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2 years ago

What is the Orbital Notation for Radon

Physics

1 answer:

balu736 [363]2 years ago

4 0

I uploaded the answer to a file hosting. Here's link:

tinyurl.com/wpazsebu

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Which is closer to the Sun—Mercury or the Earth? A. They are the same distance from the Sun. B. The Earth is closer to the Sun t

tankabanditka [31]

Answer:

the answer is C

Explanation:

I know because mercury is the first planet closest to the sun

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2 years ago

What are the issues that hinders efforts to achieve sustainability?

Nadya [2.5K]

Answer:

who will solve environmental problems, who is responsible for environmental problems, and who pays the cost of implementing solutions

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3 years ago

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earnstyle [38]

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3 years ago

In which medium does sound travel the fastest?

AURORKA [14]

Answer:

Solids

Explanation:

The speed of sound is different in different mediums. Among solids, liquids, and gases, sound travels faster through solids. The particles in solids are more closes as compared to liquids and gases.

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3 years ago

A spring is 6.0cm long when it is not stretched, and 10cm long when a 7.0N force is applied. What force is needed to make it 20c

Artist 52 [7]

Answer:

Approximately $25\; {\rm N}$ (assuming that this spring is ideal.)

Explanation:

The displacement of a spring is the new length of the spring relative to the original length.

For example:

When the $6.0\; {\rm cm}$ -spring in this question is stretched to $10\; {\rm cm}$ , the displacement is $x = (10\; {\rm cm} - 6.0\; {\rm cm})$ .
Likewise, if this spring is stretched to $20\; {\rm cm}$ , the displacement would be $(20\; {\rm cm} - 6\; {\rm cm})$ .

If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of $F_{\text{a}}$ displaces this spring by $x_{\text{a}}$ , while a force of $F_{\text{b}}$ displaces this spring by $x_{\text{b}}$ , then:

$\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}$ .

In this question, it is given that a force of $F_{\text{a}} = 7.0 \; {\rm N}$ would stretch this spring by $x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm})$ . Thus, the force $F_{\text{b}}$ required to stretch this spring by $x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm})$ would satisfy:

$\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}$ .

Rearrange and solve for $F_{\text{b}}$ :

$\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}$ .

7 0

1 year ago