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Is a quantum of light or a single packet/particle of light at a given wavelength.

Yanka [14]

Photon is a quantum of light or a single packet/particle of light at a given wavelength.

Answer: Option B

<u>Explanation: </u>

It is known that light has dual nature of wave as well as particles. Light waves can behave in wave nature as well as in particle nature depending upon the situation. So the light waves are assumed in different views to easily understand the nature of light waves.

There are several models proposed to simplify the nature of light. Among the several assumptions, one of the most prominent observations are that light waves or quantum of light are termed as photons which are made up of single packet/particles of light in a given wavelength.

4 0

2 years ago

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$

and re-arranging the equation, we find

$v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s$

7 0

2 years ago

A constant force is applied to an object, causing the object to accelerate at 10 m/s^2. What will the acceleration be if: a) The

Liula [17]

What you need to know is that the force is

F=ma

The force is the product of mass and acceleration

this means that the acceleration is

a=F/m

a) The force is halved?
this means that f will be $\frac{F}{2}$ now:

a= $\frac{F}{2m}$

So the accelaration will also he halved (it's the original acceleratation divided by 2)

b) The object's mass is halved?
a= $\frac{F}{m/2}$ =a= $\frac{F2}{m}$

which is the original acceleration times two!! so it will double

c) The force and the object's mass are both halved?
now we have

a= $\frac{F/2}{m/2}$ =a= $\frac{2F}{2m}$ =a= $\frac{F}{m}$

so they will cancel each other out and the acceleration will stay the same!

5 0

2 years ago

Multiply the following three numbers and report your answer to the correct number of significant figures: 0.020cm x 50cm x 11.1c

Neko [114]

Answer:11.1

Explanation:

Three significant figures

5 0

3 years ago

Earth travels fastest in January and slowest in July. What is the most likely explanation for this?

Keith_Richards [23]

Answer:

Earth is nearest the Sun in July and farthest away in July.

Explanation:

3 0

3 years ago

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