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2 years ago

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?

2 answers:

6 0

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.

Brums [2.3K]2 years ago

6 0

Answer:

1. 0.07

2. 1.15

Explanation:

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A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in ou

irakobra [83]

Answer:

λ = 3.2 x 10⁻⁷ m = 320 nm

Explanation:

The relationship between the velocity of electromagnetic waves (UV rays) and the their frequency is:

v = fλ

where,

v = c = speed of the electromagnetic waves (UV rays) = speed of light

c = 3 x 10⁸ m/s

f = frequency of the electromagnetic waves (UV rays) = 9.38 x 10¹⁴ Hz

λ = wavelength of the electromagnetic waves (UV rays) = ?

Therefore, substituting the values in the relation, we get:

3 x 10⁸ m/s = (9.38 x 10¹⁴ Hz)(λ)

λ = (3 x 10⁸ m/s)/(9.38 x 10¹⁴ Hz)

λ = 3.2 x 10⁻⁷ m = 320 nm

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3 years ago

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What is the power of a refrigerator with voltage 110 V and current 0.8 A?

tia_tia [17]

Answer: 88

Explanation:

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2 years ago

A balloon is ascending at a rate of +4.00 m/s to a height of 11.0 m above the ground when a package is dropped. In the absence o

o-na [289]

Answer:

Vf = 14.7 m/s

Explanation:

Vf² = Vi² + 2 * a * Δy

given:

a = 9.81 m/s²

Δy = 11m

Vi = 0 when upon release

Vf² = 0 + 2 (9.81) 11

Vf = 14.7 m/s

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2 years ago

A cheetah runs 900 meters in 30 seconds. how fast is it going?

Serga [27]

900 meters/30 seconds= 30 meters/second

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2 years ago

Two resistances, R1 and R2, are connected in series across a 12-V battery. The current increases by 0.500 A when R2 is removed,

KATRIN_1 [288]

Answer:

a) R₁ = 14.1 Ω, b) R₂ = 19.9 Ω

Explanation:

For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances

all resistors connected

V = i (R₁ + R₂)

with R₁ connected

V = (i + 0.5) R₁

with R₂ connected

V = (i + 0.25) R₂

We have a system of three equations with three unknowns for which we can solve it

We substitute the last two equations in the first

V = i ( $\frac{V}{ i+0.5} + \frac{V}{i+0.25}$ )

1 = i ( $\frac{1}{i+0.5} + \frac{1}{i+0.25}$ )

1 = i ( $\frac{i+0.5+i+0.25}{(i+0.5) \ ( i+0.25) }$ ) = $\frac{i^2 + 0.75i}{i^2 + 0.75 i + 0.125}$

i² + 0.75 i + 0.125 = 2i² + 0.75 i

i² - 0.125 = 0

i = √0.125

i = 0.35355 A

with the second equation we look for R1

R₁ = $\frac{V}{i+0.5}$

R₁ = 12 /( 0.35355 +0.5)

R₁ = 14.1 Ω

with the third equation we look for R2

R₂ = $\frac{V}{i+0.25}$

R₂ = $\frac{12}{0.35355+0.25}$

R₂ = 19.9 Ω

3 0

2 years ago