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Tema [17]
3 years ago
15

Which is equal to 4 hectometers?

Physics
1 answer:
Debora [2.8K]3 years ago
8 0

Answer:

2. 0.4 kilometers

Explanation:

Hope it helps

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Pls help me l will make it brainlest ​
nikklg [1K]

Answer:

<u>0.04 °C⁻¹</u>

Explanation:

First, we need to calculate linear expansivity, then after finding that value, we can move on to finding the area expansivity.

<u />

=============================================================

Finding Linear Expansivity :

⇒ α = Final length - Original length / (Original length × ΔT)

⇒ α = 9 - 4 / (4 × 70 - 20)

⇒ α = 5 / 5 × 50

⇒ α = <u>0.02</u>

============================================================

Finding Area Expansivity :

⇒ Area Expansivity = 2 × Linear Expansivity

⇒ β = 2 × α

⇒ β = 2 × 0.02

⇒ β = <u>0.04 °C⁻¹</u>

3 0
1 year ago
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Scientists might make a computer model of volcanic eruptions. What is the
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Answer:

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7 0
1 year ago
The term used to describe the quantity of matter that a body possesses is:
barxatty [35]
The term used to describe the quantity of matter that a body possesses is mass.
7 0
3 years ago
A disk rotates around an axis through its center that is perpendicular to the plane of the disk. The disk has a line drawn on it
natka813 [3]

Answer:

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Explanation:

The rotated angle is given by:

\theta=\omega0*t+1/2*\alpha*t^2

Since this is a quadratic equation it can be solved using:

x=\frac{-b \± \sqrt{b^2-4*a*c}  }{2*a}

Rewriting our equation:

1/2*\alpha*t^2+\omega0*t-\theta=0

t = \frac{\±\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

Since \sqrt{\omega0^2+2*\theta*\alpha} >\omega0 we discard the negative solution.

t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}

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Where is visible light located on the electromagnetic spectrum and why?
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Radio waves, gamma-rays, visible light, and all the other parts of the electromagnetic spectrum are electromagnetic ... The different types of radiation are defined by the the amount of energy found in the photons.
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