The speed of a sound wave

1 answer:

5 0

I think the correct answer from the choices listed above is the first option. The speed of a sound wave is slower than an electromagnetic wave or l<span>ight waves are </span><span>faster than sound waves. Hope this answers the question. Have a nice day.</span>

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If a nucleus decays by successive b, a, a emissions, its mass number will Group of answer choices decrease by seven. decrease by

nirvana33 [79]

Answer:

The mass number will decrease by eight (8).

Explanation:

Given;

successive beta (b), alpha (a), alpha (a) emissions.

Generally, when a radioactive element emits a beta-particle (b), its mass number doesn't increase but its atomic number increases by 1 . $(^{0}_{-1}\beta )$

Also when a radioactive element emits an alpha-particle (a), its mass number decreases by 4, while its atomic number decrease by 2. $(^4_2\alpha)$

For the given question, a successive beta (b), alpha (a), and alpha (a) emissions = (0) + (-4) + (-4) = -8

Thus, when a radioactive element emit a successive beta (b), alpha (a), alpha (a) particles, the mass number will decrease by eight (8).

<u>To learn more, please visit:</u> brainly.com/question/23276969

7 0

3 years ago

Which best describes the magnitude of the velocity of the car?

Vlad [161]

I think 30 as that’s usually average speed

6 0

3 years ago

A helix with 17 turns has height H and radius R. Charge is distributed on the helix so that the charge density increases like (i

marishachu [46]

The total charge on the helix is

Q = $\frac{221}{2}\sqrt{(\frac{H}{17})^2+(2\piR)^2$

Height of one turn = $\frac{H}{n}$

Horizontal distance in one turn = $2\pi R$

<h3>length of spring in one turn is</h3>

$l = \sqrt{h^2+(2\piR)^2}\\\\
l = \sqrt{(\frac{H}{n})^2+(2\piR)^2}
$

So, the total length of spring

$L = n*l\\\\
L = n * \sqrt{(\frac{H}{n})^2+(2\piR)^2}
$

Therefore, charge on spring

$Q = \frac{1}{2}*L*(13-0)\\\\
Q = \frac{1}{2}*n\sqrt{(\frac{H}{n})^2+(2\piR)^2}*13\\\\
Q = \frac{1}{2}*17*13\sqrt{(\frac{H}{17})^2+(2\piR)^2}\\\\
Q = \frac{221}{2}\sqrt{(\frac{H}{17})^2+(2\piR)^2\\\\
$