Answer:
The force of static friction acting on the luggage is, Fₓ = 180.32 N
Explanation:
Given data,
The mass of the luggage, m = 23 kg
You pulled the luggage with a force of, F = 77 N
The coefficient of static friction of luggage and floor, μₓ = 0.8
The formula for static frictional force is,
Fₓ = μₓ · η
Where,
η - normal force acting on the luggage 'mg'
Substituting the values in the above equation,
Fₓ = 0.8 x 23 x 9.8
= 180.32 N
Hence, the minimum force require to pull the luggage is, Fₓ = 180.32 N
Fundamental frequency,
f=v2l=T/μ−−−−√2l
=(50)/0.1×10−3/10−22×0.6−−−−−−−−−−−−−−−−−−−√
=58.96Hz
Let, n th harmonic is the hightest frequency, then
(58.93)n = 20000
∴N=339.38
Hence, 339 is the highest frequency.
∴fmax=(339)(58.93)Hz=19977Hz.
<h3>
What is frequency?</h3>
In physics, frequency is the number of waves that pass a given point in a unit of time as well as the number of cycles or vibrations that a body in periodic motion experiences in a unit of time. After moving through a sequence of situations or locations and then returning to its initial position, a body in periodic motion is said to have experienced one cycle or one vibration. See also simple harmonic motion and angular velocity.
learn more about frequency refer:
brainly.com/question/254161
#SPJ4
Answer:
<h3>
<em>2</em><em>4</em><em>7</em><em>9</em><em> </em><em>Newton</em></h3>
<em>Sol</em><em>ution</em><em>,</em>
<em>Mass</em><em>=</em><em>1</em><em>0</em><em>0</em><em> </em><em>kg</em>
<em>Accele</em><em>ration</em><em> </em><em>due</em><em> </em><em>to</em><em> </em><em>gravity</em><em>(</em><em>g</em><em>)</em><em>=</em><em>2</em><em>4</em><em>.</em><em>7</em><em>9</em><em> </em><em>m</em><em>/</em><em>s^</em><em>2</em>
<em>Now</em><em>,</em><em>.</em>
<em>
</em>
<em>hope</em><em> </em><em>this</em><em> </em><em>helps</em><em> </em><em>.</em><em>.</em>
<em>Good</em><em> </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em>
Answer:
a) 0.138J
b) 3.58m/S
c) (1.52J)(I)
Explanation:
a) to find the increase in the translational kinetic energy you can use the relation

where Wp is the work done by the person and Wg is the work done by the gravitational force
By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

the change in the translational kinetic energy is 0.138J
b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

the new speed is 3.58m/s
c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

hence, the change in Er is about 1.52J times the initial rotational energy