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3 years ago

Any ss2 here (11th Grade)

Physics

1 answer:

Sever21 [200]3 years ago

6 0

Answer:

Explanation:

Check the complete diagram n the attachment below

a) The angle of incidence on a plane surface is the angle between the incidence ray and the normal ray acting on a plane surface. The normal ray is the ray perpendicular to the surface while the incidence ray is the ray striking a plane surface.

According to the diagram, the angle of reflection r₂ on M₂ is 90°-g where g is the angle of glance.

Given angle of glance on M₂ to be 40°, r₂ = 90-40 = 50°

According the second law of reflection, the angle of incidence = angle of reflection, therefore i₂ = r₂ = 50° (on M₂)

Also ∠OO₂O₁ = ∠OO₁O₂ = 40° (angle of glance on M₁){alternate angle}

The angle of incidence on M₁ = 90° - 40° = 50°

b) The angle of incidence to the surface of M₁(∠PO₁A)will be the angle of glance on M₁ which is equivalent to 40°

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Hi there! :)

Reference the diagram below for clarification.

We must begin by knowing the following rules for resistors in series and parallel.

In series:
$R_T = R_1 + R_2 + ... + R_n$

In parallel:
$\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}$

We can begin solving for the equivalent resistance of the two resistors in parallel using the parallel rules.

$\frac{1}{R_{T, parallel}} = \frac{1}{10} + \frac{1}{10}\\\\\frac{1}{R_{T, parallel}} = \frac{2}{10} = \frac{1}{5}\\\\R_{T, parallel} = 5\Omega$

Now that we have reduced the parallel resistors to a 'single' resistor, we can add their equivalent resistance with the other resistor in parallel (15 Ohm) using series rules:
$R_T = 15 + 5\\\\\boxed{R_T = 20 \Omega}$

We can use Ohm's law to solve for the current in the circuit.

$i = \frac{V}{R_T}\\\\i = \frac{120}{20} = \boxed{6 A}$

For resistors in series, both resistors receive the SAME current.

Therefore, the 15Ω resistor receives 6A, and the parallel COMBO (not each individual resistor, but the 5Ω equivalent when combined) receives 6A.

In this instance, since both of the resistors in parallel are equal, the current is SPLIT EQUALLY between the two. (Current in parallel ADDS UP). Therefore, an even split between 2 resistors of 6 A is <u>3A for each 10Ω resistor</u>.

Since the 15.0 Ω resistor receives 6A, we can use Ohm's Law to solve for voltage.

$V = iR\\\\V = (6)(15) = \boxed{90 V}$