Answer:
d = 1.55 * 10⁻⁶ m
Explanation:
To calculate the distance between the adjacent grooves of the CD, use the formula,
..........(1)
The fringe number, m = 1 since it is a first order maximum
The wavelength of the green laser pointer,
= 532 nm = 532 * 10⁻⁹ m
Distance between the central maximum and the first order maximum = 1.1 m
Distance between the screen and the CD = 3 m
= Angle between the incident light and the diffracted light
From the setup shown in the attachment, it is a right angled triangle in which


Putting all appropriate values into equation (1)

3260÷4=815 which is you average seed
The concept required to solve this problem is related to the wavelength.
The wavelength can be defined as the distance between two positive crests of a wave.
The waves are in phase, then the first distance is

And out of the phase when

Thus the wavelength is

Here,
Wavelength
If we rearrange the equation to find it, we will have



Therefore the wavelength of the sound is 20cm.
Answer:
Explanation:
Given
magnitude of vector=8.7 m/s
Direction 
Therefore vector x component is 
=-2.104
i.e. it is acting in negative x axis
vector y component
Therefore it lies in negative y axis
Judging vector x and y component
It lies in third quadrant
Answer:
<em>D. The acceleration after it leaves the hand is 10 m/s/s downwards
</em>
Explanation:
<u>Vertical Throw
</u>
When an object is thrown upwards, it describes a special type of motion ruled only by gravity.
When the ball is launched, it has its maximum speed upwards. The acceleration of gravity is always the same because it's a constant value near our planet's surface. The object starts to lose speed since the acceleration of gravity is pointed downwards and makes the object stop in the mid-air at its maximum height, where the speed is zero. Then, the object starts to fall and regain speed, this time downwards until it reaches back the launching point at the very same speed it was launched, but in the opposite direction.
The time it takes to reach its maximum height is the same it takes to return to the catching point, 2 seconds later.
With all these concepts in mind, we state that:
<em>D. The acceleration after it leaves the hand is 10 m/s/s downwards </em>
The other options are not correct because:
A. The acceleration is never upwards
B. The acceleration is never 0
C. Both times are equal