A spring (70 N/m ) has an equilibrium length of 1.00 m. The spring is compressed to a length of 0.50 m and a mass of 2.0 kg is p
1 answer:
Answer:
Part a)
Part b)
Explanation:
Part a)
As we know that there is no friction in the path
So here we can use energy conservation to find the distance moved by the mass
Initial spring energy = final gravitational potential energy
so we will have
Part b)
Now if spring is connected to the block then again we can use energy conservation
so we will have
so we will have
so total distance moved upwards is
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Answer:
a) I= I₀ (cos²θ - cos⁴θ) b) 75.5º
Explanation:
a) For this exercise we must use Malus's law
I = I₀ cos² θ
where tea is the angle between the two polarizers.
We apply this expression to our case
* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical
I₁ = I₀ cos² θ
* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.
The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical
θ₂ = 90- θ’ = θ
fiate that in the exercise we must take a reference system and measure everything with respect to this system.
I = I₁ cos² θ'
we substitute
I = (I₀ cos² tea) cos² (θ - 90)
cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ
I = Io cos² θ sin² θ
1= cos²θ+ sin²θ
sin²θ = 1 - cos²θ
I= I₀ (cos²θ - cos⁴θ)
b) to find when the intensity is maximum,
we can use that we have an extreme point when the drift is zero
= 0
\frac{dI}{d \theta}= Io (2 cos θ - 4 cos³θ) = 0
whereby
cos θ - 2 cos³ θ = 0
cos θ ( 1 - 2 cos² θ) = 0
The zeros of this function are in
θ = 90º
1-2cos²θ =0 cos θ = 0.25 θ = 75.5º
Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.
Consequently, the angle that allows the maximum intensity to pass is 75.5º
Answer:
Number of electrons are flowing per second is 2.42 x 10¹⁹
Explanation:
The electric current flows through a wire is given by the relation :
....(1)
Here I is current, e is electronic charge, v is drift velocity of electrons and A is the Area of the wire.
But electric current is also define as rate of electrons passing through junction times their charge, i.e. ,
....(2)
Here N is the rate of electrons passing through junction.
From equation (1) and (2).
But area of wire,
Here d is diameter of wire.
So,
Substitute 2.91 x 10⁻³ m for d, 0.000191 m/s for v and 6 x 10²⁸ m⁻³ for n in the above equation.
N = 2.42 x 10¹⁹ s⁻¹