Answer:
The magnitude of angular acceleration is
.
Explanation:
Given that,
Initial angular velocity, 
When it switched off, it comes o rest, 
Number of revolution, 
We need to find the magnitude of angular acceleration. It can be calculated using third equation of rotational kinematics as :
So, the magnitude of angular acceleration is
. Hence, this is the required solution.
The wavelength of the first harmonic of the standing wave is 2L.
<h3>What is a standing wave?</h3>
A standing wave is one in which the obvious points remain fixed as the vibration continues. A standing wave occurs in a wind instrument such as a trumpet, saxophone etc.
We know from the formula of the first harmonic that the wavelength of the first harmonic of the standing wave is 2L.
Learn more about standing wave:brainly.com/question/1121886?
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Answer:
20.05 seconds
Explanation:
Given that:
v² = u² + 2as
v = final velocity
u = initial velocity
a = acceleration
s = distance
a = negative acceleration = - 0.43
s = distance = 86.5
v = 0
0 - 2(0.43)(86.5) = u²
74.39 = v²
U = sqrt(74.39)
U = 8.6249
From ;
t = Time in seconds
t = (v - u) / a
t = (0 - 8.6249) / - 0.43
t = 20.057
Answer:
The component of block weight parallel to the plane, Wₓ = W cosθ
Explanation:
Let the weight of the block due to gravitation is W
The direction of the weight is vertically down
Let θ be the angle formed with the vertical weight of the block and the incline.
Taking two components of weight one along the vertical weight and another component perpendicular to it.
Then the component `of weight long the parallel of the plane is
Wₓ = W cosθ
Answer:

Explanation:
The expression which represent the first diffraction minima by a circular aperture is given by
--------eqn 1
The angle through which the first minima is diffracted is given by
---------eqn 2
As
is very small so we can write 
So from eqn 1 and eqn 2 we can write
--------eqn 3
Here
is the position of first maxima D is the distance of screen from the circular aperture d is the diameter of aperture
It is given that diameter of circular aperture is 14.7 cm so 
Now putting all these value in eqn 3

