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3 years ago

What two models are used to describe how light behaves

Physics

1 answer:

S_A_V [24]3 years ago

5 0

Answer:

Light is one of nature's ways of moving energy from one place to another.

Explanation:

It has no substance, or mass. How does light travel? Light behaves like a traveling wave, something like waves in a string or on the surface of water.

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Which of the following is not a natural resource? a. time b. water c. land d. air Please select the best answer from the choices

AleksAgata [21]

Answer: Option (a) is the correct answer.

Explanation:

Resources which are created by human beings are known as man-made resources.

For example, glass, rayon, nylon etc are all man-made resources.

Whereas resources which are naturally created are known as natural resources.

For example, wind, air, water etc are all natural resources.

Thus, we can conclude that out of the given options time is not a natural resource.

3 0

3 years ago

49. \ A rectangular plate is rotating with a constant angular speed about an axis that passes perpendicularly through one corner

faust18 [17]

Answer:

$\frac{L_1}{L_2} = \sqrt{(n^2 - 1)}$

Explanation:

For this interesting problem, we use the definition of centripetal acceleration

a = v² / r

angular and linear velocity are related

v = w r

we substitute

a = w² r

the rectangular body rotates at an angular velocity w

We locate the points, unfortunately the diagram is not shown. In this case we have the axis of rotation in a corner, called O, in one of the adjacent corners we call it A and the opposite corner A

the distance OB = L₂

the distance AB = L₁

the sides of the rectangle

It is indicated that the acceleration in in A and B are related

$a_A = n \ a_B$

we substitute the value of the acceleration

w² r_A = n r_B

the distance from the each corner is

r_B = L₂

r_A = $\sqrt{L_1^2 + L_2^2}$

we substitute

\sqrt{L_1^2 + L_2^2} = n L₂

L₁² + L₂² = n² L₂²

L₁² = (n²-1) L₂²

4 0

3 years ago

The repulsive force between two protons has a magnitude of 2.00 N. What is the distance between them?

Aloiza [94]

Answer:

The answer to your question is letter A. r = 1.07 x 10⁻¹⁴ m

Explanation:

Data

F = 2 N

d = ?

q = 1.6 x 10 ⁻¹⁹ C

k = 8.987 Nm²/C²

Formula

$F = K\frac{q1q2}{r^{2}}$

Solve for r

$r = \sqrt{\frac{kq1q2}{F}}$

Substitution

$r = \sqrt{\frac{8.987 x 10^{9}x1.6 x 10^{-19} x 1.6 x 10x^{-19}}{2}}$

Simplification

r = $\sqrt{\frac{2.3 x 10^{-28}}{2}}$

r = $\sqrt{1.15 x 10^{-24}}$

Result

r = 1.07 x 10⁻¹⁴ m

6 0

3 years ago

Read 2 more answers

A projectile is shot from the edge of a cliff above the ground level with initial velocity of at an angle with the horizontal. (

Xelga [282]

Answer:

t = √2y/g

Explanation:

This is a projectile launch exercise

a) The vertical velocity in the initial instants ( $v_{oy}$ = 0) zero, so let's use the equation

y = $v_{oy}$ t -1/2 g t²

y= - ½ g t²

t = √2y/g

b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity

x = vox t

x = v₀ₓ √2y/g

c) Speeds before touching the ground

vₓ = vox = constant

$v_{y}$ = $v_{oy}$ - gt

$v_{y}$ = 0 - g √2y/g

$v_{y}$ = - √2gy

tan θ = Vy / vx

θ = tan⁻¹ (vy / vx)

θ = tan⁻¹ (√2gy / vox)

d) The projectile is higher than the cliff because it is a horizontal launch

6 0

3 years ago

Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat

Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e $\frac{dv}{dt} =10 cm^3/s$

Consider r be the radius of the balloon.

The volume of of a sphere is

$v=\frac{4}{3} \pi r^3$

Differentiate with respect to t

$\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}$

$\Rightarrow 10 =4\pi r^2\frac{dr}{dt}$

$\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}$

The surface of area of the balloon is(S) = $4\pi r^2$

$S=4\pi r^2$

Differentiate with respect to t

$\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}$

Putting the value of $\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}$

$\Rightarrow \frac{dS}{dt} =\frac{20}{ r}$

Given that r = 5 cm

$[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}$ =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

5 0

3 years ago