Electromagnet Fluctation

A sample of gold has a mass of 30.94 grams and density of 19.32g/cm^3. What volume of space will this sample of gold occupy?

sdas [7]

Data:
mass, m = 30.94 g
density, d = 19.32 g/cm^3

Formula: d = m / v => v = m / d = 30.94 g / 19.32 g/cm^3 = 1.60 cm^3

Then, the answer is the option C.

6 0

2 years ago

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A scientist heats flexible container full of neon gas .what will most likely happen to the container as the gas absorbs heat

Ilia_Sergeevich [38]

As the container starts to heat up, so will the neon gas. Heat is nothing but energy, and when you add energy to a gas, it will start vibrating much faster and hit the edges of the container at a higher rate and a faster velocity. Therefore, it's possible to deduce that the container will most likely rupture and/or "explode".

4 0

2 years ago

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PLEASE HELP QUICK I WILL GIVE BRAINLIEST AnswerTwo descriptions about physical quantities are given below: Quantity A: It is una

zubka84 [21]

Both are quantities of mass

5 0

3 years ago

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In a double-slit experiment, the third-order maximum for light of wavelength 510 nm is located 17 mm from the central bright spo

Marysya12 [62]

Answer:

14.9 mm

Explanation:

We know dsinθ = mλ where d = separating of slit, m = order of maximum = 3 and λ = wavelength = 510nm = 510 × 10⁻⁹ m

Also tanθ = L/D where L = distance of m order fringe from central bright spot = 17 mm = 0.017 m and D = distance of screen from slit = 1.6 m

So, sinθ = mλ/d

Since θ is small, sinθ ≅ tanθ

So,

mλ/d = L/D

d = mλD/L

Substituting the values of the variables into the equation, we have

d = 3 × 510 × 10⁻⁹ m × 1.6 m/0.017 m

d = 2448 × 10⁻⁹ m²/0.017 m

d = 144000 × 10⁻⁹ m

d = 1.44 × 10⁻⁴ m

d = 0.144 × 10⁻³ m

d = 0.144 mm

Now, for the second-order maximum, m' of the 670 nm wavelength of light,

m'λ'/d = L'/D where m' = order of maximum = 2, λ' = wavelength of light = 670 nm = 670 × 10⁻⁹ m, d = slit separation = 0.144 mm = 0.144 × 10⁻³ m, L' = distance of second order maximum from central bright spot and D = distance of screen from slit = 1.6m

So, L' = m'λ'D/d

So, substituting the values of the variables into the equation, we have

L' = 2 × 670 × 10⁻⁹ m × 1.6 m/0.144 × 10⁻³ m

L' = 2144 × 10⁻⁹ m²/0.144 × 10⁻³ m

L' = 14888.89 × 10⁻⁶ m

L' = 0.01488 m

L' ≅ 0.0149 m

L' = 14.9 mm

4 0

3 years ago

A perfectly spherical balloon of air is tethered to the floor of a stagnant body of water by cable AB, and sits just below the s

storchak [24]