A process known as fixation<span>. the majority of nitrogen is fixed by </span>bacteria<span>, most of which are </span>symbiotic<span> with plants</span>
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Answer:
A)
B)
C)
Explanation:
Given that a pendulum is suspended by a shaft with a very light thin rod.
Followed by the given information: m = 100 g, I = 0.5 m, g = 9.8 m / s²
We can determine the answer to these questions using angular kinematics.
Angular kinematics is just derived from linear kinematics but in different symbols, and expressions.
Here are the formulas for angular kinematics:
- θ = ωt
- ∆w =
- L [Angular momentum] = mvr [mass × velocity × radius]
A) What is the minimum speed required for the pendulum to traverse the complete circle?
We can use the formula v = √gL derived from
B) The same question if the pendulum is suspended with a wire?
C) What is the ratio of the two calculated speeds?
B, air blowing from across the field is as a bullet fired from a rifle
The Period of the resulting shm will be T=39.7
<u>Explanation:</u>
<u>Given data</u>
m=3kg
d=.06m
k=1200 N/m
Θ=3 °
T=?
we have the formulas,
I = (1/6)Md2
F = ma
F = -kx = -(mω2x)
k = mω2 τ = -d(FgsinΘ)
T=2 x 3.14/ √(m/k)
Solution for the given problem would be,
F=-Kx (where x= dsin Θ)
F=-k dsin Θ
F=-(1200)(.06)sin(3 °)
F=-10.16N
<u>By newton's second law.</u>
F = ma
a= F/m
a=(-10.16N)/3
a=3.38
<u>using the k=mω value</u>
k=mω
ω=k/m
ω=1200/3
ω=400
<u>Using F = -kx value</u>
x = F/-k
x=(-10.16)/1200
x=0.00847m
<u>Restoring the torque value </u>
τ = -dmgsinΘ where( τ = Iα so.).. Iα = -dmgsinΘ α = -(.06)(4)α =
α =(.06)(4)(9.81)sin(4°)
α=-1.781
<u>Rotational to linear form</u>
a = αr
r = .1131 m
a=-1.781 x .1131 m
a=-0.2015233664
<u>Time Period</u>
T=2 x 3.14/ √(m/k)
T=6.28/√(3/1200)
T=6.28/0.158
T=39.7
