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Talja [164]
3 years ago
8

A gold bar has a density of 19.3 g/mL. If the gold bar has a mass of 6.3 grams, what is the volume?

Physics
1 answer:
Natali [406]3 years ago
7 0

Answer:

The correct answer is "0.32 mL".

Explanation:

The given values are:

Density of gold bar,

d = 19.3 g/mL

Mass of gold bar,

m = 6.3 grams

Now,

The volume will be:

⇒  Density = \frac{Mass}{Volume}

or,

⇒  Volume=\frac{Mass}{Density}

On substituting the values, we get

⇒               =\frac{6.3 \ g}{19.3 \ g/mL}

⇒               =0.32 \ mL

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Answer:

B. the number of field lines on the source charge

Explanation:

As we know that electric flux is defined as the number of electric field lines passing through a given area.

So here  electric flux due to a point charge "q" is given by

so here we know that flux depends on the magnitude of charge and hence we can say that number of filed lines originating from a point charge will depends on the magnitude of the charge.

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3 years ago
What is the role of gravity when it comes to changing the velocity of objects?
Alex787 [66]

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3 0
3 years ago
If you wish to warm 100 kg of water by 20°C for your bath, how much heat is required? (Give your answer in calories and joules.)
taurus [48]
Q = mcθ

Where m = mass of water in kg.
c = specific heat capacity in kJ/kg⁰C, c for water = 4200 kJ/kg⁰C
θ = temperature rise in ⁰C

Q = 100*4200* 20    Note here the temperature rise is 20 ⁰C
Q = 8 400 000 J

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4 0
3 years ago
True or false your vocal cords act like a guitar string when it comes to sound
torisob [31]

Answer:

True

Explanation:

8 0
3 years ago
Read 2 more answers
The wavelength of the visible line in the hydrogen spectrum that corresponds to m = 5 in the Balmer equation is: A. 656 nm. B. 4
BaLLatris [955]

Answer:

The wavelength of the visible line in the hydrogen spectrum is 434 nm.

Explanation:

It is given that, the wavelength of the visible line in the hydrogen spectrum that corresponds to n₂ = 5 in the Balmer equation.

For Balmer series, the wave number is given by :

\dfrac{1}{\lambda}=R(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})

R is the Rydberg's constant

For Balmer series, n₁ = 2. So,

\dfrac{1}{\lambda}=1.097\times 10^7\times (\dfrac{1}{2^2}-\dfrac{1}{5^2})

\lambda=4.34\times 10^{-7}\ m

or

\lambda=434\ nm

So, the wavelength of the visible line in the hydrogen spectrum is 434 nm. Hence, this is the required solution.

6 0
3 years ago
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