All categories

4 years ago

Which symbol represents a type of electromagnetic radiation released during radioactive decay?

Physics

1 answer:

Tanya [424]4 years ago

8 0

The answer for <span>electromagnetic radiation released during radioactive decay i</span>s C. He

You might be interested in

What is the tangential velocity of a record player which makes 11 revolutions in 20 seconds?

Vsevolod [243]

What is the tangential velocity of a record player which makes 11 revolutions in 20 seconds? Help hello

3 0

3 years ago

Which type of energy is stored in chemical compounds and released in chemical reactions? chemical energy potential energy kineti

Vilka [71]

Answer:

The answer is potential energy.

Explanation:

The reason is because chemical energy is stored in bonds that connects atoms with other atoms and molecules. It is in form of potential energy, when a chemical reaction takes place, the stored chemical energy is released

7 0

3 years ago

A hollow non-conducting spherical shell has inner radius R1 = 5 cm and outer radius R2 = 19 cm. A charge Q = -35 nC lies at the

Gnom [1K]

Answer:

a. E = -13.8 kN/C

b. E = +8.51 kN/C

Explanation:

We will apply Gauss' Law to the regions where the electric field is asked.

Gauss' Law states that if you draw an imaginary surface enclosing a charge distribution, then the electric field through the imaginary surface is equal to the total charge enclosed by this surface divided by electric permittivity.

$\int\vec{E}d\vec{a} = \frac{Q_{\rm enc}}{\epsilon_0}$

a. For this case, we will draw the imaginary surface between the inner and outer shell of the sphere. The total charge enclosed by this surface will be equal to the sum of the charge Q at the center and charge of the shell within the volume from R1 and r.

Here, r = 0.5(R1+R2) = 12 cm.

$E4\pi r^2 = \frac{Q_{\rm enc}}{\epsilon_0}\\Q_{\rm enc} = Q + \rho V_{\rm enc} = Q + (Ar) (\frac{4}{3}\pi (r^3 - R_1^3)) = (-35\times 10^{-9}) + (16\times 10^{-6})(12\times 10^{-2})(\frac{4}{3}\pi((12\times 10^{-2})^3 - (5\times 10^{-2})^3)) = -2.21\times 10^{-8}~C$

$E = \frac{-2.21\times 10^{-8}}{4\pi (12\times10^{-2})^2 \epsilon_0} = -1.38\times 10^4~N/C\\E = -13.8~kN/C$

b. For this case, we will draw the imaginary surface on the outside of the shell.

The total charge enclosed by this surface will be equal to the sum of the charge at the center and the total charge of the shell.

$Q_{\rm enc} = Q + \rho V = Q + (Ar)[\frac{4}{3}\pi (R_2^3 - R_1^3)]\\Q_{\rm enc} = (-35\times 10^{-9}) + [(16\times 10^{-6})(38\times 10^{-2})][\frac{4}{3}\pi((19\times 10^{-2})^3 - (5\times 10^{-2})^3)]\\Q_{\rm enc} = 1.36\times 10^{-7}~C$

$E = \frac{1.36\times 10^{-7}}{4\pi (38\times10^{-2})^2 \epsilon_0} = 8.51\times 10^3~N/C\\E = 8.51~kN/C$

7 0

3 years ago

Consider two identical wave pulses on a rope having a fixed end. Suppose the first pulse reaches the end of the rope, is reflect

jek_recluse [69]

Answer:

zero

Explanation:

Consider two identical wave pulses on a rope having a fixed end. Suppose the first pulse reaches the end of the rope, is reflected back, and then meets the second pulse, both waves will be out of phase by π radians. Therefore, they form destructive interference and hence the amplitude of the resultant pulse would be zero.

4 0

3 years ago

Why are Physical changes Not represented by Equations

solmaris [256]

Badadadadadadadadaradad

6 0

3 years ago

Read 2 more answers