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
I believe you mean Tim Peake ? .Major Timothy Nigel "Tim" Peake CMG is a British Army Air Corps officer, European Space Agency astronaut and International Space Station crew member.
Born: April 7, 1972 (age 44), Chichester, United Kingdom
Space missions: Expedition 46, Expedition 47, Soyuz TMA-19M
Nationality: British
Answer: 18000 coulombs
Explanation:
Given that:
Current, I = 5.0A
Electric charge Q = ?
Time, T = 1 hour
(The SI unit of time is seconds. So, covert 1.0 hour to seconds)
If 1 hour = 60 minutes and 60 seconds = 1 minute
Then, 1.0 hour = (60 x 60)
= 3600 seconds
Since electric charge, Q = current x time
i.e Q = I x T
Q = 5.0 A x 3600 seconds
Q = 18000 coulombs
Thus, 18000 coulombs of charge flows through the lamp in this time.
Light is described as both wave as well particle because light behave as wave in phenomenon like diffraction, interference, refraction, Doppler effect, etc whereas it behave as a particle in phenomenon like photoelectric effect, etc. Hope this helps.
