Which formula represents Gay-Lussac's law?

A ray of monochromatic light ( f = 5.09 × 10^14 Hz) passes from air into Lucite at an angle

harina [27]

Let us consider the air with the index 1 and the lucite with index 2. Using the Snell's Secound Law, we have:

$\frac{sen\O_{2}}{sen\O_{1}} = \frac{n_{2}}{n_{1}} \\ sen\O_{2}= \frac{n_{2}*sen\O_{1}}{n_{1}}$

Entering the unknowns, remembering that the air refrective index is 1 and the lucite refrective index is 1.5, comes:

$sen\O_{2}= \frac{n_{2}*sen\O_{1}}{n_{1}} \\ sen\O_{2}= \frac{1.5* \frac{1}{2} }{1} \\ sen\O_{2}=0.75$

Using the arcsin properties, we get:

$sen\O_{2}=0.75 \\ arcsin(0.75)=\O_{2} \\ \boxed {\O_{2}=48.59^o}$

Obs: Approximate results, and the drawing is attached

If you notice any mistake in my english, let me know, because i am not native.

6 0

3 years ago

Which of the following would likely be the best reflector of heat energy

mr Goodwill [35]

Ok so B would be the best surface to reflect heat energy as it's polished and does not disturb the wave of energy like the others would. Also in therms of colour it's better because it's a lighter colour than navy ( the darker colour of the spectrum ). This matters as darker colours absorb light where light colour reflect it.
Hope this helps :).

6 0

3 years ago

Read 2 more answers

A charge of −20 µC is distributed uniformly over the surface of a spherical conductor of radius 11.0 cm. Determine the electric

Alex73 [517]

Answer:

(a) $-6.76\times 10^{12}\ N/C$

(b) $-1.352\times 10^{13}\ N/C$

(c) $-7.2\times 10^{11}\ N/C$

Explanation:

(a)

Given:

Charge on sphere (Q) = $-20\ \mu C=-20\times 10^{-6}\ C$

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 5 cm = 0.05 m

Coulomb's constant (k) = $9\times 10^{9}\ Nm^2/C^2$

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r ≤ R' from the center of sphere is given as:

$E=(\frac{kQ}{R^3})r$

Plug in the given values and solve for 'E'. This gives,

$E_{in}=(\frac{9\times 10^{9}\times -20}{(0.110)^3})\times 0.05\\\\E_{in}=-1.352\times 10^{14}\times 0.05\\\\E_{in}=-6.76\times 10^{12}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)$

(b)

Given:

Charge on sphere (Q) = $-20\ \mu C=-20\times 10^{-6}\ C$

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 10 cm = 0.10 m

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r ≤ R' from the center of sphere is given as:

$E=(\frac{kQ}{R^3})r$

Plug in the given values and solve for 'E'. This gives,

$E_{in}=(\frac{9\times 10^{9}\times -20}{(0.110)^3})\times 0.10\\\\E_{in}=-1.352\times 10^{14}\times 0.10\\\\E_{in}=-1.352\times 10^{13}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)$

(c)

Given:

Charge on sphere (Q) = $-20\ \mu C=-20\times 10^{-6}\ C$

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 50 cm = 0.50 m

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r > R' from the center of sphere is given as:

$E=\dfrac{kQ}{r^2}$

Plug in the given values and solve for 'E'. This gives,

$E_{out}=(\frac{9\times 10^{9}\times -20}{(0.50)^2})\\\\E_{out}=-7.2\times 10^{11}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)$

8 0

4 years ago

A car is moving at 25.5 m/s when it accelerates at 1.94 m/s^2 for 2.3 s. What is the car's final speed? (Keep in mind direction

Stolb23 [73]

Answer:

29.96m/s

Explanation:

Given parameters:

Initial speed = 25.5m/s

Acceleration = 1.94m/s²

Time = 2.3s

Unknown:

Final speed of the car = ?

Solution:

To solve this problem, we are going to apply the right motion equation:

v = u + at

v is the final speed

u is the initial speed

a is the acceleration

t is the time taken

Now insert the parameters and solve;

v = 25.5 + (1.94 x 2.3) = 29.96m/s

3 0

3 years ago

Please help asap its for anatomy

Brilliant_brown [7]

The smooth muscle in the wall of the bladder when stretched triggers the micturition reflex (urination).

<h3>What is a Bladder?</h3>

This is defined as a lined layers of muscle tissue that stretch to hold urine in organisms.

In older people the elasticity of the bladder is reduced which is why it makes it harder for them to hold urine for a long time.

Read more about Micturition here brainly.com/question/26493943

6 0

2 years ago