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suter [353]
3 years ago
10

What is the angle of incidence of a wave if its angle of reflection is 20 degrees?

Physics
1 answer:
konstantin123 [22]3 years ago
3 0

<em>Your answer is:</em>

<em>c) </em><u><em>20°</em></u>

<em />

<em>∝ </em><em>Sidhdi</em>

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Vector B has x, y, and z components of 2.4,
Fed [463]

The magnitude of the vector B is 10.9

A vector is a quantity which has magnitude as well as direction and it follows vector laws of addition.

To calculate the magnitude of the vector, we have to put the square of the components of the vector along the axes under the root.

Vector B has components,

x = 2.4

y = 9.8

z = 4.1

Applying the formula,

|B| = √x²+y²+z²

|B| = √(2.4)² + (9.8)² + (4.1)²

|B| = √5.76+96.04+16.81

|B| = √118.61

|B| = 10.9

Talking about the direction the the Vector B, it will be the line joining the origin with the points (2.4,9.8,4.1)

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Which of the following statements about planetary satellites is true? all planetary satellites are as large as our moon or bigge
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3 years ago
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what equastion do you use to solve Riders in a carnival ride stand with their backs against the wall of a circular room of diame
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Answer:

μsmín = 0.1

Explanation:

  • There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
  • This friction force has a maximum value, that can be written as follows:

       F_{frmax} = \mu_{s} *F_{n} (1)

       where  μs is the coefficient of static friction, and Fn is the normal force,

       perpendicular to the wall and aiming to the center of rotation.

  • This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
  • This force has the following general expression:

       F_{c} =  m* \omega^{2} * r (2)

       where ω is the angular velocity of the riders, and r the distance to the

      center of rotation (the  radius of the circle), and m the mass of the

      riders.

      Since Fc is actually Fn, we can replace the right side of (2) in (1), as

      follows:

     F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)

  • When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

       m* g = m* \mu_{smin} * \omega^{2} * r (4)

  • (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
  • Cancelling the masses on both sides of (4), we get:

       g = \mu_{smin} * \omega^{2} * r (5)

  • Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

      60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)

  • Replacing by the givens in (5), we can solve for μsmín, as follows:

       \mu_{smin} = \frac{g}{\omega^{2} *r}  = \frac{9.8m/s2}{(6.28rad/sec)^{2} *2.5 m} =0.1 (7)

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Answer: bbbb

Explanation:

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Answer:

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