Answer:
The minimum coefficient of friction required is 0.35.
Explanation:
The minimum coefficient of friction required to keep the crate from sliding can be found as follows:


Where:
μ: is the coefficient of friction
m: is the mass of the crate
g: is the gravity
a: is the acceleration of the truck
The acceleration of the truck can be found by using the following equation:


Where:
d: is the distance traveled = 46.1 m
: is the final speed of the truck = 0 (it stops)
: is the initial speed of the truck = 17.9 m/s
If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.

Therefore, the minimum coefficient of friction required is 0.35.
I hope it helps you!
Answer:
The magnitude of the force of friction equals the magnitude of my push
Explanation:
Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.
Let F = push and f = frictional force and f' = net force
F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0
So, F - f = 0
Thus, F = f
So, the magnitude of the force of friction equals the magnitude of my push.
Answer:
a) C.M 
b) 
Explanation:
The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"
The center of mass on a two dimensional plane is defined with the following formulas:


Where M represent the sum of all the masses on the system.
And the center of mass C.M 
Part a
represent the masses.
represent the coordinates for the masses with the units on meters.
So we have everything in order to find the center of mass, if we begin with the x coordinate we have:


C.M 
Part b
For this case we have an additional mass
and we know that the resulting new center of mass it at the origin C.M
and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

If we solve for a we got:




And solving for b we got:

So the coordinates for this new particle are:

1. When an object is moving away from us, the light from the object is known as redshift, and when an object is moving towards us, the light from the object is known as blueshift.
2. A wavelength increases in size, and its frequency, and energy decrease.
3. The frequency of a wave increases, and its wavelength decreases.
Redshift is an important term for astronomers. The term can be taken literally. The wavelengths of light are stretched and perceived as shifting toward the red portion of the spectrum. The same thing happens to sound waves when the source moves relative to the observer.
As the wave frequency decreases, the wavelength increases as long as the wave velocity remains constant. If the wave speed stays the same as the frequency decreases, it means that fewer wave peaks or troughs pass through a given point in a given time period. The number of complete wavelengths in a given unit of time is called frequency. Frequency and energy decrease with increasing wavelength.
Learn more about A wavelength here:-brainly.com/question/24452579
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