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3 years ago

How does the speed of visible light compare with the speed of gamma rays

Physics

1 answer:

DochEvi [55]3 years ago

3 0

Answer:

They are the same

Explanation:

Electromagnetic waves consist of perpendicular oscillations of electric and magnetic field, which oscillate perpendicularly to the direction of motion of the wave (transverse wave). One property of the electromagnetic waves is that they travel in a vacuum always at the same speed, called speed of light:

$c=3.0\cdot 10^8 m/s$

Electromagnetic waves are classified into 7 different types according to their frequency; from highest to lowest frequency, we have:

Gamma rays

X-rays

Ultraviolet

Visible light

Infrared

Microwaves

Radio waves

We see that both visible light and gamma rays are electromagnetic waves, so they both travel in a vacuum at the same speed, the speed of light.

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Una onda sonora se produce durante 1,5 s. Posee una longitud de onda de 2,4 m y una velocidad de 340 m/s.

cestrela7 [59]

Answer:

Una secadora de cabello tiene una resistencia de 10Ω al circular una corriente de 6 Amperes, si está conectado a una diferencia de potencial de 120 V, durante 18 minutos ¿Qué cantidad de calor produce?, expresado en calorías

Explanation:

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3 years ago

A gas of helium atoms at 273 k is in a cubical container with 25.0 cm on a side. (a) what is the minimum uncertainty in momentum

qwelly [4]

wave function of a particle with mass m is given by ψ(x)={ Acosαx −

2α

≤x≤+

2α

0 otherwise , where α=1.00×1010/m.

(a) Find the normalization constant.

(b) Find the probability that the particle can be found on the interval 0≤x≤0.5×10−10m.

(d) Find its average momentum.

(e) Find its average kinetic energy −0.5×10−10m≤x≤+0.5×10−10m.

6 0

3 years ago

A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30° angle. The block’s initial speed is 10 m/s. What vert

Romashka [77]

Answer:

Vertical Height = 0.784 meter, Speed back at starting point = 10 m/s

Explanation:

Given Data:

V is the overall velocity vector, $Vi$ and $Ui$ are its initial vertical and horizontal components

$R = 10 m/s\\ Projection Angle (theta) = 30 degrees\\Vi = 10*sin(30) = 5 m/s\\Ui = 10*cos(30) = 8.66 m/s$

To find:

Max Height $h$ achieved

Calculation:

1) Using the $3^{rd}$ equation of motion, we know

$2*a*s = Vf^{2} - Vi^{2}$

2) In terms of gravity $g$ height $h$ and the vertical component of Velocity $Vf , Vi$ .

3) As $Vf = 0$ as at maximum height the vertical component of velocity is zero maximum height achieved

$2*g*h = Vf^{2} -Vi^{2}$

putting values

4) $h = 0.784 m/s$

5) As for the speed when it reaches back its starting point, it will have a speed similar to its launching speed, the reason being the absence of air friction (Air drag)

3 0

3 years ago

For this problem, imagine that you are on a ship that is oscillating up and down on a rough sea. Assume for simplicity that this

ikadub [295]

Answer:

no idea

Explanation:

7 0

3 years ago

While water skiing behind her father’s boat, Letty is pulled at constant speed by a force of 164 N from the tow rope that makes

kap26 [50]

Answer:

0.265

Explanation:

Draw a free body diagram. There are four forces:

Normal force Fn pushing up.

Weight force mg pulling down.

Tension force T at an angle θ.

Friction force Fn μ pushing left.

Sum the forces in the y direction:

∑F = ma

Fn + T sin θ − mg = 0

Fn = mg − T sin θ

Sum the forces in the x direction:

∑F = ma

T cos θ − Fn μ = 0

Fn μ = T cos θ

μ = T cos θ / Fn

μ = T cos θ / (mg − T sin θ)

Given T = 164 N, θ = 10.0°, m = 65.0 kg, and g = 9.8 m/s²:

μ = (164 N cos 10.0°) / (65.0 kg × 9.8 m/s² − 164 N sin 10.0°)

μ = 0.265

7 0

3 years ago