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alisha [4.7K]
3 years ago
13

Which waves in the electromagnetic spectrum are considered high energy waves?. . A.radio waves and microwaves. B.X–rays and gam

ma rays. C.sound waves and light waves. D.microwaves and X–rays
Physics
2 answers:
LenaWriter [7]3 years ago
8 0

Answer:

The answer would be<u> Gamma rays and X-rays </u>

Explanation:

<em>The different types of radiation are defined by the the amount of energy found in the photons. Radio waves have photons with low energies, microwave photons have a little more energy than radio waves, infrared photons have still more, then visible, ultraviolet, X-rays, and, the most energetic of all, gamma-rays.</em>

<em> hope it helps </em>

<em>Mark as branliest please</em><em> </em>

Sergeu [11.5K]3 years ago
3 0
The answer is B. X-rays and gamma rays. The electromagnetic spectrum consists of all types of electromagnetic radiation. The EM spectrum is arranged from the longest wavelength with the lowest frequency and lowest energy which are the radio waves and microwaves to the shortest wavelength with the highest frequency and highest energy which are the x-rays and gamma rays. Visible light that can be seen by a humans is a small portion in between the ultraviolet and infrared rays.
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A wave has a wavelength of 15 m and a frequency of 10 Hz. What is the speed of this wave?
choli [55]
M/s=hz•m 10hz•15m=150m/s
5 0
3 years ago
The wires in a household lamp cord are typically 3.5 mm apart center to center and carry equal currents in opposite directions.
zhuklara [117]

Explanation:

It is given that,

Distance between wires, d = 3.5 mm = 0.0035 m

Power of light bulb, P = 100 W

Potential difference, V = 120 V

(a) We need to find the force per unit length each wire of the cord exert on the other. It is given by :

\dfrac{F}{l}=\dfrac{\mu_o I^2}{2\pi r}

Power, P = V × I

I=\dfrac{P}{V}=\dfrac{100}{120}=0.83\ A

This gives, \dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times (0.83)^2}{2\pi \times 0.0035}

\dfrac{F}{l}=0.0000393\ N/m

\dfrac{F}{l}=3.93\times 10^{-5}\ N/m

(b) Since, the two wires carry equal currents in opposite directions. So, teh force is repulsive.

(c) This force is negligible.

Hence, this is the required solution.

8 0
3 years ago
A tank of volume 0.25 ft is designed to contain 50 standard cubic feet of air. The temperature is 80° F.Calculate the pressure i
slavikrds [6]

Answer:

The inside Pressure of the tank is 4499.12 lb/ft^{2}

Solution:

As per the question:

Volume of tank, V = 0.25 ft^{3}

The capacity of tank, V' = 50ft^{3}

Temperature, T' = 80^{\circ}F = 299.8 K

Temperature, T = 59^{\circ}F = 288.2 K

Now, from the eqn:

PV = nRT                      (1)

Volume of the gas in the container is constant.

V = V'

Similarly,

P'V' = n'RT'                       (2)

Also,

The amount of gas is double of the first case in the cylinder then:

n' = 2n

\]frac{n'}{n} = 2

where

n and n' are the no. of moles

Now, from eqn (1) and (2):

\frac{PV}{P'V'} = \frac{nRT}{n'RT'}

P' = 2P\frac{T'}{T} = 2\times 2116\times \frac{299.8}{288.2} = 4499.12 lb/ft^{2}

7 0
3 years ago
When the distance between two charges is halved, the electrical force between them?
Llana [10]
If the distance between two charges is halved, the electrical force between them increases by a factor 4.

In fact, the magnitude of the electric force between two charges is given by:
F= k \frac{q_1 q_2}{r^2}
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:
r'= \frac{r}{2}
the magnitude of the force changes as follows:
F'=k \frac{q_1 q_2}{r'^2}=k \frac{q_1 q_2}{( \frac{r}{2})^2 }=k \frac{q_1 q_2}{ \frac{r^2}{4} } =4k  \frac{q_1 q_2}{r^2}=4 F
so, the force increases by a factor 4.
3 0
3 years ago
Skylar goes to a pumpkin patch and pick out a pumpkin that has a mass of 6000 grams how many kilograms is the pumpkin
Alex
1000g = 1kg
6000g =6000 × 1/1000
=6kg
6 0
2 years ago
Read 2 more answers
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