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inysia [295]
2 years ago
12

Which of these best describes the relationship between the incident ray, the reflected ray, and the normal for a curved mirror?(

1 point)
The angle that the incident ray makes with the normal is different than the angle that the reflected ray makes with the normal. All points on a curved mirror have the same normal.
The angle that the incident ray makes with the normal is different than the angle that the reflected ray makes with the normal. All points on a curved mirror have the same normal.

The angle that the incident ray makes with the normal is the same as the angle that the reflected ray makes with the normal. Different points on a curved mirror have a different normal.
The angle that the incident ray makes with the normal is the same as the angle that the reflected ray makes with the normal. Different points on a curved mirror have a different normal.

The angle that the incident ray makes with the normal is different than the angle that the reflected ray makes with the normal. Different points on a curved mirror have a different normal.
The angle that the incident ray makes with the normal is different than the angle that the reflected ray makes with the normal. Different points on a curved mirror have a different normal.

The angle that the incident ray makes with the normal is the same as the angle that the reflected ray makes with the normal. All points on a curved mirror have the same normal.
Physics
1 answer:
lions [1.4K]2 years ago
8 0

For a curved mirror, all points have the same normal and the angle of incidence is also equal to the angle of reflection.

According to the laws of reflection, the incident ray, reflected ray and normal all lie on the same plane. For a curved mirror, the normal remains the same at all points along the curved mirror.

Again, the angle made between the incident ray and the normal is the same as the angle made between the reflected ray and the normal. Therefore, the angle of reflection is equal to the angle of incidence.

Learn more: brainly.com/question/17638582

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(URGENT) A ball rolls off a ledge. Its velocity is 7.70m/s in a horizontal direction. It falls on the floor 1.60m below the ledg
elena-s [515]

Answer:

the horizontal distance is 4.355 meters

Explanation:

The computation of the horizontal distance while travelling in the air is shown below:

Data provided in the question is as follows

Velocity = u = 7.70 m/s

H = 1.60 m

R = horizontal direction

Based on the above information

As we know that

R = u × time

where,

Time = \sqrt{\frac{2H}{g} }

So,

= 7.70\times \sqrt{\frac{2\times 1.60}{10} }

= 4.355 meters

hence, the horizontal distance is 4.355 meters

6 0
3 years ago
What element x is most likely to react to form the compound xf5?
djyliett [7]
<span>antimony. It has +3,+5,-3 so yeah. the others carbon+2,+4,-4, chlorine +1,+5,+7,-1 but -1 is the most often so it isn't Cl, calcium +2.</span>
8 0
3 years ago
in a circuit, a 10 resistor is connected in series to a parallel group containing a 60 resistor and a 5 resistor. what is the to
natka813 [3]
The 60 and the 5 in parallel have an equivalent resistance of 4.615 ohms. (rounded). The 10 in series makes it 14.615...
5 0
3 years ago
Read 2 more answers
If the distance d (in meters) traveled by an object in time t (in seconds) is given by the formula d = A + Bt^2, the SI units of
Yuliya22 [10]

Answer:

The SI units of the “A” is m (meters)

The SI units of the “B” is m/s^2

Explanation:

Given the distance = d meters.

Time taken to travel = t (seconds)

Function of the distance, d = A + Bt^2

Now we have given the above information and from the given distance function, we have to find the SI units of the A and B. Here, below are the SI units.

Thus, the SI units of the “A” is = m (meters)

The SI units of the “B” is = m/s^2

6 0
3 years ago
A hot-air balloon is rising upward with a constant speed of 2.51 m/s. When the balloon is 3.16 m above the ground, the balloonis
icang [17]

Answer:

  t = 1.099 s

Explanation:

given,

constant speed = 2.51 m/s

height of balloon above ground = 3.16 m

time elapsed before it hit the ground = ?

Applying equation of motion to the compass

y = u t + \dfrac{1}{2}at^2

-3.16 = 2.51 t + \dfrac{1}{2}\times (-9.8)t^2

4.9 t^2 - 2.51 t - 3.16 = 0

using quadratic formula to solve the equation

t = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}

t = \dfrac{-(-2.51)\pm \sqrt{2.51^2-4(4.9)(-3.16)}}{2\times 4.9}

  t = 1.099 s, -0.586 s

hence, the time elapses before the compass hit the ground is equal to 1.099 s.

8 0
3 years ago
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