Physics College

Which one of the following types of electromagnetic radiation is produced by the sudden deceleration of high speed electrons?
a.x-rays
b.microwaves
c.infrared radiation
d.visible light
e.gamma rays

Answers

Answer 1

The correct answer is a. x-rays is produced by the sudden deceleration of high speed electrons.

What is x-rays?

When high-speed electrons are suddenly decelerated or slowed down, they release energy in the form of electromagnetic radiation. This process is known as bremsstrahlung or "braking radiation". The energy of the emitted radiation depends on the initial speed of the electrons and the degree of deceleration.

In the case of bremsstrahlung, the emitted radiation can range from radio waves to gamma rays, but the highest energy radiation produced by bremsstrahlung is x-rays. Therefore, the sudden deceleration of high-speed electrons produces x-rays.

X-rays are ionizing radiation, meaning that they have enough energy to remove electrons from atoms or molecules, which can cause damage to living tissue. Therefore, exposure to X-rays should be limited and controlled to minimize health risks.

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Complete question is: x-rays is produced by the sudden deceleration of high speed electrons.

Related Questions

3. Large amplitude vibrations produced when the of receiver of the applied forced vibration matches the

Answers

An object's amplitude dramatically increases when the frequency of the applied forced vibrations matches the object's natural frequency. Resonance describes this behavior.

Theory A wave's amplitude directly relates to the quantity of energy it can carry. A wave with a high amplitude carries a lot of energy, whereas one with a low amplitude carries only a little. A wave's strength is determined by the typical energy that moves through a given area in a certain amount of time and in a particular direction.The sound wave's amplitude grows in proportion to its strength. We perceive louder noises to be of higher intensity. Comparative sound intensities are frequently expressed using decibels (dB)

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A metal wire, fixed at one end, has length l and cross-sectional area A. The wire extends a distance e which mass m is hung from the other end of the wire.What is an expression for the Young Modulus E of the metal?

Answers

The expression for the Young Modulus E of the metal is E = mgl / Ae. The Young Modulus E of the metal is calculated using the equation  E = (F l) / (A e2 m), where F is the force applied to the wire.


To find the expression for the Young modulus E of a metal wire with length l, cross-sectional area A, and mass m hung from the other end of the wire, we need to use the following formula:Stress (σ) = Load (F) / Area (A)Strain (ε) = Extension (Δl) / Original length (l)Young Modulus (E) = Stress (σ) / Strain (ε)We know that the metal wire is fixed at one end and the wire extends a distance e when a mass m is hung from the other end of the wire. Therefore, the extension Δl is equal to e.

Let's assume that g is the acceleration due to gravity. Therefore, the load F is equal to m * g.Substituting the values of F, A, and Δl in the above formula, we get:Stress (σ) = F / A = (m * g) / AStrain (ε) = Δl / l = e / lYoung Modulus (E) = Stress (σ) / Strain (ε)= (m * g) / (A * e / l) = mgl / AeTherefore, an expression for the Young Modulus E of the metal is E = mgl / Ae.

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125cm³ of a gas was collected at 15 °C and 755 mm of mercury pressure. Calculate the volume of the gas that will be collected at standard temperature and pressure

Answers

Answer:

119,2 см³

Explanation:

по формуле Клопейрона (P1×V1):T1=(P2×V2):T2

если из этой формулы найти V2, ответ будет равен примерно на 119,2 см³

Estimat the number and wattage of lamps. which would be required to illuminate a workshop space 60x1.5 meteres by means of lamps mounted 5 metres above the working Plane The average illumination required is about 100 wt. coefficient of utilisation = 0.4 luminous efficiency 16 lumens per watt. Assume a space-height ratio of unity and a cundle Power depreciation of 20%​

Answers

The number and wattage of lamps required to illuminate the workshop would be approximately 8 lamps and 70 watts respectively.

Wattage calculation

To estimate the number and wattage of lamps required to illuminate a workshop space of 60x1.5 meters, we can follow these steps:

Calculate the area of the workshop:

Area = length x widthArea = 60m x 1.5mArea = 90 square meters

Determine the total lumens required:

Lumens = area x average illuminationLumens = 90 sq m x 100 luxLumens = 9000 lumens

Adjust for the coefficient of utilization and luminous efficiency:

Effective lumens = lumens / (coefficient of utilization x luminous efficiency)Effective lumens = 9000 / (0.4 x 16)Effective lumens = 1406.25 lumens

Adjust for space-height ratio and candle power depreciation:

Effective lumens per lamp = effective lumens x space-height ratio x (1 - depreciation)Effective lumens per lamp = 1406.25 x 1 x (1 - 0.2)Effective lumens per lamp = 1125 lumens

Determine the number of lamps required:

Number of lamps = total lumens required / effective lumens per lampNumber of lamps = 9000 / 1125Number of lamps = 8 lamps (rounded up)

Determine the wattage of each lamp:

Wattage per lamp = effective lumens per lamp / luminous efficiencyWattage per lamp = 1125 / 16Wattage per lamp = 70.3 watts (rounded up)

Therefore, approximately 8 lamps with a wattage of 70 watts each would be required to illuminate the workshop space.

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