Radio waves have low electromagnetic energy compared to X-rays and ultraviolet waves.
Electromagnetic energy refers to the energy associated with electromagnetic waves, which are a form of energy that can travel through empty space. The energy of an electromagnetic wave is directly proportional to its frequency. Therefore, the frequency determines the energy level of the wave.
In the given options, radio waves would have the lowest electromagnetic energy. Radio waves have the longest wavelength and lowest frequency among the three options. X-rays, on the other hand, have a higher frequency and shorter wavelength, making them more energetic. Ultraviolet waves have an even higher frequency and shorter wavelength, making them the most energetic among the three options.
Answer:
Explanation:
a)
d = separation of the slits = 0.30 mm = 0.30 x 10⁻³ m
λ = wavelength of the light = 496 nm = 496 x 10⁻⁹ m
n = order of the bright fringe
D = screen distance = 130 cm = 1.30 m
= Position of nth bright fringe
Position of nth bright fringe is given as
For n = 1
For n = 2
For n = 3
b)
Position of nth dark fringe is given as
For n = 1
For n = 2
For n = 3
infrared radiant energy is sensed by human skin as warmth. Hence option C is correct.
Radiation in physics is the emission or transmission of energy as waves, particles, or both, via space or a material medium.[1][2] This comprises:
electromagnetic radiation, which includes gamma radiation, x-rays, microwaves, infrared, visible light, and ultraviolet radiation
Particle radiation includes beta radiation, proton radiation, neutron radiation, and other particles with non-zero rest energies.
ultrasonography, sound, and seismic waves (reliant on a physical transmission medium) are examples of acoustic radiation.
gravity radiation, which manifests as gravitational waves or ripples in spacetime's curvature
Depending on the energy of the emitted particles, radiation is frequently divided into ionising and non-ionizing categories. More than 10 eV is carried by ionising radiation, which is sufficient to ionise atoms, molecules.
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i think it is infared
Answer:
liquids
Explanation
Answer:
A. The wires exert equal magnitude attractive forces on each other.
Explanation:
Magnetic field due to current i on current 2i
B₁ = 10⁻⁷ x 2 i / r where r is distance between the two wires
Force on wire II due to wire I per unit length
= magnetic field x current in wire II
= B₁ x 2 i
= [ 10⁻⁷ x 2 i / r ] x 2i
= 4 x 10⁻⁷ i² / r
Magnetic field due to current 2i on current i
B₂ = 10⁻⁷ x 4 i / r where r is distance between the two wires
Force on wire I due to wire II per unit length
= magnetic field x current in wire I
= B₂ x i
= [ 10⁻⁷ x 4 i / r ] x i
= 4 x 10⁻⁷ i² / r
So final forces on each wire are same .
This force will be attractive in nature . The direction of force can be known from fleming's right hand rule .
Answer:
An electric current passing through a coil of wire gives a strong form of magnetism called electromagnetism. When the electric current passes through a single straight piece of wire the electromagnetism is weak.
Explanation:
Passing an electric current through a coil of wire generates a magnetic field. The strength of this field can be modified by changing the amount of current or the number of turns in the coil.
When an electric current passes through a coil of wire, as opposed to a straight piece, it creates a magnetic field around the coil. This is the principle behind electromagnets and many electrical appliances we use on a daily basis. The strength of the magnetic field depends on the amount of current and the number of turns in the coil. For example, the more turns the wire has, the stronger the magnetic field.
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B. 1700 Hz, 5100 Hz
C. 3400 Hz, 6800 Hz
D. 3400 Hz, 10,200 Hz
Answer:
B. 1700 Hz, 5100 Hz
Explanation:
Parameters given:
Length of ear canal = 5.2cm = 0.052 m
Speed of sound in warm air = 350 m/s
The ear canal is analogous to a tube that has one open end and one closed end. The frequency of standing wave modes in such a tube is given as:
f(m) = m * (v/4L)
Where m is an odd integer;
v = velocity
L = length of the tube
Hence, the two lowest frequencies at which a dog will have increased sensitivity are f(1) and f(3).
f(1) = 1 * [350/(4*0.052)]
f(1) = 1682.69 Hz
Approximately, f(1) = 1700 Hz
f(3) = 3 * [350/(4*0.052)]
f(3) = 5048 Hz
Approximately, f(3) = 5100 Hz