Answer:
i think its the third one or second one
When one object is rubbed against another, static electricity can be created. This is because the rubbing creates a negative charge that is carried by electrons.
Answer:
x(t)=0.337sin((5.929t)
Explanation:
A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 2 m/sec, find the position of the mass after t seconds.
Solution. Let x(t) denote the position of the mass at time t. Then x satisfies the differential equation
Definition of parameters
m=mass 3kg
k=force constant
e=extension ,m
ω =angular frequency
k=90/1.6=56.25N/m
ω^2=k/m= 56.25/1.6
ω^2=35.15625
ω=5.929
General solution will be
differentiating x(t)
dx(t)=-5.929c1sin(5.929t)+5.929c2cos(5.929t)
when x(0)=0, gives c1=0
dx(t0)=2m/s gives c2=0.337
Therefore, the position of the mass after t seconds is
x(t)=0.337sin((5.929t)
Answer:
<em>The correct answer is (a) 312.5nm and (b) 125nm</em>
Explanation:
<em>The first step to take is to find The minimum thickness of the slick of the oil</em>
<em>Given that,</em>
<em>(a) tmin λ/2n </em>
<em>We substitute 750nm ( in air) for λ and 1.20 for n for the expression of minimum thickness t of the oil slick at that spot</em>
<em>thus,</em>
<em>tmin = (750nm)/2(1.2) = 312.5nm</em>
<em>The minimum thickness of the oil slick at that spot is =312.5nm</em>
<em>(B) we find the minimum thickness t </em>
<em>The minimum thickness of the oil slick at the spot will be calculated by,</em>
<em>tmin = λ/4n</em>
<em>we then 750nm ( in air) for λ and 1.50 for n in the expression for the minimum thickness of the slick of the oil.</em>
<em>tmin = (750nm)/4 (1.5) = 125nm</em>
<em>Therefore the minimum thickness t will now be = 125nm</em>
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