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Galina-37 [17]
3 years ago
15

explain why radio waves can travel through space,but sound waves cannot.explain the behavior of sound waves and radio waves in s

pace.
Physics
1 answer:
LUCKY_DIMON [66]3 years ago
5 0
Radio waves can travel through space because it is a radio wave.But sound waves cannot because it is called sounds waves .The behavior of the sounds wave and radio waves in space iws because chocalte are brown.spngebob squarepants are blue.Furthermore,dog can poop and we can poop.Also,this is how you can also explain like this:Radio Waves and Sounds Waves is Science.How i know this because i used my brain and i'm smart'' this is perfect explaination

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Two large, parallel, conducting plates are 1.00 cm apart and having electric charges of equal magnitudes (80.0 nC/m^2) and oppos
boyakko [2]

(a) 9040 V/m

The magnitude of the electric field between two parallel, oppositely charged plates is given by

E=\frac{\sigma}{\epsilon_0}

where

\sigma is the charge surface density on each plate

\epsilon_0 = 8.85\cdot 10^{-12} is the vacuum permittivity

In this problem, the magnitude of the charge density on each plate is

\sigma = 80.0 nC/m^2 = 80.0\cdot 10^{-9} C/m^2

Substituting into the formula, we find the magnitude of the electric field:

E=\frac{80.0\cdot 10^{-9}}{8.85\cdot 10^{-12}}=9040 V/m

(b) 90.4 V

The electric potential difference between the two plates is given by

\Delta V=Ed

where

E is the magnitude of the electric field

d is the separation between the two plates

In  this problem, we have

E = 9040 V/m

d = 1.00 cm = 0.01 m

Substituting into the formula, we find

\Delta V=(9040)(0.01)=90.4 V

6 0
3 years ago
A diver is is pushed upwards by a diving board. She weighs 739 N and accelerates from rest to a speed of 4.60 m/s while moving 0
gavmur [86]

Answer:

Answer A

Explanation:

It seems more likely

8 0
3 years ago
Based on this electric field diagram, which statement best compares the charge of A with B?
lilavasa [31]

Based on this electric field diagram, the statement which best compares the charge of A with B is "A is negatively charged and B is positively charged. The charge on A is greater than that on B".

<u>Answer:</u> Option A

<u>Explanation:</u>

The charge is quantized represented as elementary charge, about 1.602×10−19 coulombs. Their are two kinds of electric charging: positive and negative (usually transported, separately, by protons and electrons). Like charges repel each other, while attraction occurs among unlike charges. An entity without net charge is considered neutral. If a piece of matter comprises more electrons than protons, it has a negative charge, when there are fewer, it'll have a positive charge and when there are equal amounts, this will be neutral.

4 0
4 years ago
The rigid beam is supported by the three suspender bars. bars ab and ef are made of aluminum and bar cd is made of steel. if eac
faltersainse [42]

Answer:

Pmax = 67.5 KN

Explanation:

We need to calculate the maximum allowable value of P for both aluminum and steel bars.

<u>FOR STEEL BARS</u>:

Since,

(σallow)st = (Pmax)st/A

where,

(σallow)st = maximum allowable stress of steel bar = 200 MPa = 2 x 10⁸ Pa

A = Cross-sectional area of steel bar = 450 mm² = 0.45 x 10⁻³ m²

(Pmax)st = Maximum allowable force for steel bar = ?

Therefore,

2 x 10⁸ Pa = (Pmax)st/0.45 x 10⁻³ m²

(Pmax)st = (2 x 10⁸ Pa)(0.45 x 10⁻³ m²)

(Pmax)st = 9 x 10⁴ N = 90 KN

<u>FOR Aluminum BARS</u>:

Since,

(σallow)al = (Pmax)al/A

where,

(σallow)al = maximum allowable stress of Al bar = 150 MPa = 1.5 x 10⁸ Pa

A = Cross-sectional area of Aluminum bar = 450 mm² = 0.45 x 10⁻³ m²

(Pmax)al = Maximum allowable force for Aluminum bar = ?

Therefore,

1.5 x 10⁸ Pa = (Pmax)al/0.45 x 10⁻³ m²

(Pmax)al = (1.5 x 10⁸ Pa)(0.45 x 10⁻³ m²)

(Pmax)al = 6.75 x 10⁴ N = 67.5 KN

Since,

(Pmax)al < (Pmax)st

Therefore,

The maximum allowable force will be:

Pmax = (Pmax)al

<u>Pmax = 67.5 KN</u>

3 0
4 years ago
Find the wavelength (in nm) of a 2.52 eV photon. a. 492.34127 nm.b. 585.88611 nm.c. 541.5754 nm.d. 418.49008 nm.
elena-14-01-66 [18.8K]

Answer:

The correct option is (a).

Explanation:

Given that,

The energy of photon, E = 2.52 eV

We need to find the wavelength of the photon in nm. The formula for the energy of a photon is given by :

E=\dfrac{hc}{\lambda}\\\\\lambda=\dfrac{hc}{E}\\\\\lambda=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{2.52\times 1.6\times 10^{-19}}\\\\=4.93\times 10^{-7}\ m\\\\=493\ nm

The nearest option is a) i.e. 492.34127 nm.

6 0
3 years ago
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