<h3><u>Answer</u>;</h3>
= 22°
<h3><u>Explanation</u>;</h3>
- According to Snell's law, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant. The constant value is called the refractive index of the second medium with respect to the first.
- Therefore; Sin i/Sin r = η
In this case; Angle of incidence = 90° -60° =30°, angle of refraction =? and η = 1.33
Thus;
Sin 30 / Sin r = 1.33
Sin r = Sin 30°/1.33
= 0.3759
r = Sin^-1 0.3759
= 22.08
<u>≈ 22°</u>
Answer:
Explanation:
Electric field due to charge at origin
= k Q / r²
k is a constant , Q is charge and r is distance
= 9 x 10⁹ x 5 x 10⁻⁶ / .5²
= 180 x 10³ N /C
In vector form
E₁ = 180 x 10³ j
Electric field due to q₂ charge
= 9 x 10⁹ x 3 x 10⁻⁶ /.5² + .8²
= 30.33 x 10³ N / C
It will have negative slope θ with x axis
Tan θ = .5 / √.5² + .8²
= .5 / .94
θ = 28°
E₂ = 30.33 x 10³ cos 28 i - 30.33 x 10³ sin28j
= 26.78 x 10³ i - 14.24 x 10³ j
Total electric field
E = E₁ + E₂
= 180 x 10³ j +26.78 x 10³ i - 14.24 x 10³ j
= 26.78 x 10³ i + 165.76 X 10³ j
magnitude
= √(26.78² + 165.76² ) x 10³ N /C
= 167.8 x 10³ N / C .
This can be answered using trigonometric analysis. This sloped path that is 150 m long is the hypotenuse of the triangle. The adjacent angle would then be 65 degrees. Given these:
sin 65 = h / 150
Where: h = vertical displacement = 150 (sin 65)
h = 135.95 meters
Answer:
Terminal velocity of object = 12.58 m/s
Explanation:
We know that the terminal velocity is attained when drag force and gravitational force are of the same magnitude.
Gravitational force = mg = 80 * 9.8 = 784 N
Drag force = 
Equating both, we have

So v = 12.58 m/s or v = -15.58 m/s ( not possible)
So terminal velocity of object = 12.58 m/s
Answer:
Two possible points
<em>x= 0.67 cm to the right of q1</em>
<em>x= 2 cm to the left of q1</em>
Explanation:
<u>Electrostatic Forces</u>
If two point charges q1 and q2 are at a distance d, there is an electrostatic force between them with magnitude

We need to place a charge q3 someplace between q1 and q2 so the net force on it is zero, thus the force from 1 to 3 (F13) equals to the force from 2 to 3 (F23). The charge q3 is assumed to be placed at a distance x to the right of q1, and (2 cm - x) to the left of q2. Let's compute both forces recalling that q1=1, q2=4q and q3=q.





Equating


Operating and simplifying

To solve for x, we must take square roots in boths sides of the equation. It's very important to recall the square root has two possible signs, because it will lead us to 2 possible answer to the problem.

Assuming the positive sign
:




Since x is positive, the charge q3 has zero net force between charges q1 and q2. Now, we set the square root as negative



The negative sign of x means q3 is located to the left of q1 (assumed in the origin).