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3 years ago

What does the superscript in a chemical formula tell you?

Physics

1 answer:

Sunny_sXe [5.5K]3 years ago

4 0

The superscripts in a chemical formula are the overall charge of an ion

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The ___ of a gas is due to the force exerted on the walls of the

Furkat [3]

Answer:

its molecules

Explanation:

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2 years ago

Vector of magnitude 15 is added to a vector of magnitude 25. The magnitude of this sum

loris [4]

Explanation:

Given that,

Magnitude of vector A, |A| = 15

Magnitude of vector B, |B| = 25

We need to find the magnitude of this sum.

The maximum sum of the resultant vector,

$R_{max}=|A_1|+|A_2|\\\\=15+25\\\\=45$

The minimum sum of the resultant vector,

$R_{min}=|A_1|-|A_2|\\\\=15-25\\\\=-10$

So, the magnitude of this sum either 45 or -10.

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2 years ago

Ag,Au and Cu are called coinage metals why plzzzz hurry its urgent plzzz

Ronch [10]

Answer:

This is because these metals are used for minting (making) coins.

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3 years ago

Which law is used to find the magnitude of a magnetic force?

Talja [164]

Answer:

The Flemings left hand rule is used to find the magnitude of a magnetic force

Explanation:

Fleming's left hand rule states that if the first three fingers are held mutually at right angles to one another, then the fore finger points into the direction of magnetic field the middle finger in the direction of current while the thumb points in the direction of force.

Mathematically

Magnetic Force F= BILsinθ

Where

B= magnetic field density Tesla

I= current

L= length of conductor

θ= angle of conductor with field

3 0

3 years ago

At what wavelength would a star radiate the greatest amount of energy if the star has a surface temperature of 60,000 K?

kompoz [17]

Answer:

$\lambda=4.81\times 10^{-8}\ m$

Explanation:

We have,

The surface temperature of the star is 60,000 K

It is required to find the wavelength of a star that radiated greatest amount of energy. Wein's displacement law gives the relation between wavelength and temperature such that :

$\lambda T=2.89\times 10^{-3}$

Here,

$\lambda$ = wavelength

$\lambda=\dfrac{2.89\times 10^{-3}}{60000}\\\\\lambda=4.81\times 10^{-8}\ m$

So, the wavelength of the star is $4.81\times 10^{-8}\ m$ .

7 0

3 years ago