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marin [14]
3 years ago
6

A wave is incident on the surface of a mirror at an angle of 41° with the normal. What can you say about its angle of reflection

?
a.)It is less than the angle of incidence.

b.)It is equal to the angle of incidence.

c.)It is greater than the angle of incidence.

d.)It is zero, as it reflects along the normal
Physics
2 answers:
Oksana_A [137]3 years ago
4 0

It's angle of reflection must be 41 degrees

we know, by the first law of reflection that angle of incidence is always equal to angle of reflection..........

Viktor [21]3 years ago
4 0

Answer:

b.)It is equal to the angle of incidence.

Explanation:

As we know that as per law of reflection of light angle of incidence with respect to the normal must be equal to the angle of reflection with normal in other direction.

So we can say that

angle of incidence = angle of reflection

so here we can say that the correct answer would be

b.)It is equal to the angle of incidence.

so the light ray will reflect at the same angle as that of angle of incidence given in it

so it must be

\theta_i = \theta_r = 41^o

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The barricade at the end of a subway line has a large spring designed to compress 2.00 m when stopping a 1.10 ✕ 105 kg train mov
Mrac [35]

Answer:

(a) k = 1684.38 N/m = 1.684 KN/m

(b) Vi = 0.105 m/s

(c) F = 1010.62 N = 1.01 KN

Explanation:

(a)

First, we find the deceleration of the car. For that purpose we use 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = ?

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = 0.35 m/s

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Therefore,

a = [(0 m/s)² - (0.35 m/s)²]/(2)(2 m)

a = 0.0306 m/s²

Now, we calculate the force applied on spring by train:

F = ma

F = (1.1 x 10⁵ kg)(0.0306 m/s²)

F = 3368.75 N

Now, for force constant, we use Hooke's Law:

F = kΔx

where,

k = Force Constant = ?

Δx = Compression = 2 m

Therefore.

3368.75 N = k(2 m)

k = (3368.75 N)/(2 m)

<u>k = 1684.38 N/m = 1.684 KN/m</u>

<u></u>

<u>(</u>c<u>)</u>

Applying Hooke's Law with:

Δx  = 0.6 m

F = (1684.38 N/m)(0.6 m)

<u>F = 1010.62 N = 1.01 KN</u>

<u></u>

(b)

Now, the acceleration required for this force is:

F = ma

1010.62 N = (1.1 kg)a

a = 1010.62 N/1.1 x 10⁵ kg

a = 0.0092 m/s²

Now, we find initial velocity of train by using 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = -0.0092 m/s² (negative sign due to deceleration)

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = ?

s = distance covered by train before stopping = 0.6 m

Therefore,

-0.0092 m/s² = [(0 m/s)² - Vi²]/(2)(0.6 m)

Vi = √(0.0092 m/s²)(1.2 m)

<u>Vi = 0.105 m/s</u>

4 0
4 years ago
A 1500kg car traveling at 25m/s skids to a stop. The force of friction between the tires and the road is 10500N. How far does th
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Given parameters:

Mass of the car = 1500kg

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Unknown:

Distance moved by the car after brake is applied = ?

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The frictional force is a force that  opposes motion of a body.

To solve this problem, we need to find the acceleration of the car. After this, we apply the appropriate motion equation to solve the problem.

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the negative sign is because the frictional force is in the opposite direction

m is the mass of the car

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Now using;

   V² = U² + 2as

   V is the final velocity

   U is the initial velocity

   a is the acceleration

   s is the distance moved

  0² = 25² + 2 x 7 x s

  0 = 625 - 14s

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learn more:

Velocity problems brainly.com/question/10932946

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