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3 years ago

Which statement correctly represents the law of the conservation of mass?

Physics

1 answer:

alisha [4.7K]3 years ago

6 0

Answer:

The statement H is the correct one: The number of atoms on the reactant side equals the number of atoms on the product side.

Explanation:

The law of conservation of mass states that mass can not be created or destroyed, just can be transformed. The mass of a system can not be added or removed, thereby the transfer of mass in a reaction must remain constant.

This implies that the mass of the molecules on the side of the reactants before the reaction must be equal to the mass of the atoms on the side of the products after the reaction.

Therefore, the statement H is the correct one: The number of atoms on the reactant side equals the number of atoms on the product side.

I hope it helps you!

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Plz plz answer quick...

Kruka [31]

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3 years ago

A child on a swing sweeps out a distance of 45 ft on the first pass. If she is allowed to continue swinging until she stops, an

earnstyle [38]

Answer:

d = 90 ft

Explanation:

Here in each swing the distance sweeps by the swing is half of the initial distance that it will move

So here we can say that total distance in whole motion is given as

$d = 45 + \frac{45}{2} + \frac{45}{4} + \frac{45}{8}..........$

since it is half of the distance that it will move in each swing so it would be a geometric progression with common ratio of 1/2

so sum of such GP is given by the formula

$S = \frac{a}{1 - r}$

$d = \frac{45}{1 - \frac{1}{2}}$

$d = 90 ft$

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4 years ago

Which of these are part of the electromagnetic spectrum?

bazaltina [42]

Answer: A, C, E

Explanation:

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3 years ago

Read 2 more answers

A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc

marta [7]

This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so:

$v_{d} = \frac{J}{n\left | q \right |}$

<span>The current I is any motion of charge from one region to another, so this is given by:

</span> $I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)$

The magnitude of the current density is:

$J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})$

Being:

$A=2mm^{2} = 2x10^{-6}m^{2}$
<span>
Finally, for the drift velocity magnitude vd, we find:

</span> $v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)$

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd

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3 years ago

Why the distance between two loudspeakers have to be large??

tangare [24]

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Explanation:

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