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denis-greek [22]
2 years ago
5

What is the speed of light (in m/s) in air? (Enter your answer to at least four significant figures. Assume the speed of light i

n a vacuum is 2.997 ✕ 108 m/s.) m/s What is the speed of light (in m/s) in polystyrene? m/s
Physics
1 answer:
jenyasd209 [6]2 years ago
8 0

Answer:

The speed of light in air is 2.996x10⁸ m/s, and polystyrene is 1.873x10⁸ m/s.

Explanation:

To find the speed of light in air and in polystyrene we need to use the following equation:

c_{m} = \frac{c}{n}

Where:

c_{m}: is the speed of light in the medium

n: is the refractive index of the medium

In air:

c_{a} = \frac{c}{n_{a}} = \frac{2.997 \cdot 10^{8} m/s}{1.0003} = 2.996 \cdot 10^{8} m/s

In polystyrene:

c_{p} = \frac{c}{n_{p}} = \frac{2.997 \cdot 10^{8} m/s}{1.6} = 1.873 \cdot 10^{8} m/s  

Therefore, the speed of light in air is 2.996x10⁸ m/s, and polystyrene is 1.873x10⁸ m/s.

I hope it helps you!

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Jules is conducting an experiment involving friction. He is measuring the temperature of various objects and surfaces after quic
Semenov [28]

Answer:

The correct prediction will be:  

The temperature of the surface of the ball bearing when rubbed over glass will be the least.

The incorrect prediction:

The tennis ball over the linoleum floor will have no friction, as the temperatures will not change.

6 0
2 years ago
The Orion nebula is one of the brightest diffuse nebulae in the sky (look for it in the winter, just below the three bright star
Oxana [17]

Answer:

T=183.21K

Explanation:

We have to take into account that the system is a ideal gas. Hence, we have the expression

PV=nRT

where P is the pressure, V is the volume, n is the number of moles, T is the temperature and R is the ideal gas constant.

Thus, it is necessary to calculate n and V

V is the volume of a sphere

V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi (5.9*10^{15}m)^3=8.602*10^{47}m^3

V=8.86*10^{50}L

and for n

n=\frac{(4000M_s)/(2*mH)}{6.022*10^{23}mol^{-1}}=3.95*10^{36}mol

Hence, we have (1 Pa = 9.85*10^{-9}atm)

T=\frac{PV}{nR}=\frac{(6.8*10^{-9}*9.85*10^{-6}atm)(8.86*10^{50}L)}{(0.0820\frac{atm*L}{mol*K})(3.95*10^{36}mol)}\\\\T=183.21K

hope this helps!!

8 0
2 years ago
In a movie, Tarzan evades his captors by hiding under water for many minutes while breathing through a long, thin reed. Assume t
gladu [14]

Answer: 0.98m

Explanation:

P = -74 mm Hg = 9605 Pa = 9709N/m^2

= 9605 kg m/s^2/m^2

density of water: rho = 1 g/cc = 1 (10^-3 kg)/(10^-2 m)^-3 = 1000 kg/m^3

Pressure equation: P = rho g h

h = P/(rho g)

h = (9605 kg/m/s^2) / (1000 kg/m^3) / (9.8 m/s^2)

h = 0.98 m

0.98m is the maximum depth he could have been.

8 0
2 years ago
Which of the following most accurately represents John Dalton’s model of the atom? A. a tiny, solid sphere with an unpredictable
aleksley [76]
A and c are the answersss
6 0
2 years ago
In the late 19th century, great interest was directed toward the study of electrical discharges in gases and the nature of so-ca
gizmo_the_mogwai [7]

Answer:

A) He finds the same value of q / m for different materials , B)      y = ½ (q / m) E L² / v₀ₓ² , C) v = E / B , D)   B = 2.13 10⁻⁶ T, E) For the first part I have two off-center points., For the second part I can center one point but the other is off center

Therefore the third statement is correct

Explanation:

Part A

Thomson's experiments are the first proof that the atoms that until now were considered indivisible were constituted by different elements, in these experiments Thomson himself the ratio q / m of several cathodes and always found the same value, which allowed to establish that In atoms there are two types of particles, some of which are mobile and others are still.

When examining his statements the correct one is: He finds the same value of q / m for different materials

Part B

For this part let's use Newton's second law

        F = ma

        q E = m a

        a = q / m E

We use the kinematic relationship

          y = voy t - ½ to t²

          x = v₀ₓ t

The initial vertical velocity of electrons is zero

           y = ½ a (x / v₀ₓ)²

We replace

           y = ½ (q / m) E L² / v₀ₓ²

Part C

If there is no deflection, the electric and magnetic forces are the same and in the opposite direction

         Fm = Fe

         q v B = q E

          v = E / B

Part D

       

        We replace

        y = ½ (q / m) E L² / (E / B)²

         y = ½ (q / m) L² B² / E

If we do not want any deflection the magnetic field has to return the electrons the amount that they lower y = -4.12 cm

      -4.12 10⁻² = ½ q / m 0.12² B² / 1.1 10³

       -16.97 10⁻⁴ = 6.54 10⁻³ B² q / m

      B² = -2.59 10⁻¹ q / m

      q / m = -1.758 10¹¹ C/ kg

      B = √ 0.259 1.758 10¹¹ = √ 4.55 10⁻¹²

      B = 2.13 10⁻⁶ T

Part E

As the charge that the two particles is different

For the first part I have two off-center points.

For the second part I can center one point but the other is off center

Therefore the third statement is correct

8 0
2 years ago
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