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3 years ago

As a rocket rises, its kinetic energy changes

Physics

2 answers:

djverab [1.8K]3 years ago

5 0

The answer is True
hope i helped

Bond [772]3 years ago

3 0

<span>Transformed into potential energy</span>

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What is the reflectivity of a glass surface (n =1.5) in air (n = 1) at an 45° for (a) S-polarized light and (b) P-polarized ligh

Goshia [24]

Answer:

a) $R_s = 0.092$

b) $R_p = 0.085$

Explanation:

given,

n =1.5 for glass surface

n = 1 for air

incidence angle = 45°

using Fresnel equation of reflectivity of S and P polarized light

$R_s=\left | \dfrac{n_1cos\theta_i-n_2cos\theta_t}{n_1cos\theta_i+n_2cos\theta_t} \right |^2\\R_p=\left | \dfrac{n_1cos\theta_t-n_2cos\theta_i}{n_1cos\theta_t+n_2cos\theta_i} \right |^2$

using snell's law to calculate θ t

$sin \theta_t = \dfrac{n_1sin\theta_i}{n_2}=\dfrac{sin45^0}{1.5}=\dfrac{\sqrt{2}}{3}$

$cos \theta_t =\sqrt{1-sin^2\theta_t} = \dfrac{sqrt{7}}{3}$

a) $R_s=\left | \dfrac{\dfrac{1}{\sqrt{2}}-\dfrac{1.5\sqrt{7}}{3}}{\dfrac{1}{\sqrt{2}}+\dfrac{1.5\sqrt{7}}{3}} \right |^2$

$R_s = 0.092$

b) $R_p=\left | \dfrac{\dfrac{\sqrt{7}}{3}-\dfrac{1.5}{\sqrt{2}}}{\dfrac{\sqrt{7}}{3}+\dfrac{1.5}{\sqrt{2}}} \right |^2$

$R_p = 0.085$

3 0

3 years ago

You can see your image in a shiny, flat surface because the light waves are bouncing off the surface back at you. This is an exa

Alexxx [7]

I think it would be Reflection because the light in the ray reflects off of any flat or shiny object.... Hope this helps ^-^....

4 0

2 years ago

Read 2 more answers

Explain why steel is mor dense than water

natta225 [31]

Steel is more dense because it’s heavy while water is light

7 0

3 years ago

Read 2 more answers

Two particles in a high-energy accelerator experiment are approaching each other head-on, each with a speed of 0.9520c as measur

Over [174]

Answer:

the magnitude of the velocity of one particle relative to the other is 0.9988c

Explanation:

Given the data in the question;

Velocities of the two particles = 0.9520c

Using Lorentz transformation

Let relative velocity be W, so

v $_r$ = ( u + v ) / ( 1 + ( uv / c²) )

since each particle travels with the same speed,

u = v

v $_r$ = ( u + u ) / ( 1 + ( u×u / c²) )

v $_r$ = 2(0.9520c) / ( 1 + ( 0.9520c )² / c²) )

we substitute

v $_r$ = 1.904c / ( 1 + ( (0.906304 × c² ) / c²) )

v $_r$ = 1.904c / ( 1 + 0.906304 )

v $_r$ = 1.904c / 1.906304

v $_r$ = 0.9988c

Therefore, the magnitude of the velocity of one particle relative to the other is 0.9988c

5 0

3 years ago

A ball is dropped from rest at the top of a 6.10 m

natita [175]

Answer:

n = 5 approx

Explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

$\frac{v_1}{v}$ = e ( coefficient of restitution ) = $\frac{1}{\sqrt{10} }$

and

$\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }$

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

$e = \sqrt{\frac{h_1}{6.1} }$

$e = \sqrt{\frac{h_2}{h_1} }$

So on

$e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }$

$(\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}$ = .00396

Taking log on both sides

- n / 2 = log .00396

n / 2 = 2.4

n = 5 approx

3 0

2 years ago