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Hatshy [7]
4 years ago
7

If an object has a volume of 2.5mL and a mass of 10g than what is the density of an object?

Physics
1 answer:
Hoochie [10]4 years ago
3 0
Density is equal to mass divided by volume, so the density of the object is 10/2.5= 4<span> g/cm</span>
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What do all types of electromagnetic radiation have in common?
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<span>They are emitted by the unstable nuclei of certain atoms.
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How does the law of conservation of mass apply to this reaction: C2H4 + O2 → H2O + CO2?
kicyunya [14]

The answer is a

the equation needs to be balanced. There are fewer oxygen atoms in the equation than hydrogen or a carbon

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A woman climbs up a ladder in 1.37 s at 2.20 m/s. How tall is the ladder?
ArbitrLikvidat [17]

Answer:

The ladder is 3.014 m tall.

Explanation:

To solve this problem, we must use the following formula:

v = x/t

where v represents the woman’s velocity, x represents the distance she climbed (the height of the ladder), and t represents the time it took her to move this distance

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v = x/t

2.20 = x/1.37

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x = (2.20 * 1.37)

x = 3.014 m

Therefore, the ladder is 3.014 m tall.

Hope this helps!

6 0
3 years ago
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What combines to form bonds in compounds?
Mila [183]
A chemical compound consists of 2 or more chemical elements. They can be separated into simpler substances. Atoms of elements can be combined to form compounds. Molecules are groups of 2 or more atoms held together by chemical bonds. Minerals are chemical compound that can be found in nature. Valence is a measure of the combining power of an element.Answer: A. Atoms of elements.

8 0
3 years ago
The parallel plates in a capacitor, with a plate area of 8.00 cm2 and an air-filled separation of 2.70 mm, are charged by a 8.70
MrRa [10]

Answer:

a)  ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J,  d)  W = 150 10⁻¹² J

Explanation:

Let's find the capacitance of the capacitor

         C = \epsilon_o \frac{A}{d}

         C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³

         C = 2.62 10⁻¹² F

for the initial data let's look for the accumulated charge on the plates

          C = \frac{Q}{\Delta V}

          Q₀ = C ΔV

           Q₀ = 2.62 10⁻¹² 8.70

           Q₀ = 22.8 10⁻¹² C

a) we look for the capacity for the new distance

          C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³

          C₁ = 1.04 10⁻¹² F

       

          C₁ = Q₀ / ΔV₁

          ΔV₁ = Q₀ / C₁

          ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²

          ΔV₁ = 21.9 V

b) initial stored energy

          U₀ = \frac{Q_o}{ 2C}

          U₀ = (22.8 10⁻¹²)²/(2  2.62 10⁻¹²)

          U₀ = 99.2 10⁻¹² J

c) final stored energy

          U_f = (22.8 10⁻¹²) ² /(2  1.04 10⁻⁻¹²)

          U_f = 249.9 10⁻¹² J

d) the work of separating the plates

as energy is conserved work must be equal to energy change

          W = U_f - U₀

          W = (249.2 - 99.2) 10⁻¹²

          W = 150 10⁻¹² J

note that as the energy increases the work must be supplied to the system

6 0
3 years ago
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