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3 years ago

How do we feel the energy of infrared waves?

Physics

2 answers:

seropon [69]3 years ago

8 0

We feel infrared waves from the heat that they produce

maks197457 [2]3 years ago

6 0

When an object is not quite hot enough to radiate visible light, it will emit most of its energy in the infrared. For example, hot charcoal may not give off light but it does emit infrared radiation which we feel as heat. The warmer the object, the more infrared radiation it emits.

hope this helped :)
alisa202

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A sledgehammer strikes an anvil with a velocity of 50 ft/sec (Fig. 2.59). The hammer and the anvil weigh 12 lb and 100 lb, respe

saveliy_v [14]

Hard question thx for the points give me brainlest points plz

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3 years ago

An x-ray photon is scattered by an originally stationary electron. how does the frequency of the scattered photon compare relati

Viefleur [7K]

The frequency of the scattered photon decreases or it will be lower compare to the frequency of incident photon. An x-ray photon scatters in one direction after a collision and some energy is transferred to the electron as it recoils in another direction resulting to have less energy in the scattered photon. In addition, the frequencies will also depend on the differences of the angle at which the scattered photon leaves the collision and this incident is called Compton Effect.

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4 years ago

A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w

trapecia [35]

To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

A) Conservation of Energy,

$KE = PE$

$\frac{1}{2} mv ^2 = \frac{1}{2} k A^2$

Here,

m = Mass

v = Velocity

k = Spring constant

A = Amplitude

Rearranging to find the Amplitude we have,

$A = \sqrt{\frac{mv^2}{k}}$

Replacing,

$A = \sqrt{\frac{(0.750)(31*10^{-2})^2}{13}}$

$A = 0.0744m$

(B) For this part we will begin by applying the concept of Period, this in order to find the speed defined in the mass-spring systems.

The Period is defined as

$T = 2\pi \sqrt{\frac{m}{k}}$

Replacing,

$T = 2\pi \sqrt{\frac{0.750}{13}}$

$T= 1.509s$

Now the velocity is described as,

$v = \frac{2\pi}{T} * \sqrt{A^2-x^2}$

$v = \frac{2\pi}{T} * \sqrt{A^2-0.75A^2}$

We have all the values, then replacing,

$v = \frac{2\pi}{1.509}\sqrt{(0.0744)^2-(0.750(0.0744))^2}$

$v = 0.2049m/s$

7 0

3 years ago

Many web sites describe how to add wires to your clothing to keep you warm while riding your motorcycle. The wires are added to

Tomtit [17]

Answer:

$P=42.075W$

Explanation:

The power provided by a resistor (wire in this case) is given by:

$P=\frac{V^2}{R}$ .

The resistance of a wire is given by:

$R=\frac{\rho L}{A}$

Where for the resistivity the one of the copper should be used: $\rho=1.68\times10^{-8}\Omega m$ .

The area A is that of a circle, which written in terms of its diameter is:

$A=\pi r^2=\pi (d/2)^2=\frac{\pi d^2}{4}$

Putting all together:

$P=\frac{AV^2}{\rho L}=\frac{\pi d^2V^2}{4\rho L}$

Which for our values is:

$P=\frac{\pi (0.00025m)^2(12V)^2}{4(1.68\times10^{-8}\Omega m)(10m)}=42.075W$

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4 years ago

The electric field everywhere on the surface of a thin, spherical shell of radius 0.770 m is of magnitude 860 N/C and points rad

11111nata11111 [884]

Answer:

(a) $Q = 7.28\times 10^{14}$

(b) The charge inside the shell is placed at the center of the sphere and negatively charged.

Explanation:

Gauss’ Law can be used to determine the system.

$\int{\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E4\pi r^2 = \frac{Q_{enc}}{\epsilon_0}\\(860)4\pi(0.77)^2 = \frac{Q_{enc}}{8.8\times 10^{-12}}\\Q_enc = 7.28\times 10^{14}$

This is the net charge inside the sphere which causes the Electric field at the surface of the shell. Since the E-field is constant over the shell, then this charge is at the center and negatively charged because the E-field is radially inward.

The negative charge at the center attracts the same amount of positive charge at the surface of the shell.

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4 years ago