energy gain = translational kinetic energy + rotational kinetic energy

So, mgh = 1/2mv² + 1/2Iω² where m = mass of object = 2.0 kg, g =acceleration due to gravity = 9.8 m/s², h = height of incline = 1.2 m, v = translational velocity of object, I = moment of inertia of object and ω = angular speed = v/r where r = radius of object.

i. translational speed of thin-walled cylinder upon reaching the bottom

So, For the thin-walled cylinder, I = mr², we find its translational velocity, v

So, mgh = 1/2mv² + 1/2Iω²

mgh = 1/2mv² + 1/2(mr²)(v/r)²

mgh = 1/2mv² + 1/2mv²

mgh = mv²

v² = gh

v = √gh

v = √(9.8 m/s² × 1.2 m)

v = √(11.76 m²/s²)

v = 3.43 m/s

ii. translational speed of solid cylinder upon reaching the bottom

So, For the solid cylinder, I = mr²/2, we find its translational velocity, v'

So, mgh = 1/2mv'² + 1/2Iω²

mgh = 1/2mv² + 1/2(mr²/2)(v'/r)²

mgh = 1/2mv'² + mv'²

mgh = 3mv'²/2

v'² = 2gh/3

v' = √(2gh/3)

v' = √(2 × 9.8 m/s² × 1.2 m/3)

v' = √(23.52 m²/s²/3)

v' = √(7.84 m²/s²)

v' = 2.8 m/s

b. Determine which cylinder has the greatest translational speed upon reaching the bottom.

Since v = 3.43 m/s > v'= 2.8 m/s,

the thin-walled cylinder has the greatest translational speed upon reaching the bottom.

3 0

3 years ago

Corrosive substances are rarely harmful to human skin.

Shtirlitz [24]

Corrosive substances are rarely harmful to human skin. This statement is False. Corrosive substances are ALMOST ALWAYS harmful to the skin.

Most problems addressed by the technological design process have only one solution. This statement is also False. There is more than one way to skin a cat.

5 0

3 years ago

Read 2 more answers

Consider a situation where two point charges of charge Q = 16 nC and mass m = 33 g are 27 cm apart. Then, one of these point cha

N76 [4]

Given that,

Charge = 16 nC

Mass = 33 g

Distance = 27 cm

We need to calculate the acceleration

Using formula of electrostatic force

$F=\dfrac{kqQ}{r^2}$

$ma=\dfrac{kq^2}{r^2}$

Put the value into the formula

$33\times10^{-3}\times a=\dfrac{9\times10^{9}\times(16\times10^{-9})^2}{(27\times10^{-2})^2}$

$a=\dfrac{9\times10^{9}\times(16\times10^{-9})^2}{(27\times10^{-2})^2\times33\times10^{-3}}$

$a=0.0009577\ m/s^2$

$a=9.577\times10^{-4}\ m/s^2$

We need to calculate the speed of charge

Using equation of motion

$v^2=u^2+2as$

Where, v= speed

u = initial speed

a = acceleration

s = distance

$v^2=0+2\times9.577\times10^{-4}\times16\times10^{-2}$

$v=\sqrt{0+2\times9.577\times10^{-4}\times16\times10^{-2}}$

$v=0.0175\ m/s$

Hence, The speed of the charge is 0.0175 m

8 0

3 years ago

A uniform electric field of magnitude 442 N/C pointing in the positive x-direction acts on an electron, which is initially at re

Brums [2.3K]

A) The electric field is constant and in the same direction as the electron's displacement, so the work done is given by:

W = Fd

W = work, F = electric force, d = displacement

The electric force on the electron is given by:

F = Eq

F = electric force, q = electron charge

Substitute F:

W = Eqd

Given values:

E = 442N/C, q = 1.6×10⁻¹⁹C, d = 3.50×10⁻²m

Plug in and solve for W:

W = 442(1.6×10⁻¹⁹)(3.50×10⁻²)

W = 2.48×10⁻¹⁸J

B) The electric field does work to move the electron. Apply the conservation of energy and you'll see that the electron's potential energy loss is equal to the work done by the field in moving the electron.

The electric potential energy change is -2.48×10⁻¹⁸J

C) Apply the work-energy theorem; the electron's kinetic energy equals the work done on it by the field.

KE = 0.5mv² = W

m = electron mass, v = velocity, W = work

Given values:

W = 2.48×10⁻¹⁸J, m = 9.11×10⁻³¹

Plug in and solve for v:

0.5(9.11×10⁻³¹)v² = 2.48×10⁻¹⁸

v = 2.33×10⁶m/s

3 0

3 years ago

Define rectilinear propagation of light.??​

Define rectilinear propagation of light.??