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defon
3 years ago
7

La luz roja visible tiene una longitud de onda de 680 nanómetros (6,8 x 10-7 m). La velocidad de la luz es de 3.0 x108 m / s. ¿C

uál es la frecuencia de la luz roja visible?
Physics
1 answer:
Lina20 [59]3 years ago
3 0

Answer:

Frequency, f=4.41\times 10^{14}\ Hz

Explanation:

Visible red light has a wavelength of 680 nanometers (6.8 x 10⁻⁷ m). The speed of light is 3.0 x 10 ⁸ m / s. What is the frequency of visible red light?

It is given that,

Wavelength of a visible red light is, \lambda=6.8\times 10^{-7}\ m

Speed of light is, c=3\times 10^8\ m/s

We need to find the frequency of visible red light. It can be calculated using below relation.

c=f\lambda\\\\f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{6.8\times 10^{-7}}\\\\f=4.41\times 10^{14}\ Hz

So, the frequency of visible red light is 4.41\times 10^{14}\ Hz.

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Phobos's Orbit. Phobos orbits Mars at a distance of 9,380 km from the center of the planet and has a period of 0.3189 days. Assu
Elenna [48]

Answer:

Explanation:

The relation between time period of moon in the orbit around a planet can be given by the following relation .

T² = 4 π² R³ / GM

G is gravitational constant , M is mass of the planet , R is radius of the orbit and T is time period of the moon .

Substituting the values in the equation

(.3189 x 24 x 60 x 60 s)²  = 4 x 3.14² x ( 9380 x 10³)³ / (6.67 x 10⁻¹¹ x M)

759.167 x 10⁶ = 8.25 x 10²⁰ x 39.43 / (6.67 x 10⁻¹¹ x M )

M = .06424  x 10²⁵

= 6.4 x 10²³ kg .

4 0
3 years ago
DESCRIBE THE REQUIREMENTS OF AN INTERNET CONNECTION?<br>please tell me the answer​
AlekseyPX

Answer: The basic requirements for connecting to the Internet are a computer device, a working Internet line, and the right modem for that Internet line. In addition, software programs such as Internet browsers, email clients, Usenet clients, and other special applications are needed in order to access the Internet.

Explanation: brainleist pls :)

4 0
3 years ago
H. A truck starts to move from rest. If it gains the acceleration of 3 m/s2 in 5 sec,
SashulF [63]

Answer:

1. The final velocity of the truck is 15 m/s

2. The distance travelled by the truck is 37.5 m

Explanation:

1. Determination of the final velocity

Initial velocity (u) = 0 m/s

Acceleration (a) = 3 m/s²

Time (t) = 5 s

Final velocity (v) =?

The final velocity of the truck can be obtained as follow:

v = u + at

v = 0 + (3 × 5)

v = 0 + 15

v = 15 m/s

Therefore, the final velocity of the truck is 15 m/s

2. Determination of the distance travelled

Initial velocity (u) = 0 m/s

Acceleration (a) = 3 m/s²

Time (t) = 5 s

Distance (s) =?

The distance travelled by the truck can be obtained as follow:

s = ut + ½at²

s = (0 × 5) + (½ × 3 × 5²)

s = 0 + (½ × 3 × 25)

s = 0 + 37.5

s = 37.5 m

Therefore, the distance travelled by the truck is 37.5 m

6 0
2 years ago
We have an Atwood device, two blocks connect by a string strung over a pulley, but the twist this time is that both blocks are o
Zanzabum

The Acceleration of the system is 6.41 m/s².

Given,

α= 15°, m₁ = 7kg

β= 65°, m₂ = 11 kg

Let, a be the acceleration and T is the tensions at the end it's the cord.

Let, the mass m₂ be coming down along the inclined plane along the inclined surface towards downward m₂g sin β and the tension in the upward direction,

Resultant force, m₂a=m₂g sin β -T

11a=((11) ×g sin 65°) -T  ...(i)

Now, considering the motion of m₁ which moves downwards, the forces are m₁g sinα, and T both are acting downwards.

Resultant force m₁a = m₁g sin α+T

7a =7g sin 15°+T  ...(ii)

Solving both the equations by adding them,

18a=11gsin 65°+7g sin 15°-T+T

18a=11gsin 65°+7g sin 15°=115.45

a=115.45/18=6.41 m/s²

Hence, the Acceleration of the system is 6.41 m/s².

Learn more about the acceleration here:

brainly.com/question/22048837

#SPJ10

6 0
2 years ago
A two-stage rocket is traveling at 1200 mis with respect to the earth when the first stage runs out of fuel. Explosive bolts rel
Aloiza [94]

Answer:

The speed of the second stage after separation is 4905 m/s

Explanation:

Hi there!

Due to conservation of momentum, the momentum of the system first stage - second stage must remain constant before and after the separation. The momentum of the system is calculated by adding the momentums of each object:

initial momentum = final momentum

m₁₊₂ · v = m1 · v1 + m2 · v2

Where:

m₁₊₂ = mass of the two stage rocket

v = velocity of the rocket

m1 = mass of stage 1

v1 =  velocity of stage 1

m2 = mass of stage 2

v2 = velocity of stage 2

We have the following data:

m1 = 3 · m2

m₁₊₂ = m1 + m2 = 3 · m2 + m2 = 4 · m2

v = 1200 m/s

v1 = -35 m/s  (let´s consider the backward direction as negative)

v2 = ?

Then, replacing these data in the equation of momentum of the system:

m₁₊₂ · v = m1 · v1 + m2 · v2

4 m2 · 1200 m/s = 3 m2 · (-35 m/s) + m2 · v2

Let´s solve the equation for v2:

divide both sides of the equation by m2:

4 · 1200 m/s = 3 · (-35 m/s) + v2

4800 m/s = -105 m/s + v2

v2 = 4800 m/s + 105 m/s = 4905 m/s

The speed of the second stage after separation is 4905 m/s

6 0
3 years ago
Read 2 more answers
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