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andrew11 [14]
3 years ago
7

What does saturns core look like?

Physics
1 answer:
HACTEHA [7]3 years ago
7 0

Answer:

At Saturn's center is a dense core of metals like iron and nickel surrounded by rocky material and other compounds solidified by the intense pressure and heat. It is enveloped by liquid metallic hydrogen inside a layer of liquid hydrogen—similar to Jupiter's core but considerably smaller

Explanation:

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Boyle's law says that the volume of a gas varies inversely with the pressure. When the volume of a certain gas is 4l , the press
Anna11 [10]

Boyle's law says that the volume of a gas varies inversely with the pressure. When the volume of a certain gas is 4l , the pressure is 720 kpa (kilopascals). What is the pressure when the volume is 10l ?

7 0
3 years ago
Find acceleration. Will give brainliest!
zalisa [80]

Answer:

16200 km/s

270 km/min

4.5 km/h

Explanation:

Acceleration Formula: Average Acceleration = Δv/Δt (change in velocity over change in time)

Simply plug in our known variables and solve:

a = (45.0 - 0)/10

a = 45.0/10

a = 4.5 km/h

7 0
3 years ago
Read 2 more answers
What is the power through a device with a resistance of 100 ohms if a current of 8 A is running through it?
Kazeer [188]

Answer:

6400 W (or) 6.4 KW

Explanation:

Formula we use,

→ P = I²R

Let's solve for the power of device,

→ P = I²R

→ P = (8)² × 100

→ P = 64 × 100

→ [ P = 6400 W ]

Hence, the power is 6400 W.

5 0
2 years ago
Which of the following does each different kind of atom represent?
r-ruslan [8.4K]

Answer:

B

Explanation:

because atoms make up an element.

5 0
3 years ago
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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

8 0
3 years ago
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