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egoroff_w [7]
4 years ago
5

What type of waves can only travel through a medium?

Physics
2 answers:
scoray [572]4 years ago
7 0

<u>Answer:</u> The waves that travel through a medium is electromagnetic waves.

<u>Explanation:</u>

There are two types of waves:

  • <u>Electromagnetic waves:</u> These are waves which can travel through vacuum. These have electrical and magnetic component associated with them. They travel with the speed of light. They does not require a medium to travel. For Example: Infrared waves, Microwaves
  • <u>Mechanical waves:</u> These are the waves that require a medium to travel so that they can transport their energy from one location to another. For Example:  Sound waves

Hence, the waves that travel through a medium is electromagnetic waves.

katovenus [111]4 years ago
5 0

Answer:

The energy of a mechanical wave can travel only through matter. The matter through which the wave travels is called the medium (plural, media). The medium in the water wave pictured above is water, a liquid. But the medium of a mechanical wave can be any state of matter, even a solid.Nov 1, 2012

Explanation:

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To transfer the visitor's city map into a two dimensional map,
Dovator [93]

Answer:

Imagine you have an orange. This is your imaginary Earth. When you look at it in any direction, you won’t be able to see all sides of it. But when you peel the orange, flatten and stretch it out, you can begin to see everything.

Similarly, a map projection is a method by which cartographers translate a sphere or globe into a two-dimensional representation. In other words, a map projection systematically renders a 3D ellipsoid (or spheroid) of Earth to a 2D map surface.

Explanation:

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1 year ago
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miss Akunina [59]

Answer:

Explanation:

The rocky planets include Mercury, Venus, Earth, and Mars. they are made up of rocks and metals and have solid surfaces.They are the closest four planets to the Sun. The gas giants are  Neptune, Uranus, Saturn, and Jupiter. they are further from the sun and are in the outer part of the solar system. I hope this helps!

6 0
3 years ago
Water enters a house at 1.5 m/s through a pipe with an inside radius of 1.0cm and at a pressure of 400,000 Pa. The water then tr
kupik [55]

Answer:

pressure of the water = 3.3 × 10^{5} pa

Explanation:

given data

velocity v1 = 1.5 m/s

pressure P = 400,000 Pa

inside radius r1 = 1.00 cm

pipe radius r2 = 0.5 cm  

h1 = 0 (datum at inlet)

h2 = 5.0 m (datum at inlet)

density of water ρ = 1000 kg/m³

to find out

pressure of the water

solution

we consider here flow speed in bathroom that is =  v2 and Pressure in bathroom is =  P2  

here we will use both continuity and Bernoulli equations

because here we have more than one unknown so that  

v1 × A1 = v2 × A2 × P1 +  ρ g h1 + (0.5)ρ v1² = P2 + ρ g h2 + (0.5) ρ v2²

now we use here first continuity equation for get v2

v2 = v1 \frac{A1}{A2}    

v2 = 1.5 \frac{\pi * 0.01^2}{\pi * 0.005^2}    

v2 = 6 m/s

and now we use here bernoulli eqution for find here p2 that is

P2 = P1 - 0.5× ρ ×(v2² - v1²) - ρ g (h2- h1 )

P2 = 400000  - 0.5× 1000 ×(6² - 1.5²) - 1000 × 9.81 × (5-0 )

P2 = 3.3 × 10^{5} pa

3 0
3 years ago
The 1.5-kg collar C starts from rest at A and slides with negligible friction on the fixed rod in the vertical plane. Determine
arlik [135]

Answer:

Explanation:

check attached image for figure, there is supposed to be a figure for this question containing a distance(height of collar at position A) but i will assume 0.2m or 200mm

Consider the energy equilibrium of the system

U_{A-B}=\bigtriangleup T\\\\F\cos 30°\times h_A - F\sin30°\times h_A + Wh_A=\frac{1}{2}m(v^2_B-v^2_A)\\\\v_B=\sqrt{\frac{2Fh_A(\cos 30° - \sin30°)+mgh_A}{m+v^2_A}}

Here, F is the force acting on the collar, h_A is the height of the collar at position A, m is the mass of the collar C, g is the acceleration due to gravity, v_B is the velocity of the collar at position B, and v_A is the velocity of the collar at A

Substitute 14.4N for F, 0.2m for h_A, 1.5kg for m, 9.81m/s^2 for g and 0 for v_A

v_B=\sqrt{\frac{2(14.4\times 0.2(\cos 30° - \sin30°)+1.5\times 9.81\times 0.2}{1.5+0}}\\\\=\sqrt{\frac{6.618}{1.5}}\\\\=4.412m/s

Therefore, the velocity at which the collar strikes the end B is 4.412m/s

8 0
3 years ago
A potential difference of 3.00 nV is set up across a 2.00 cm length of copper wire that has a radius of 2.00 mm. How much charge
Anvisha [2.4K]

The number of charge drifts are 3.35 X 10⁻⁷C

<u>Explanation:</u>

Given:

Potential difference, V = 3 nV = 3 X 10⁻⁹m

Length of wire, L = 2 cm = 0.02 m

Radius of the wire, r = 2 mm = 2 X 10⁻³m

Cross section, 3 ms

charge drifts, q = ?

We know,

the charge drifts through the copper wire is given by

q = iΔt

where Δt = 3 X 10⁻³s

and i = \frac{V}{R}

where R is the resistance

R = \frac{pL}{r^{2} \pi }

ρ is the resistivity of the copper wire = 1.69 X 10⁻⁸Ωm

So, i = \frac{\pi(r)^{2}V  }{pL}

q = \frac{\pi(r^{2} )Vt }{pL}

Substituting the values,

q = 3.14 X (0.02)² X 3 X 10⁻⁹ X 3 X 10⁻³ / 1.69 X 10⁻⁸ X 0.02

q = 3.35 X 10⁻⁷C

Therefore, the number of charge drifts are 3.35 X 10⁻⁷C

3 0
3 years ago
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