Answer:
the vertical acceleration of the case is 1.46 m/s
Explanation:
Given;
mass of the clarinet case, m = 3.07 kg
upward force applied by the man, F = 25.60 N
Apply Newton's second law of motion;
the upward force on the clarinet case = its weight acting downwards + downward force due to its downward accelaration
F = mg + m(-a)
the acceleration is negative due to downward motion from the top of the piano.
F = mg - ma
ma = mg - F

Therefore, the vertical acceleration of the case is 1.46 m/s²
The magnitude of their resultant vector is 4.6 meters/seconds
Since we are to add the velocity vectors in order to find the magnitude of their resultant vector.
Hence:
Resultant vector magnitude=5.8 meters/seconds + (1.2 meters/seconds)
Resultant vector magnitude=5.8 meters/seconds-1.2 meters/seconds
Resultant vector magnitude 4.6 meters/seconds
Inconclusion The magnitude of their resultant vector is 4.6 meters/seconds
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To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

Here,
M = Mass
R = Radius of the hoop
The precession frequency is given as

Here,
M = Mass
g= Acceleration due to gravity
d = Distance of center of mass from pivot
I = Moment of inertia
= Angular velocity
Replacing the value for moment of inertia


The value for our angular velocity is not in SI, then


Replacing our values we have that


The precession frequency is




Therefore the precession period is 5.4s
The indicated data are of clear understanding for the development of Airy's theory. In optics this phenomenon is described as an optical phenomenon in which The Light, due to its undulatory nature, tends to diffract when it passes through a circular opening.
The formula used for the radius of the Airy disk is given by,

Where,
Range of the radius
wavelength
f= focal length
Our values are given by,
State 1:



State 2:



Replacing in the first equation we have:


And also for,


Therefor, the airy disk radius ranges from
to 
Answer:
The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
Explanation:
Given;
coefficient of kinetic friction, μ = 0.84
speed of the automobile, u = 29.0 m/s
To determine the the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;
v² = u² + 2ax
where;
v is the final velocity
u is the initial velocity
a is the acceleration
x is the shortest distance
First we determine a;
From Newton's second law of motion
∑F = ma
F is the kinetic friction that opposes the motion of the car
-Fk = ma
but, -Fk = -μN
-μN = ma
-μmg = ma
-μg = a
- 0.8 x 9.8 = a
-7.84 m/s² = a
Now, substitute in the value of a in the equation above
v² = u² + 2ax
when the automobile stops, the final velocity, v = 0
0 = 29² + 2(-7.84)x
0 = 841 - 15.68x
15.68x = 841
x = 841 / 15.68
x = 53.64 m
Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m