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Alexxandr [17]
2 years ago
15

The wavelength of some red light is 700.5 nm. what is the frequency of this red light?

Physics
1 answer:
Alborosie2 years ago
4 0
The frequency of the red light is 428 terahertz. To get the value of the red light's frequency, use the formula F = velocity/wavelength. The velocity of light is 3.00 x 10^8 m/s. For easier computation, convert 700.5 nanometers to meter. 1 nanometer is equal to 1 x 10^-9 meters. 700.5 nanometers is equal to 7.005 x 10^-7 meters. Divide the velocity 3.00 x 10^8m/s by wavelength 7.005 x 10^-7 meters. The result will be 4.28 x 10^14 Hertz or 428 terahertz.
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Given the indices of refraction n1 and n2 of material 1 and material 2, respectively, rank these scenarios on the basis of the p
lisov135 [29]

Answer:

c>d>f=a>b>e

Explanation:

When a pair of medial has greater difference between the their individual refractive indices with respect to vacuum then it has a greater deviation between the refracted ray and the incident ray.

According to the Snell's law:

\rm refractive\ index\ (n)=\frac{speed\ of\ light\ in\ the\ incident\ medium}{speed\ of\ light\ in\ the\ refracted\ medium}

a)

n_1-n_2=1.33-1.00\\=0.33

b)

n_2-n_1=1.46-1.33

=0.23

c)

n_2-n_1=2.42-1.33\\=1.09

d)

n_2-n_1=1.46-1.00\\=0.46

e)

n_1-n_2=1.50-1.33\\=0.17

f)

n_2-n_1=1.33-1.00\\=0.33

c>d>f=a>b>e

5 0
2 years ago
A point charge q1 = 4.50 nC is located on the x-axis at x = 1.95 m , and a second point charge q2 = -6.80 nC is on the y-axis at
Vinvika [58]

Answer:

Explanation:

One charge is situated at x = 1.95 m . Second charge is situated at y = 1.00 m

These two charges are situated outside sphere as it has radius of .365 m with center at origin. So charge inside sphere = zero.

Applying Gauss's theorem

Flux through spherical surface = charge inside sphere / ε₀

= 0 / ε₀

= 0 Ans .

3 0
2 years ago
A boy throws a 15 kg ball at 4.7 m/s to a 65 kg girl who is stationary and standing on a skateboard. After catching the ball, th
jek_recluse [69]

Answer:

a)v_{f}=0.88m/s

Explanation:

To solve this problem we use the Momentum's conservation Law, before and after the girl catch the ball:

\\ p_{1}=p_{2}\\m_{ball}*v_{o.ball}+m_{girl}*v_{o.girl} = m_{ball}*v_{f.ball} + m_{girl}*v_{f.girl}        (1)

At the beginning the girl is  stationary:

v_{o.girl}=0m/s       (2)

If the girl catch the ball, both have the same speed:

v_{f.girl}=v_{f.ball}=v_{f}       (3)

We replace (2) and (3) in (1):

m_{ball}*v_{o.ball} = (m_{ball}+m_{girl})*v_{f} \\

We can now solve the equation for v_{f}:

v_{f}=\frac{m_{ball}*v_{o.ball}}{(m_{ball}+m_{girl})}=\frac{15*4.7}{15+65}=0.88m/s

4 0
2 years ago
You drop a 6.0x10^-2 kg ball from height of 1.0m above hard flat surface. Ball strikes surface and energy decreases by 0.14J, th
harina [27]
If you drop a <span>6.0x10^-2 kg ball from height of 1.0m above hard flat surface, and a</span>fter the ball had bounce off the flat surface, the kinetic energy of the ball would be mgh - 0.14 = 0.45. 
3 0
2 years ago
Read 2 more answers
A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) th
zheka24 [161]

Answer:

(a) A = 0.650 m

(b) f = 1.3368 Hz

(c) E = 17.1416 J

(d)  K = 11.8835 J

     U = 5.2581 J

Explanation:

Given

m = 1.15 kg

x = 0.650 cos (8.40t)

(a) the amplitude,

A = 0.650 m

(b) the frequency,

if we know that

ω = 2πf = 8.40    ⇒   f = 8.40 / (2π)

⇒   f = 1.3368 Hz

(c) the total energy,

we use the formula

E = m*ω²*A² / 2

⇒  E = (1.15)(8.40)²(0.650)² / 2

⇒  E = 17.1416 J

(d) the kinetic energy and potential energy when x = 0.360 m.

We use the formulas

K = (1/2)*m*ω²*(A² - x²)       (the kinetic energy)

and

U = (1/2)*m*ω²*x²              (the potential energy)

then

K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)

⇒  K = 11.8835 J

U = (1/2)*(1.15)*(8.40)²*(0.360)²

⇒  U = 5.2581 J

4 0
2 years ago
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