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4 years ago

When a potassium atom forms an ion, it loses one electron. What is the electrical charge of the potassium ion? *

2 answers:

5 0

+1 An electron has a negative charge so losing a charge of -1 from an uncharged, or neutral, atom will leave an ion with a positive charge.

IceJOKER [234]4 years ago

5 0

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Answer:

The correct answer is answer choice D. +1

Explanation:

Since electrons have negative charges, losing one electron will cause the atom to have a positive charge of 1. This charge comes from the protons, which, until one electron was lost, balanced out the negative charge of the electrons and caused the atom to be neutral.

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Hope I helped!!!

Best of luck deary...

<3

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If a sound is 30 dB and its absolute pressure was 66 x 10-9 Pa, what must have been the reference pressure?

Ksenya-84 [330]

Given:

I = 30dB

P = 66 × $10^{-9}$ Pa

Solution:

Formula used:

I = $20\log_{10}(\frac{P}{P_{o}})$ (1)

where,

I = intensity of sound

P = absolute pressure

$P_{o}$ = reference pressure

Using Eqn (1), we get:

$30 = 20\log _{10}\frac{66\times 10^{-9}}{P^{o}}$

$P_{o}$ = $\frac{66\times 10^{-9}}{10^{1.5}}$

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3 years ago

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A block attached to a spring with an unknown spring constant oscillates with a period of 2.0 s. What is the period if

Zigmanuir [339]

Answer:

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

Explanation:

The statement is incomplete. We proceed to present the complete statement: A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if a. The mass is doubled? b. The mass is halved? c. The amplitude is doubled? d. The spring constant is doubled?

We have a block-spring system, whose angular frequency ( $\omega$ ) is defined by the following formula: