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4 years ago

For a plane mirror,how is the object s related to the distance s?

1 answer:

3 0

Light is a very complex phenomenon, but in many situations its behavior can be understood with a simple model based on rays and wave fronts. A ray is a thin beam of light that travels in a straight line. A wave front is the line (not necessarily straight) or surface connecting all the light that left a source at the same time. For a source like the Sun, rays radiate out in all directions; the wave fronts are spheres centered on the Sun. If the source is a long way away, the wave fronts can be treated as parallel lines.

Rays and wave fronts can generally be used to represent light when the light is interacting with objects that are much larger than the wavelength of light, which is about 500 nm. In particular, we'll use rays and wave fronts to analyze how light interacts with mirrors and lenses.

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What is magnum effect?

iragen [17]

An observable phenomenon that is commonly associated with a spinning object moving through a fluid.

5 0

3 years ago

Two forces,one of 12 N and another of 24 N,act on a body in such a way that they make an angle of 90degree with each other.Find

jok3333 [9.3K]

Answer:

26.83 N.

Explanation:

If the angle between two vector is 90°, to get the resultant, we use Pythagoras theorem.

a² = b²+c²......................... Equation 1

Where a = R = Resultant, b = 12 N, c = 24 N.

Substitute these values into equation 1

R² = 12²+24²

R² = 144+576

R² = 720

√R² = √720

R = 26.83 N.

Hence, the result of the two force is 26.83 N.

4 0

3 years ago

As viewed from above in this picture, what direction will the current be in the coil of wire that will cause the loop to rotate

Gala2k [10]

Answer:

When viewed from above, the current in the coil should point towards the top-right corner of the picture.

Explanation:

The current in this coil have only two possible directions: clockwise or counter-clockwise. However, since the diagram shows the coil from above, not from a cross-section, just saying clockwise or counter-clockwise might be ambiguous. The statement that the current is directed towards the top-right corner of the picture is equivalent to saying that when viewed from the lower-right corner of this diagram, the current in the coil is moving clockwise.

Note that at the center of this picture, the current is parallel to the magnetic field- there will be no force on the coil at that position. On the other hand, (also when viewed from above,) at the top-right corner and the lower-left corner of the coil, the current in the coil will be perpendicular to the magnetic field. That's where the force on the coil will be the strongest.

With that in mind, apply the right-hand rule to find the direction of the force on the coil in each of the two possibilities.

Assume that when viewed from above, the current is flowing towards the top-right corner of the picture. Consider the wire near the top-right corner of this coil (as viewed above on this picture.) The current will be going into the picture into the magnetic field. By the right-hand rule, the current on the wire near that point should be pointing towards the bottom of this picture. (Point fingers on the right hand in the direction of the current $I$ . Rotate the right hand such that when curling the fingers, they point in the direction of the magnetic field $B$ . The direction of the right thumb should now point in the direction of the force on the wire $F$ .)

Based on the same assumption, the current in the wires near the bottom left corner of this coil will be pointing out of the picture. By the right hand rule, the magnetic force on the coil in that region should be pointing towards the top of this picture. Combing these two forces, the coil would indeed be rotating around the center of this picture in the direction shown in the diagram.

It can also be shown that if the current points towards the bottom left corner of the picture when viewed from above, the coil will be rotating about the center of this picture in the opposite direction.

7 0

3 years ago

A force gives a 5.0 kg object an acceleration of 2.0 m/s 2. The same force would give a 20 kg object an acceleration of _____. 0

Oksi-84 [34.3K]

m = 5 kg

a = 2 m/s²

to find the force that accelerates the 4 kg object @ 2 m/s²

F = ma = 5 kg x 2 m/s² = 10 N

To find what acceleration 10 N would give a 20 kg object

a = F/m = 10 N/20 kg = 0.5 m/s

6 0

3 years ago

Read 2 more answers

Calculate the kinetic energy of a 0.032 kg ball as it leaves a hand to be thrown upwards at 6.2 m/s

AnnZ [28]

Answer:

The ball will have a kinetic energy of 0.615 Joules.

Explanation:

Use the kinetic energy formula

$E_k = \frac{1}{2}mv^2 = \frac{1}{2}0.032kg\cdot 6.2^2 \frac{m^2}{s^2}= 0.615J$

The kinetic energy at the moment of leaving the hand will be 0.615 Joules. (From there on, as it ball is traveling upwards, this energy will be gradually traded off with potential energy until the ball's velocity becomes zero at the apex of the flight)

3 0

3 years ago