PHYSICS COLLEGE

What occurs when a light wave enters a substance and its speed suddenly slowsdown?
refraction
Olightening
Oreflection
the vacuum effect

Answers

Answer 1
Answer:

Final answer:

Refraction occurs when a light wave enters a substance and its speed suddenly slows down.


Explanation:

When a light wave enters a substance and its speed suddenly slows down, it undergoes a phenomenon known as refraction. Refraction occurs due to the change in speed and direction of light as it passes from one medium to another with a different refractive index. The change in speed causes the light wave to bend, resulting in a change in its path.


Learn more about Refraction here:

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What is the force of gravity acting on a 1-kg m mass? (g = 9.8 m/s ^ 2)

Answers

Answer: Use this F=Ma.

Explanation: So your answer will be

F=1 Kg+9.8 ms-2

So the answer will be

F=9.8N

How'd I do this?

I just used Newton's second law of motion.

I'll also put the derivation just in case.

Applied force α (Not its alpha, proportionality symbol) change in momentum

Δp α p final- p initial

Δp α mv-mu (v=final velocity, u=initial velocity and p=v*m)

or then

F α m(v-u)/t

So, as we know v=final velocity & u= initial velocity and v-u/t =a.

So F α ma, we now remove the proportionality symbol so we'll add a proportionality constant to make the RHS & LHS equal.

So, F=kma (where k is the proportionality constant)

k is 1 so you can ignore it.

So, our equation becomes F=ma

A 2.00 kg cart on a frictionless track is pulled by force of 3.00 N. What is the acceleration of the cart?

Answers

1.5 will be its acceleration

If the_____of a wave increases, its frequency must decrease.a. period
b. energy
c. amplitude
d. velocity

Answers

Answer:wrong. Jts not velocity. Its period.

Explanation:

Took the test

A screen is placed 60.0 cm from a single slit, which is illuminated with light of wavelength 690 nm. If the distance between the first and third minima in the diffraction pattern is 3.10 mm, what is the width of the slit?

Answers

Answer:

slit width, b = 0.2671 mm

Given:

distance of screen from the slit, x = 60.0 cm

wavelength of light, \lambda = 690 nm = 690* 10^(- 9) m

distance between 1st and 3rd minima, t = 3.10 mm = 3.10* 10^(-3) m

Solution:

Calculation of the distance between 1st and 3rd minima:

t = ((3 - 1)\lambda x)/(b)

3.10* 10^(- 3) = (2* 690* 10^(-9)* 60.0* 10^(-2))/(b)

b = 0.2671 mm

slit width, b = 0.2671 mm

An airplane is traveling 835 km/h in a direction 41.5 ∘ west of north. Find the components of the velocity vector in the northerly and westerly directions. How far north and how far west has the plane traveled after 2.20 h ?

Answers

I assume the graph is looking like in the picture bellow.

North component:
cos(41.5) * 835 = 625.37 km/h

West component of speed:
sin(41.5) * 835 = 553.29 km/h

After 2.2 hours plane will fly:
2.2*625.37 = 1375.81 km north
2.2*553.29 = 1217.23 km  west

Final answer:

To find the components of the velocity vector, you can use trigonometry. The north component is calculated using the sine function and the west component is calculated using the cosine function. After 2.20 hours, the distance traveled north and west can be found by multiplying the velocity components by the time.

Explanation:

To find the components of the velocity vector in the northerly and westerly directions, we can use trigonometry. The velocity vector is 835 km/h and is traveling in a direction 41.5° west of north. To find the north component, we can use the sine function: North component = velocity * sin(angle). To find the west component, we can use the cosine function: West component = velocity * cos(angle).

After 2.20 hours, we can find the distance traveled north and west by multiplying the velocity components by the time: Distance north = North component * time and Distance west = West component * time.

Let's calculate the values:

  1. North component = 835 km/h * sin(41.5°)
  2. West component = 835 km/h * cos(41.5°)
  3. Distance north = North component * 2.20 h
  4. Distance west = West component * 2.20 h

Learn more about Velocity components here:

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An electron in a vacuum chamber is fired with a speed of 7400 km/s toward a large, uniformly charged plate 75 cm away. The electron reaches a closest distance of 15 cm before being repelled. What is the plate's surface charge density?

Answers

Answer:

2.29e-9C/m²

Explanation:

Using E = σ/ε₀ means the force on the electron is F = eE = eσ/ε₀.

The work done on the electron is W = Fd = deσ/ε₀. This equals the kinetic energy lost, ½mv².

½mv² = deσ/ε₀

d = 75cm – 15cm = 60cm = 0.6m

σ = mv²ε₀/(2de)

. .= 9.11e-31 * (7.4e6)² * 8.85e-12 / (2 * 0.6 * 1.6e-19)

. .= 2.29e-9 C/m² (i.e. 2.29x10^-9 C/m²)
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