Answer:
COMPLETE QUESTION
A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?
Explanation:
Given that,
Extension of spring
x = 0.0208m
Mass attached m = 3.39kg
Additional mass to have a frequency f
Let the additional mass be m
Using Hooke's law
F= kx
Where F = W = mg = 3.39 ×9.81
F = 33.26N
Then,
F = kx
k = F/x
k = 33.26/0.0208
k = 1598.84 N/m
The frequency is given as
f = ½π√k/m
Make m subject of formula
f² = ¼π² •(k/m
4π²f² = k/m
Then, m4π²f² = k
So, m = k/(4π²f²)
So, this is the general formula,
Then let use the frequency above
f = 3Hz
m = 1598.84/(4×π²×3²)
m = 4.5 kg
<u>First Symbol </u>: Cobalt (Co)
Its Group Number - 9
Its Period Number - 4
Its Family Name - Transition Metal
<u>Second Symbol</u> : Silicon (Si)
Its Group Number - 14
Its Period Number - 2
Its Family Name - Semiconductor
<u>Third Symbol</u> : Astatine (At)
Its Group Number - 17
Its Period Number - 6
Its Family Name - Halogen
<u>Fourth Symbol </u>: Magnesium (Mg)
Its Group Number - 2
Its Period Number - 3
Its Family Name - Alkaline Earth Metal
<u>Fifth Symbol</u> : Xenon (Xe)
Its Group Number - 18
Its Period Number - 5
Its Family Name - Noble Gas
Answer:
a) 567J
b) 283.5J
c)850.5J
Explanation:
The expression for the translational kinetic energy is,

Substitute,
14kg for m
9m/s for v

The translational kinetic energy of the center of mass is 567J
(B)
The expression for the rotational kinetic energy is,

The expression for the moment of inertia of the cylinder is,

The expression for angular velocity is,

substitute
1/2mr² for I
and vr for w
in equation for rotational kinetic energy as follows:



The rotational kinetic energy of the center of mass is 283.5J
(c)
The expression for the total energy is,

substitute 567J for E(r) and 283.5J for E(R)

The total energy of the cylinder is 850.5J
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