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3 years ago

In the virtual lab, calculations were made of the predicted diffraction angle, using the formula

Physics

1 answer:

djyliett [7]3 years ago

6 0

Answer:

1. The predicted angle of diffraction with a 3 cm wavelength was degrees.

the answer is 57.3

2. The predicted angle of diffraction with a 4 cm wavelength was degrees.

the answer is 76.4

3. The predicted angle of diffraction with a 5 cm wavelength was degrees.

the answer is 95.5

Explanation:

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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

Lina20 [59]

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 $4.41\times 10^{14}\ Hz$ .

3 0

2 years ago

Choose the law each sentence describes. This law relates a planet's orbital period and its average distance to the Sun. The orbi

hram777 [196]

These are the Kepler's laws of planetary motion.

This law relates a planet's orbital period and its average distance to the Sun. - Third law of Kepler.

The orbits of planets are ellipses with the Sun at one focus. - First law of Kepler.

The speed of a planet varies, such that a planet sweeps out an equal area in equal time frames. - Second law of Kepler.

7 0

3 years ago

Read 2 more answers

A box is sliding down an incline tilted at a 11.1° angle above horizontal. The box is initially sliding down the incline at a sp

raketka [301]

Answer:s=0.68 m

Explanation:

Given

Inclination $\theta =11.1^{\circ}$

Speed of block(u)=1.6 m/s

Coefficient of kinetic Friction $\mu _k=0.39$

deceleration provided by friction=g\sin \theta -\mu _kg\cos \theta [/tex]

Using $v^2-u^2=2as$

Final velocity v=0

$0-1.6^2=2(g\sin \theta -\mu _kg\cos \theta )s$

$s=\frac{-1.6^2}{2\cdot (9.8\sin 11.1-0.39\times 9.8\times \cos 11.1)}$

s=0.68 m

5 0

3 years ago

A 62 kg box is lifted 12 meters off the ground. How much work was done?

OleMash [197]

7291.2 I hope this is right

6 0

3 years ago

Read 2 more answers

The electric motor of a model train accelerates the train from rest to 0.620 m/s in 21ms. The total mass of the train is 825 g.

Rudiy27

Explanation:

Average power = change in energy / change in time

P = ΔE / Δt

P = (½ mv²) / t

P = (½ (0.825 kg) (0.620 m/s)²) / (0.021 s)

P = 7.55 Watts

7 0

2 years ago