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Ymorist [56]
1 year ago
15

Which phenomena support only the wave theory of light? select 2 options. reflection refraction diffraction interference photoele

ctric effect
Physics
1 answer:
attashe74 [19]1 year ago
3 0

Interference and diffraction are the phenomena that support only the wave theory of light. Options 2 and 3 are correct.

<h3 /><h3>What is the interference of waves?</h3>

The result of two or more wave trains flowing in opposite directions on a crossing or coinciding pathways. This phenomenon is known as the interference of waves.

The phenomenon of interference occurs when two wave pulses are traveling along a string toward each other.

The light wave hypothesis states that light behaves like a wave. Since light is an electromagnetic wave, it may be transmitted without a physical medium.

Light has magnetic and electric fields, much like electromagnetic waves do.

Transverse waves, such as those seen in light waves, oscillate in the same direction as the wave's path. A wave of light may experience interference as well as diffraction as a result of these properties.

All of the remaining options are the light phenomenon.

Hence, options 2 and 3 are correct.

To learn more about the interference of waves refer to the link;

brainly.com/question/16098226

#SPJ1

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I need help with questions 6-8. Thank you!! Image is attached
Digiron [165]

6) b) 2.7 m/s

7) b) DCA

8) b) B

Explanation:

6)

In a displacement-time plot, the slope of the line is given by

m=\frac{\Delta y}{\Delta x}

where

\Delta y is the change in the y-variable, so it is the displacement

\Delta x is the change in the x-variable, so it is the time elapsed

So, the slope of the line in a displacement-time plot corresponds to the velocity:

v=m=\frac{d}{t}

Therefore, to find the velocity of the object, we have to estimate the slope of its curve.

To estimate the velocity of object B, we have to estimate the slope of the line tangent to curve B at 10 seconds.

By doing an estimate by eye, we see that the displacement of object B changes from -10 m to 0 m when time increases from about 8 s to 12 s, so the velocity is about:

v=\frac{0-(-10)}{12-8}\sim 2.5 m/s

So the closest option is b) 2.7 m/s.

7)

As we said in part A, the velocity of each object is given by the slope of each curve.

Therefore:

- The steeper the curve, the higher the velocity

- The less steep the curve, the lower the velocity

From the graph, we observe that, among A, C and D:

- Curve D has the largest slope (in absolute value), so object D has the largest magnitude of the velocity

- Curve C is less steep than curve C, so object C has the second largest magnitude of velocity

- Curve A is flat, so the slope is zero, so its velocity is zero

So, from greatest magnitude to lowest magnitude of velocity, we have:

b) DCA

8)

In the graph, the overall displacement of each object is given by the change in the y-variable, \Delta y.

This means that the object with largest displacement is the object whose curve has the largest variation in y.

From the graph, we see that:

- Object b has the largest variation in y,  from -15 m to 30 m, so

\Delta y=30-(-15)=45 m

- Then, object D has the second largest displacement (in magnitude), from -15 m to 25 m,

|\Delta y| = 25 -(-15)=40 m

Finally, object C has displacement

\Delta y = 20-(-5)=25 m

While object A has displacement zero. Therefore, the correct option is

b) B

3 0
3 years ago
QUESTION 3
Alisiya [41]

The force of frictions is opposed to relative motion.

The acceleration of the crate is approximately <u>2.937 m/s²</u>.

Reason:

The given parameters are;

The mass of the wood, m = 100 kg

The force which can move the wood, F = 588 N

Wood on wood static friction, \mu_s = 0.5

Wood on wood kinetic friction, \mu_k = 0.3

Solution;

The force of friction, F_f, acting when the crate is moving is given as

follows;

F_f = m × g × \mu_k

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

Therefore, we have;

F_f = 100 × 9.81 × 0.3 = 294.3

The force of friction, F_f = 294.3 N

The force with which the crate moves, F = 588 - 294.3 = 293.7

The force with which the crate moves, F = 293.7 N

Force = Mass, m × Acceleration, a

a = \dfrac{F}{m}

Therefore;

a = \dfrac{293.7 \ N}{100 \ kg} = 2.937

The acceleration of the crate, a ≈ <u>2.937 m/s²</u>.

Learn more about friction here:

brainly.com/question/94428

8 0
2 years ago
What is the path of a projectile called? Friction Track Trajectory acceleration
sergiy2304 [10]
It is trajectory acceleration. A friction track is a device to study motion in low friction environments, I believe. Does this help?
3 0
3 years ago
Read 2 more answers
True or False: A change of 1 pH unit represents a tenfold change in the<br> acidity of the solution.
alisha [4.7K]

Answer:

I was also going to ask same question edited:ok i found its true

7 0
3 years ago
A 75.0kg bicyclist (including the bicycle) is pedaling to the right, causing her speed to increase at a rate of 2.20m/s^2, despi
malfutka [58]

1) 4 forces

2) 165 N

3) 225 N

Explanation:

1)

There are in total 4 forces acting on the bicylist:

- The gravitational force on the byciclist, acting vertically downward, of magnitude mg, where m is the mass of the bicyclist and g is the acceleration due to gravity

- The normal force exerted by the floor on the bicyclist and the bike, N, vertically upward, and of same magnitude as the gravitational force

- The force of push F, acting horizontally forward, given by the push exerted by the bicylist on the pedals

- The air drag, R, of magnitude R = 60.0 N, acting horizontally backward, in the direction opposite to the motion of the bicyclist

2)

The magnitude of the net force on the bicyclist can be calculated by considering separately the two directions.

- Along the vertical direction, we have the gravitational force (downward) and the normal force (upward); these two forces are equal in magnitude, since the acceleration of the bicyclist along this direction is zero, therefore the net force in this direction is zero.

- Along the horizontal direction, the two forces (forward force of push and air drag) are balanced, since the acceleration is non-zero, so we can use Newton's second law of motion to find the net force on the bicylist:

F_{net}=ma

where

F_{net} is the net force

m = 75.0 kg is the mass of the bicyclist

a=2.20 m/s^2 is its acceleration

Solving, we find the net force:

F_{net}=(75.0)(2.20)=165 N

3)

In this part, we basically want to find the forward force of push, F.

We can rewrite the net force acting on the bicyclist as

F_{net}=F-R

where:

F is the forward force of push

R is the air drag

We know that:

F_{net}=165 N is the net force on the bicyclist

R = 60.0 N is the magnitude of the air drag

Therefore, by re-arranging the equation, we can find the force generated by the bicylicst by pedaling:

F=F_{net}+R=165+60=225 N

6 0
3 years ago
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