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katrin [286]
4 years ago
15

Suppose a sound wave and an electromagnetic wave have the same frequency. Which has the longer wavelength?

Physics
1 answer:
ivann1987 [24]4 years ago
3 0

Answer:

Electromagnetic wave has longer wavelength.

Explanation:

The velocity (v) of a wave is the product between frequency (f) and wavelength (λ):

v=\lambda f (1)

Electromagnetic waves travel on air near the velocity of light 299792458 \frac{m}{s}, and sound waves travel on air at approximately 343,2 \frac{m}{s}, so if the frequency is equal for both, and the velocity is directly proportional to wavelength as shown in (1) then higher velocities imply higher wavelengths, that is, electromagnetic wave has longer wavelength.

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What is kepler's law??​
Elan Coil [88]
<h2>QUESTION:- </h2>

➜what is kepler's law??

\huge\red{\boxed{\huge\mathbb{\red A \pink{N}\purple{S} \blue{W}\orange{ER}}}}

Kepler gave the three laws or theorems of motion of the orbitals bodies

{\huge {\bold{ \red{ \star}}}}{ \blue{ \bold{FIRST \: \: \: LAW}}}

This law state that the celestial bodies revolves around the stars in elliptical orbit and star as a single focus.

Example :- Earth revolves around the Sun as assuming it as single focus

This also shows that earth revolves around the sun in elliptical orbit.

{\huge {\bold{ \blue{ \star}}}}{ \green{ \bold{SECOND \: \: \: LAW}}}

Area covered by the planet is equal in equal duration of time irrespective of the position of the planet.

It also states that Angular momentum is constant

As Angular momentum is constant it means areal velocity is also constant.

\frac{ \triangle \: A}{ \triangle \: T} = \frac{L}{2m}△T△A=2mL

where:-

A is the area.

T is the time.

L is the angular momentum.

M is the mass of the body.

{\huge {\bold{ \green{ \star}}}}{ \purple{ \bold{THIRD \: \: \: LAW}}}

square of the time of the revolution is directly proportional to the cube of the distance between the planet and star in Astronomical unit.

{T}^{2} = {a}^{3}T2=a3

where:-

T = time of revolution

a is the distance between the planet and star.

\purple\star \: {Thanks \: And \: Brainlist} \blue \star \\ {\orange{ \star}}{if \: U \: Like d \: My \: Ans} {\green{ \star }}

8 0
3 years ago
A skydiver of mass 80kg jumps from a slow moving aircraft and reach a terminal speed of 50 m/s. What's her acceleration when her
miskamm [114]

Answer:

a = g = 9.81[m/s^2]

Explanation:

This problem can be solve using the second law of Newton.

We know that the forces acting over the skydiver are only his weight, and it is equal to the product of the mass by the acceleration.

m*g = m*a

where:

g = gravity = 9.81[m/s^2]

a = acceleration [m/s^2]

Note: If the skydiver will be under air resistance forces his acceleration will be different.

7 0
3 years ago
If a cars velocity is slowing down is it considered a positive or negative acceleration?
Natali [406]

Answer:

Negative

Explanation:

Observe that the object below moves in the positive direction with a changing velocity. An object which moves in the positive direction has a positive velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion (in this case, a negative acceleration).

4 0
3 years ago
The resultant of 2 forces at right angles is 100 lbs. If one of the forces makes an angle of 30 degress with the resultant, comp
Ipatiy [6.2K]

Answer:

86.6 lbs

Explanation:

Let the force is X.

Resultant force, R = 100 lbs

Other force is Y. Angle between resultant force and force X is 30°.

According to the diagram

Cos30=\frac{X}{R}

0.866=\frac{X}{100}

X = 86.6 lbs

Other force Y

Sin30=\frac{X}{R}

0.5=\frac{Y}{100}

Y = 50 lbs

5 0
4 years ago
Show that the electric potential along the axis of a uniformly charged disk of radius R and charge density sigma is given by by
OlgaM077 [116]

Explanation:

Area of ring \ 2{\pi} a d a

Charge of on ring d q=-(\ 2{\pi} a d a)

Charge on disk

Q=-\left(\pi R^{2}\right)

\begin{aligned}d v &=\frac{k d q}{\sqrt{x^{2}+a^{2}}} \\&=2 \pi-k \frac{a d a}{\sqrt{x^{2}+a^{2}}} \\v(1) &=2 \pi c k \int_{0}^{R} \frac{a d a}{\sqrt{x^{2}+a^{2}}} \cdot_{2 \varepsilon_{0}}^{2} R \\&=2 \pi \sigma k[\sqrt{x^{2}+a^{2}}]_{0}^{2} \\&=\frac{2 \pi \sigma}{4 \pi \varepsilon_{0}}[\sqrt{z^{2}+R^{2}}-(21)] \\&=\frac{\sigma}{2}(\sqrt{2^{2}+R^{2}}-2)\end{aligned}

Note: Refer the image attached

8 0
4 years ago
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