Answer:
Below
Explanation:
● What makes conductors special is that they have some free electrons that can move.
● An insulator has no free electrons wich explains why current isn't transmitted through it.
PLEASE HELP 50 pts + brainliest. 3. A laser light travels from air (n=1) through water (n=1.33), and then a piece of glass (n=1.
55). The light first enters the water at an angle of 25.0 degrees (from air to water). Draw a diagram. Use Snell’s Law to find the angle of refraction in the water and then in the glass. I don't know how to do this.
1 answer:
Answer: the angle of refraction in the water is 34.2 degrees and then in the glass is 41 degrees approximately
Explanation:
A laser light travels from air (n=1) through water (n=1.33), and then a piece of glass (n=1.55).
If the light first enters the water at an angle of 25.0 degrees (from air to water). That means
The incident angle = 25 degrees
Using Snell' law
n(air)/n(water) = sin 25/sin R
Where R is the angle of refraction
1/1.33 = sin 25/sin R
Make sin R the subject of formula
Sin R = sin 25 / 0.7518
Sin R = 0.5620
R = sin^-1(0.5620)
R = 34.2 degrees
This angle will be an incident angle when the light is travelling from the water to glass. Using Snell law agains,
1.33/1.55 = sin 34.2/sin R
Cross multiply
1.33 sin R = 0.87122
Make R the subject of formula
Sin R = 0.87122/1.33
Sin R = 0.6550
R = sin^-1(0.6550)
R = 40.9 degrees
Therefore, the angle of refraction in the water is 34.2 degrees and then in the glass is 41 degrees approximately.
Find the attached file for the figure
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Answer:
(a) 0.17 m
(b) 5.003 m
(c) 6.38 × N
(d) 7.37 × N
Explanation:
(a) The minimum value of will occur when q3 = 0 m or at origin and q1, q2 are at 0.17 m so the distance between q3 and q1, q2 is 0.17 m, therefore the <em>minimum value of x= 0.17 m</em>.
(b) The maximum value of x will occur when q3 = 5 m because it is said in the question that 5 is the maximum distance travelled by q3. To find the hypotenuse i.e. the distance between q3 and q1,q2, we use Pythagoras theorem.
<em>Hence, the maximum distance is 5.002 m</em>
(c) For minimum magnitude we use the minimum distance calculated in (a)
Minimum Distance = 0.17 m
For electrostatic force=
×
(d) For maximum magnitude, we use the maximum distance calculated in (b)
Maximum Distance = 5.002 m
Using the formula for electrostatic force again:
F =
F= 7.37× N
Answer:
The magnitude of the lift force L = 92.12 kN
The required angle is ≅ 16.35°
Explanation:
From the given information:
mass of the airplane = 9010 kg
radius of the airplane R = 9.77 mi
period T = 0.129 hours = (0.129 × 3600) secs
= 464.4 secs
The angular speed can be determined by using the expression:
ω = 2π / T
ω = 2 π/ 464.4
ω = 0.01353 rad/sec
The direction
θ = 16.35°
The magnitude of the lift force L = mg ÷ Cos(θ)
L = (9010 × 9.81) ÷ Cos(16.35)
L = 88388.1 ÷ 0.9596
L = 92109.32 N
L = 92.12 kN