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Oksi-84 [34.3K]
4 years ago
10

Which material is a gas

Physics
1 answer:
Aneli [31]4 years ago
6 0

h, it has the lowest density so it makes sense it is the gas.

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A transformer is designed to provide 6 V from a 150 V supply. If the primary coil has 1000 turns, how many turns does the second
zavuch27 [327]
The applicable relationship is N1/N2 = V1/V2, meaning the ratio of primary voltage to secondary voltage is equal to the ratio of primary turns to secondary turns.

Here N1 = 1000, V1 = 250, V2 = 400V and N2 = TBD.

Rewriting the above relationship, N2 = N1 V2/V1 = 1000 x 400/250 = 1600 turns.
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3 years ago
Please help
NikAS [45]
D, Metamorphism
-Metamorphic rocks are any rocks that change into another rock when subjected to high heat and pressure.
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Which statement best describes the formation of igneous rocks?
Cerrena [4.2K]

b. describes it best because a form of molten cools down and makes igneous rock. Good luck! Please mark me brainliest!


8 0
4 years ago
Read 2 more answers
From the deepest to the surface, what are the parts of the earth's interior?
shepuryov [24]

The answer is A: Core --> Mantle --> Crust.

Core: The earth's core is the center of the earth, which would ultimately be the deepest. The core is made up of alloy, which is a mixture of many medals, such as iron and nickel.

Mantle: The earth's mantle is the layer between the earths crust and core. Often made of silicate rocks.

Crust: The earth's crust is the outer-most of the three options. Usually made of up different types of rocks.

3 0
3 years ago
How much charge can be added to each of the plates before a spark jumps between the two plates? For such flat electrodes, assume
Svetach [21]

Answer:

9.56\cdot 10^{-7} C

Explanation:

A parallel-plate capacitors consist of two parallel plates charged with opposite charge.

Since the distance between the plates (1 cm) is very small compared to the side of the plates (19 cm), we can consider these two plates as two infinite sheets of charge.

The electric field between two infinite sheets with opposite charge is:

E=\frac{\sigma}{\epsilon_0}

where

\sigma=\frac{Q}{A} is the surface charge density, where

Q is the charge on the plate

A is the area of the plate

\epsilon_0 = 8.85\cdot 10^{-12}F/m is the vacuum permittivity

In this problem:

- The side of one plate is

L = 19 cm = 0.19 m

So the area is

A=L^2=(0.19)^2=0.036m^2

Here we want to find the maximum charge that can be stored on the plates such that the value of the electric field does not overcome:

E=3\cdot 10^6 N/C

Substituting this value into the previous formula and re-arranging it for Q, we find the charge:

E=\frac{Q}{A\epsilon_0}\\Q=EA\epsilon_0 = (3\cdot 10^6)(0.036)(8.85\cdot 10^{-12})=9.56\cdot 10^{-7} C

7 0
4 years ago
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