Exposure to what type of radiant energy is sensed by human skin as warmth? x-rays ultraviolet infrared gamma rays

Answers

Answer 1
Answer:

infrared radiant energy is sensed by human skin as warmth. Hence option C is correct.

What is radiation ?

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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Answer 2
Answer:

i think it is infared



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Consider the interactions involved when you use a TV remote control to change the channel. Classify each interaction as long range or short range.

Answers

Explanation:

Following are two interactions that are generally involved when we use a TV remote control to change the channel :

1. Figure touches remote buttons, and its a short range interaction.

2. Now remote sends signal to Television, this is a long range interaction.

Final answer:

The interactions of a TV remote and the TV involve short-range infrared communication, while the TV receives signals from long-range electromagnetic waves broadcasted for channels in frequency ranges for VHF and UHF.

Explanation:

When you use a TV remote control to change the channel, two main interactions are involved. The first interaction is the infrared communication between the remote and the TV, which is a form of electromagnetic radiation. Infrared signals require a direct line of sight, operating over a relatively short range. On the other hand, the TV itself receives broadcast signals through antennas that capture electromagnetic waves broadcasted over a long range - these signals can be VHF or UHF TV channels.

Additionally, the TV channels are broadcasted on frequencies ranging from 54 to 88 MHz and 174 to 222 MHz for VHF, while UHF channels utilize frequencies from 470 to 1000 MHz. These signals are sent over a significant distance to your TV’s antenna, showing that television broadcast interaction is long range. These broadcast signals are part of electromagnetic spectrum and carry a large range of frequencies due to the variety of content (audio and visual information) that needs to be transmitted.

In an experiment to measure the acceleration of g due to gravity, two values, 9.96m/s (s is squared) and 9.72m/s (s is squared), are determined. Find (1) the percent difference of the measurements, (2) the percent error of each measurement, and (3) the percent error of their mean. (Accepted value:g=9.80m/s (s is squared))

Answers

Answer:

a)2.46 %

b)For 1 :101.52 %

For 2 : 99.08 %

c)100..4 %

Explanation:

Given that

g₁ = 9.96 m/s²

g₂ = 9.72 m/s²

The actual value of  g = 9.8 m/s²

a)

The difference Δ g =  9.96 -9.72 =0.24  m/s²

The\ percentage\ difference=(0.24)/(9.72)* 100=2.46\ percentage\n

b)

For first one :

Error\ in\ the\ percentage =(9.96)/(9.81)* 100 =101.52\ perncetage

For second  :

Error\ in\ the\ percentage =(9.72)/(9.81)* 100 =99.08\ perncetage

c)

The mean g(mean )

g(mean )=(9.96+9.72)/(2)\ m/s^2\ng(mean)=9.84\ m/s^2

The\ percentage=(9.84)/(9.8)* 100=100.40\ percentage

a)2.46 %

b)For 1 :101.52 %

For 2 : 99.08 %

c)100..4 %

Final answer:

The percent difference between the two measurements is 2.44%. The percent error for the first measurement is 1.63%, and for the second measurement is 0.82%. The percent error of their mean is 0.41%.

Explanation:

In physics, the percent difference is calculated by subtracting the two values, taking the absolute value, dividing by the average of the two values, and then multiplying by 100. Therefore, the percent difference between the two measurements 9.96m/s² and 9.72m/s² is:

|(9.96-9.72)|/((9.96+9.72)/2)*100 = 2.44%

The percent error involves taking the absolute difference between the experimental value and the accepted value, divided by the accepted value, then multiplied by 100. So, the percent error for the two measurements with accepted value of 9.80m/s² are:

For 9.96m/s²: |(9.96-9.80)|/9.80*100 = 1.63%

For 9.72m/s²: |(9.72-9.8)|/9.8*100 = 0.82%

The percent error of the mean involves doing the above but using the mean of the experimental measurements:

|(Mean of measurements - Accepted value)|/Accepted value * 100 |(9.96+9.72)/2-9.8|/9.8*100 = 0.41%

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The position of a particle changes linearly with time, i.e. as one power of t, as given by the following: h(t) = ( 4.1 t + 5.5 ) meters. Find the speed of the particle, in meters per second.

Answers

Answer:

   v = 4.1 m / s

Explanation:

Velocity is defined by the relation

          v =(dx)/(dt)

 we perform the derivative

         v = 4.1 m / s

Another way to find this magnitude is to see that the velocity on the slope of a graph of h vs t

        v = (\Delta x)/(\Delta t)

        Δx = v Δdt + x₀

        h=   4.1 t + 5.5

       

         v = 4.1 m / s

         x₀ = 5.5 m

The Speed of a Particle is 4.1 meters per second.

The position of a particle can be represented by a linear equation of the form h(t) = (at + b) where a and b are constants.

In this case, the equation is h(t) = (4.1t + 5.5).

To find the speed of the particle, we can take the derivative of the position equation with respect to time.

The derivative of h(t) is the rate of change of position with respect to time, which represents the velocity of the particle.

In this case, the derivative is 4.1 meters per second.

Therefore, the speed of the particle is 4.1 meters per second.

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As the Moon revolves around the Earth, it also rotates on its axis. Why is it that the same side of the Moon is always visible from Earth?

Answers

Answer: The speed of the moon's rotation keeps the same side always facing Earth.

Explanation: Please mark me brainiest

Answer:

The speed of the Moon's rotation keeps the same side always facing Earth.

Explanation:

got it right on study island :)

A Van de Graaff generator is one of the original particle accelerators and can be used to accelerate charged particles like protons or electrons. You may have seen it used to make human hair stand on end or produce large sparks. One application of the Van de Graaff generator is to create x-rays by bombarding a hard metal target with the beam. Consider a beam of protons at 1.90 keV and a current of 4.95 mA produced by the generator. (a) What is the speed of the protons (in m/s)?

Answers

Answer:

603383.67253 m/s

Explanation:

m = Mass of proton = 1.67* 10^(-27)\ kg

K = Kinetic energy = 1.9 keV

1\ ev=1.6* 10^(-19)\ J

Velocity of proton is given by

v=\sqrt{(2K)/(m)}\n\Rightarrow v=\sqrt{(2* 1.9* 10^3* 1.6* 10^(-19))/(1.67* 10^(-27))}\n\Rightarrow v=603383.67253\ m/s

The speed of the protons is 603383.67253 m/s

An object of mass 10.0kg is released at point A, slidesto the bottom of the 30° incline, then collides with ahorizontal massless spring, compressing it a maximumdistance of 0.750m. (See below.) The spring constant is 500N/m, the height of the incline is 2.0 m, and the horizontalsurface is frictionless. (a) What is the speed of the object atthe bottom of the incline? (b) What is the work of frictionon the object while it is on the incline? (c) The springrecoils and sends the object back toward the incline. Whatis the speed of the object when it reaches the base of theincline? (d) What vertical distance does it move back up theincline?

Answers

Final answer:

The final speed at the bottom of the incline can be calculated using the conservation of energy principle. There is no work done against friction as the object is moving on a frictionless surface. The speed does not change when the spring pushes it back towards the base of the incline due to lack of friction and it moves to a certain height given the angle of the incline and the initial speed.

Explanation:

  1. "Speed at the bottom of the incline:" This can be calculated using conservation of energy. The potential energy at the top (m*g*h) will convert into kinetic energy at the bottom (1/2*m*v^2). Here, m is the mass, g is acceleration due to gravity, h is the height, and v is the velocity. Using this, we can solve for v.
  2. Work of friction on the incline: As per the question, the surface is frictionless. Therefore, the work done by friction is automatically 0 as there is no force of friction.
  3. Speed of the object when it reaches the base of the incline again: As the surface is frictionless, the object reaches the incline with the same speed with which it left as there are no opposing forces to reduce its momentum.
  4. Vertical distance it moves back up the incline: This can be calculated using the principles of conservation of energy and kinematic equations, taking into account the angle of the incline and the velocity of the object.

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