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liubo4ka [24]
2 years ago
12

You are moving at a speed 2/3 c toward Randy when shines a light toward you. At what speed do you see the light approaching you

Physics
1 answer:
lord [1]2 years ago
5 0

Answer:

The speed of light will be c=3x10^8m/s

Explanation:

This is the same as the speed of light because your speed does not affecttje speed of light so you will see the light approaching you at the same speed of light c

You might be interested in
This table shows Wayne’s weight on four different planets. Planet Wayne’s weight (pounds) Mars 53 Neptune 159 Venus 128 Jupiter
oee [108]
The correct order is (in decreasing order of gravity strength)
Jupiter - Neptune - Venus - Mars

In fact, Wayne's weight on each planet is given by
W=mg
where m is Wayne's mass, which is a constant value, and g is the gravity strength at the surface of the planet. Therefore, the Wayne's weight W on each planet is directly proportional to the gravity strength of that planet: so the planet with the strongest gravity is the one where Wayne's weight is the greatest (Jupiter, 333 pounds), followed by Neptune (159),  Venus (128) and Mars (53).
8 0
2 years ago
La luz pasa del medio A al medio B formando un ángulo de 35° con la frontera horizontal entre ambos. Si el ángulo de refracción
zaharov [31]

Answer:

Índice de refracción entre los dos medios = 1,43

Refractive index between the two media = 1.43

Explanation:

El índice de refracción entre dos medios se explica mejor entendiendo primero la refracción.

Cuando las olas se mueven de un medio a otro, a menudo experimentan un cambio de dirección con respecto al medio en el que viajan.

Por lo tanto, el índice de refracción se expresa como el seno del ángulo de incidencia dividido por el seno del ángulo de refracción.

El seno del ángulo de incidencia y la refracción utilizados en esta fórmula de índice de refracción se miden respectivamente con respecto a la vertical.

En esta pregunta Ángulo de incidencia = 35° a la horizontal = (90° - 35°) a la vertical = 55° a la vertical.

Ángulo de refracción = 35°

Índice de refracción entre los dos medios

= (Sin 55°) ÷ (Sin 35°)

= 0.8192 ÷ 0.5736

= 1.428 = 1.43 a 2 d.p.

¡¡¡Espero que esto ayude!!!

English Translation

The light passes from medium A to medium B at an angle of 35 ° with the horizontal border between the two. If the angle of refraction is also 35 °, what is the relative refractive index between the two media?

Solution

The refractive index between two media is best explained by first understanding refraction.

When waves move from one medium to another, they often experience a change in direction with respect to the medium in which they are travelling.

Hence, refractive index is expressed as the sine of angle of incidence dibided by the sine of angle of refraction.

The sine of angle of incidence and refraction used in this refractive index formula are both respectively measured with respect to the vertical.

In this question,

Angle of incidence = 35° to the horizontal = (90° - 35°) to the vertical = 55° to the vertical.

Angle of refraction = 35°

Refractive index between the two media

= (Sin 55°) ÷ (Sin 35°)

= 0.8192 ÷ 0.5736

= 1.428 = 1.43 to 2 d.p.

Hope this Helps!!!

3 0
2 years ago
The movie "The Gods Must Be Crazy" begins with a pilot dropping a bottle out of an airplane. A surprised native below, who think
Sergio039 [100]

Answer:

⇔⇔⇔↑∑∑∩∅¬⊕║⊇↔∴∉∵

Explanation:

8 0
2 years ago
: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at
ser-zykov [4K]

Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

a

The  expression is   F_{No} =   A [P_{No} - \frac{P_{sea}}{2}]

b

F_{No}= 7771.125 \ N

c

 F_p = 2.2*10^{6} N

d

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

          The altitude of container in Denver is  d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m

           The surface area of the container lid is A = 0.0155m^2

           The altitude of container in New Orleans  is sea-level

           The air pressure in Denver is  P_D = 79000 \ Pa

            The air pressure in new Orleans is P_{ro} = 100250 \ Pa

Generally force is mathematically represented as

            F_{No} = \Delta P A

  So we are told the pressure inside is  is half the pressure the at sea level so the  the pressure acting on the container would

   The  pressure at sea level is a constant with a  value of  

               P_{sea} = 101000 Pa

So the \Delta P which is the difference in pressure within and outside the container is  

           \Delta P = P_{No} - \frac{P_{sea}}{2}

Therefore

                F_{No} =   A [P_{No} - \frac{P_{sea}}{2}]

Now substituting values

                F_{No} =   0.0155 [100250 - \frac{101000}{2}]

                       F_{No}= 7771.125 \ N

The force to remove the lid in Denver is  

           F_p = \Delta P_d A

So we are told the pressure inside is  is half the pressure the at sea level so the  the pressure acting on the container would

 The  pressure at sea level is a constant with a  value of  

               P_{sea} = 101000 Pa    

 At  sea level the air pressure in Denver is mathematically represented as

              P_D = \rho g h

     =>     g = \frac{P_D}{\rho h}      

Let height at sea level is h = 1

  The air pressure at height d_D

             P_d__{D}} = \rho gd_D

    =>     g = \frac{P_d_D}{\rho d_D}

  Equating the both

                 \frac{P_D}{\rho h}  = \frac{P_d_D}{\rho d_D}

                 P_d_D =  P_D * d_D

Substituting value  

                   P_d__{D}} = 1828.2 * 79000

                    P_d__{D}} = 1.445*10^{8} Pa

    So

              \Delta P_d  = P_{d} _D - \frac{P_{sea}}{2}

=>          \Delta P_d  = 1.445 *10^{8} - \frac{101000}{2}    

                        \Delta P_d = 1.44*10^{8}Pa

  So

               F_p = \Delta P_d A

                  = 1.44*10^8 * 0.0155

              F_p = 2.2*10^{6} N

               

                 

             

             

6 0
3 years ago
Which formula is used to find fluctuation of the shape of body
Sladkaya [172]

Answer:

varn=n1+1ehvkT–1

Explanation:

This is Einstein's equation.

5 0
2 years ago
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