Answer:
The sign of the charge is negative
The magnitude of the charge is 3.33 x 10⁻¹³ C
Explanation:
Given;
potential difference, V = -1.5 V
distance of the point charge, r = 2 mm = 2 x 10⁻³ m
The magnitude of the charge is calculated as follows;
A 55 kg block of ice slides down a frictionless ramp that is 2.0 m long and 0.91 m high. A worker pushes up against the ice, par
allel to the ramp, so that the block slides down the ramp at constant speed. a. Find the force exerted by the worker.
1 answer:
Answer:
223.4N
Explanation:
Hello!
To solve this problem follow the steps below, the complete procedure is in the attached image
1. draw a complete outline of the problem involving forces and geometries
2. Find the angle of the ramp using the tangent function
3. Find the component parallel to the ice block weight ramp using the sine function. Remember that the magnitude of the weight is the product of mass and acceleration.
4. For a body to move with constant speed its acceleration must be zero, remembering the second law of Newton's movement: force is equal to the product of mass and acceleration. we infer that the sum of the forces in the direction parallel to the ramp must be zero.
whereby the force that the worker makes is equal to the component of the weight of the ice block in the direction of the ramp.
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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