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3 years ago

90 percent if amphibians are...? (Choose all that apply) a. Frogs b. Toads c. Newts d. Salamanders e. Caecillans

Physics

1 answer:

Vika [28.1K]3 years ago

7 0

Answer:

frogs, toads, salamanders, and caecillans

Explanation:

About 90 percent of all amphibians are anurans (Table 1). Table 1. Number of genera, species, and world distribution of frogs, toads, salamanders, and caecilians.

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Net force = ?

Snezhnost [94]

Answer:

A. the total net force is 14

B. the total net force is 15

Explanation:

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3 years ago

Suppose you could fit 100 dimes, end to end, between your card with the pinhole and your dime-sized sunball. how many suns could

Naddika [18.5K]

Answer: 100 suns

Explanation:

We can solve this with the following relation:

$\frac{d}{x_{sunball-pinhole}}=\frac{D}{x_{sun-pinhole}}$

Where:

$d=17.91 mm =17.91(10)^{-3} m$ is the diameter of a dime

$D$ is the diameter of the Sun

$x_{sun-pinhole}=150,000,000 km=1.5(10)^{11} m$ is the distance between the Sun and the pinhole

$x_{sunball-pinhole}=100 d=1.791 m$ is the amount of dimes that fit in a distance between the sunball and the pinhole

Finding $D$ :

$D=\frac{d}{x_{sunball-pinhole}}x_{sun-pinhole}$

$D=\frac{17.91(10)^{-3} m}{1.791 m} 1.5(10)^{11} m$

$D=1.5(10)^{9} m$ This is roughly the diameter of the Sun

Now, the distance between the Earth and the Sun is one astronomical unit (1 AU), which is equal to:

$1 AU=149,597,870,700 m$

So, we have to divide this distance between $D$ in order to find how many suns could it fit in this distance:

$\frac{149,597,870,700 m}{1.5(10)^{9} m}=99.73 suns \approx 100 suns$

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3 years ago

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Answer:

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2 years ago