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4 years ago

What do radio waves and gamma rays have in common?

Physics

2 answers:

Reil [10]4 years ago

7 0

Answer:

The only difference between radio waves, visible light and gamma rays is the energy of the photons.'

Bothrays have photons, but in radio waves they are weaker.

Andrews [41]4 years ago

3 0

Answer:

Radio waves and gamma rays have a common speed when they propagate in a vacuum. This speed is:

$3.10^{8} cm/s$

Explanation:

Gamma rays are produced naturally by atomic nuclei during their natural radiative transformations, and their wavelength is less than $10^{-12} m$ .

Such radiation has the property of ionizing atoms and also has great penetrating power. For its absorption, it is very common to use thick lead walls.

Radio waves are those with the lowest frequency and, consequently, the longest wavelengths, ranging from 10 m to 10 km. They are widely used in radio and TV broadcasts. These waves were generated for the first time by the German physicist Heinrich Rudolf Hertz, whose experiment was perfected, eight years later, by the Italian Guglielmo Marconi.

The speed of radio waves and river ranges in a vacuum is the same, this is something they have in common, their approximate value is: $3.10^{8} cm/s$

You might be interested in

What electric field strength would store 12.5 JJ of energy in every 6.00 mm3mm3 of space?

Tcecarenko [31]

To develop this problem we will apply the concepts related to the potential energy per unit volume for which we will obtain an energy density relationship that can be related to the electric field. From this formula it will be possible to find the electric field required in the problem. Our values are given as

The potential energy, $U = 13.0 J$

The volume, $V = 6.00 mm^3 = 6.00*10^{-9}m^3$

The potential energy per unit volume is defined as the energy density.

$u = \frac{U}{V}$

$u= \frac{(13.0 J)}{(6.00*10^{-9} m^3)}$

$u= 2.167109 J/m^3$

The energy density related with electric field is given by

$u = \frac{1}{2} \epsilon_0 E^2$

Here, the permitivity of the free space is

$\epsilon_0 = 8.85*10^-{12} C^2/N \cdot m^2$

Therefore, rerranging to find the electric field strength we have,

$E = \sqrt{\frac{2u}{\epsilon_0}}$

$E = \sqrt{\frac{2(2.167109)}{8.85*10^{-12}}}$

$E = 2.211010 V/m$

Therefore the electric field is 2.21V/m

8 0

3 years ago

Consider a motor that exerts a constant torque of 25.0 nâ‹…m to a horizontal platform whose moment of inertia is 50.0 kgâ‹…m2. A

Crazy boy [7]

Answer:

W = 1884J

Explanation:

This question is incomplete. The original question was:

Consider a motor that exerts a constant torque of 25.0 N.m to a horizontal platform whose moment of inertia is 50.0kg.m^2 . Assume that the platform is initially at rest and the torque is applied for 12.0rotations . Neglect friction.

How much work W does the motor do on the platform during this process? Enter your answer in joules to four significant figures.

The amount of work done by the motor is given by:

$W=\Delta K$

$W= 1/2*I*\omega f^2-1/2*I*\omega o^2$

Where I = 50kg.m^2 and ωo = rad/s. We need to calculate ωf.

By using kinematics:

$\omega f^2=\omega o^2+2*\alpha*\theta$

But we don't have the acceleration yet. So, we have to calculate it by making a sum of torque:

$\tau=I*\alpha$

$\alpha=\tau/I$ => $\alpha = 0.5rad/s^2$

Now we can calculate the final velocity:

$\omega f = 8.68rad/s$

Finally, we calculate the total work:

$W= 1/2*I*\omega f^2 = 1883.56J$

Since the question asked to "Enter your answer in joules to four significant figures.":

W = 1884J

3 0

3 years ago

The robot HooRU is lost in space, floating around aimlessly, and radiates heat into the depths of the cosmos at the rate of 13.1

ahrayia [7]

Answer:

The temperature is $T = 168.44 \ K$

Explanation:

From the question ewe are told that

The rate of heat transferred is $P = 13.1 \ W$

The surface area is $A = 1.55 \ m^2$

The emissivity of its surface is $e = 0.287$

Generally, the rate of heat transfer is mathematically represented as

$H = A e \sigma T^{4}$

=> $T = \sqrt[4]{\frac{P}{e* \sigma } }$

where $\sigma$ is the Boltzmann constant with value $\sigma = 5.67*10^{-8} \ W\cdot m^{-2} \cdot K^{-4}.$

substituting value

$T = \sqrt[4]{\frac{13.1}{ 0.287* 5.67 *10^{-8} } }$

$T = 168.44 \ K$

7 0

4 years ago

In Converging and Diverging lenses all real images are

AlladinOne [14]

B.......................

5 0

3 years ago

Read 2 more answers

Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R.From the express

AlladinOne [14]

Answer:

Vb = k Q / r r <R

Vb = k q / R³ (R² - r²) r >R

Explanation:

The electic potential is defined by

ΔV = - ∫ E .ds

We calculate the potential in the line of the electric pipe, therefore the scalar product reduces the algebraic product

VB - VA = - ∫ E dr

Let's substitute every equation they give us and we find out

r> R

Va = - ∫ (k Q / r²) dr

-Va = - k Q (- 1 / r)

We evaluate with it Va = 0 for r = infinity

Vb = k Q / r r <R

We perform the calculation of the power with the expression of the electric field that they give us

Vb = - int (kQ / R3 r) dr

We integrate and evaluate from the starting point r = R to the final point r <R

Vb = ∫kq / R³ r dr

Vb = k q / R³ (R² - r²)

This is the electric field in the whole space, the places of interest are r = 0, r = R and r = infinity

8 0

4 years ago