Ask question

Ask question

All categories

photoshop1234 [79]

4 years ago

13

A flat sheet of ice has a thickness of 2.20 cm. It is on top of a flat sheet of crystalline quartz that has a thickness of 1.50

cm. Light strikes the ice perpendicularly and travels through it, and then through the quartz. In the time it takes the light to travel through the two sheets, how far would it have traveled in a vacuum?

1 answer:

kolezko [41]4 years ago

6 0

Answer:

$Distance_{vaccum}=5.19cm$

Explanation:

The speed of light in these mediums shall be lower than that in vacuum thus the total time light needs to cross both the media are calculated as under

Total time = Time taken through ice + Time taken through quartz

Time taken through ice = Thickness of ice / (speed of light in ice)

$T_{ice}=\frac{2.20\times 10^{-2} \times \mu _{ice}}{V_{vaccum}}$

$T_{quartz}=\frac{1.50\times 10^{-2} \times \mu _{quartz}}{V_{vaccum}}$

Thus in the same time the it would had covered a distance of

$Distance_{vaccum}=Totaltime\times V_{vaccum}\\\\Distance_{vaccum}=10^{-2}[2.20\mu _{ice+1.50\mu _{quartz}}]$

we have

$\mu _{ice}=1.309\\\\\mu _{quartz}=1.542$

Applying values we have

$Distance_{vaccum}=10^{-2}[2.20\times 1.309+1.50\times 1.542]$

$Distance_{vaccum}=5.19cm$

You might be interested in

A skier leaves the horizontal end of a ramp with a velocity of 31.0 m/s and lands 156.3 m from the base of a ramp how high is th

BartSMP [9]

<u>Answer:</u>

The height of ramp = 124.694 m

<u>Explanation:</u>

Using second equation of motion,

$s = ut + \frac{1}{2}at^2$

From the question,

u = 31 m/s; s = 156.3 m, a=0

substituting values

$156.3 = 31\times t + 0$

t = $\frac{156.3}{31 }$

= 5.042 s

Similary, for the case of landing

t = 5.042 s; initial velocity, u =0

acceleration = acceleration due to gravity, g = 9.81 $m/s^2$

Substituting in $h = ut + \frac{1}{2}gt^2$

$h = 0 + \frac{1}{2} \times 9.81 \times (5.042)^2$

h = 124.694 m

So height of ramp = 124.694 m

3 0

3 years ago

What is Force & how many types of Force express with example <br><br><br><br>

ehidna [41]

Answer:

The force is pull or push acting on the body which tends to change its state of rest or of motion is called force.

There are two types of force:

1.Contact force

2. Non-Contact

7 0

3 years ago

A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the

zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

$E= \frac{kqr}{R^3}$

The potential at a distance can be represented as:

V(r) - V(0) = $-\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2$

V(r) - V(0) = $[\frac{qr^2}{8 \pi E_0R^3 }]$ ₀

V(r) = $-[\frac{qr^2}{8 \pi E_0R^3 }]$ ₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

= 0.56 × 10⁻² m

R = 1.29 cm

= 1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

$V(r)$ $= -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}$

$V(r)$ = -2.15 × 10⁻⁴ V

The difference between the radial distance and center can be expressed as:

V(r) - V(0) = $-\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2$

V(r) - V(0) = $[\frac{qr^2}{8 \pi E_0R^3 }]^R$

V(r) = $-\frac{qR^2}{8 \pi E_0R^3 }$

V(r) = $-\frac{q}{8 \pi E_0R }$

V(r) $= -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}$

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0

3 years ago

HELP SCIENCE WORK ON MOTION DUE!

Anna71 [15]

Huh? Wheres the question at?

7 0

3 years ago

The reactive force developed by a jet engine to push an airplane forward is called thrust, and the thrust developed by the engin

finlep [7]

The value of the thrust developed by the engine of a boeing 777 in N and Kgf are ;

i) 376616N

ii) 38430Kgf

<h3>What is force?</h3>

Force is a push or a pull. The reactive force always serve to balance the applied force. We are here asked to convert the the thrust developed by the engine of a boeing 777 which is about 85400 lbf to the following units;

i) N

ii)kgf

Thus;

1 Ib = 0.45 Kg

1 lbf = 0.45 Kg * 9.8 m/s^2 = 4.41 N

We know that;

1 lbf = 4.41 N

85400 lbf = 85400 lbf. * 4.41 N/1 lbf

= 376616N

Again;

1 lbf = 0.45 Kgf

85400 lbf = 85400 lbf * 0.45 Kgf/1 lbf

= 38430Kgf

Learn more about force:brainly.com/question/13191643

#SPJ1

3 0

1 year ago