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Kamila [148]
2 years ago
9

Calculate the energy in electron volts of X-rays that have a frequency of 3.00 x 100 Hz.

Physics
1 answer:
inn [45]2 years ago
8 0

Answer:

E = 124 eV

Explanation:

Given,

The frequency of the X-rays, ν = 3.0 x 10¹⁰ Hz

The formula for calculating the energy of the electromagnetic waves of a single photon of having frequency 'ν' is given by the relation

                                    E = hν      joules

Where,

                       h - Planck's constant (6.626 x 10⁻³⁴  Js)

Substituting the values in the above equation

                                    E = 6.626 x 10⁻³⁴  Js   x  3.0 x 10¹⁰ Hz

                                       = 1.9878 x 10⁻²³  J

Converting it into eV

                                    E = 1.9878 x 10⁻²³  x 6.242 x 10¹⁸    eV

                                       = 124 eV

Therefore, the energy in electron volts of X-rays is, E = 124 eV

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scZoUnD [109]

Answer:

v = 2.5 mi/h

Explanation:

The velocity of an object is defined as the time rate of change of its position. In other words the velocity of an object is just a ration of displacement covered by the object to the time taken by the object, to cover that distance. So, in this case we will use the same formula or equation to calculate the velocity of the man. The formula is as follows:

v = s/t

where,

v = velocity of the man = ?

s = distance of the trail = 5 miles

t = time required to complete the trail = 2 h

Therefore,

v = 5 miles/ 2 h

<u>v = 2.5 mi/h</u>

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3 years ago
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Elanso [62]
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4 0
2 years ago
the positive particle has a charge of 31.7 mC and the particles are 2.80 mm apart, what is the electric field at point A located
vichka [17]

Answer:

the electric field at point A is

E = 5.5 ×10¹³N/C(-x direction)

Explanation:

given

electrostatics constant k = 9.0×10⁹

charge q = 31.7mC= 31.7×10⁻³C

distance r = 2.80mm

distance from midpoint to point A = 2.00mm

attached is the diagram of the solution, describing the position of the charge

note x = r/2, where x is the distance from midpoint of r to the particle

using Pythagoras theorem as in the attachment, x = 2.44mm= 2.44×10⁻³m

the electric field at point A is given as

vector <em>E </em>= 2E×cos θ( -x direction)

recall E =kq/x²

where k is the electrostatics constant = 1/4πε₀

where ε₀ is permittivity of free space

therefore using E =2{kq/x²}cosθ

∴cosθ = adjacent/hypotenuse

cosθ=1.40/2.44

E =2 {(9.0×10⁹ × 31.7×10⁻³) ÷ (2.44×10⁻³)²}×(1.40/2.44)(-x)

E=2{4.79×10¹³}×(0.574)(-x)

E = 2×2.75 ×10¹³N/C(-x direction)

Vector <em>E= </em>5.5 ×10¹³N/C(-x direction)

3 0
3 years ago
A common physics demonstration is to drop a small magnet down a long, vertical aluminum pipe. Describe the motion of the magnet
Rzqust [24]

Answer and Explanation:

This experiment is known as Lenz's tube.

The Lenz tube is an experiment that shows how you can brake a magnetic dipole that goes down a tube that conducts electric current. The magnet, when falling, along with its magnetic field, will generate variations in the magnetic field flux within the tube. These variations create an emf induced according to Faraday's Law:

\varepsilon =-\frac{d\phi_B}{dt}

This emf induced on the surface of the tube generates a current within it according to Ohm's Law:

V=IR

This emf and current oppose the flux change, therefore a field will be produced in such a direction that the magnet is repelled from below and is attracted from above. The magnitude of the flux at the bottom of the magnet increases from the point of view of the tube, and at the top it decreases. Therefore, two "magnets" are generated under and above the dipole, which repel it below and attract above. Finally, the dipole feels a force in the opposite direction to the direction of fall, therefore it falls with less speed.

7 0
3 years ago
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The force experienced by a unit test charge is a measure of the strength of an electric:
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