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weqwewe [10]
3 years ago
5

The water is reflecting light, Is this specular or diffuse reflection? explain your answer​

Physics
1 answer:
FinnZ [79.3K]3 years ago
8 0
Yes the answer is true
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A bicycle on the top of the hill has the highest potential energy, and when the bike goes down, it transfers to kinetic because it is moving

Explanation:

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How are gas particles arranged in a container?
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3. Tightly Packed

Explanation:

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According to diagonal rule, the orbital with lowest energy in the given following is ________.
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The orbital with the lowest energy is 3s.
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3 years ago
Suppose that a wind is blowing in the direction S45°E at a speed of 30 km/h. A pilot is steering a plane in the direction N60°E
Kay [80]

Answer:

The true course: 40.29^\circ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

  • V_w = velocity of wind = 30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h
  • V_p = velocity of plane in still air = 100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h

Assume:

  • V_r = resultant velocity of the plane
  • \theta = direction of the plane with the east

Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h

Let us find the direction of this resultant velocity with respect to east direction:

\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ

This means the the true course of the plane is in the direction of 40.29^\circ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h

Hence, the ground speed of the plane is 96.68 km/h.

5 0
3 years ago
. A 2.0-kg block is on a perfectly smooth ramp that makes an angle of 30° with the horizontal. (a) What is the block’s accelerat
AfilCa [17]

Answer:

a) a = 4.9 m / s²,  N = 16.97 N   and b)   F = 9.8 N

Explanation:

a) For this exercise we will use Newton's second law, we write a reference system with the x axis parallel to the plane, see attached, in this system the only force we have to break down is weight, let's use trigonometry

        sin 30 = Wx / W

        cos 30 = Wy / W

        Wx = W sin30

        Wy = W cos 30

Let's write the equations on each axis

X axis

        Wx = ma

Y Axis  

       N- Wy = 0

       N = Wy = mg cos 30

       N = 2.0 9.8 cos 30

       N = 16.97 N

We calculate the acceleration

       a = Wx / m

       a = mg sin 30 / m

       a = g sin 30

       a =9.8 sin 30

       a = 4.9 m / s²

b) For the block to move with constant speed, the acceleration must be zero, so the force applied must be equal to the weight component

      F -Wx = 0

      F = Wx

      F = m g sin 30

      F = 2.0 9.8 sin 30

      F = 9.8 N

5 0
3 years ago
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