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11 days ago

The length of the side of a cube having a density of 12.6 g/ml and a mass of 7.65 g is __________ cm.

1 answer:

4 0

Density serves as a key characteristic of any substance, defined as the mass of the substance divided by its volume (density = mass/volume). Through manipulation of this formula, we determine volume. So volume equals the mass of a substance divided by its density (Vol = mass/density). With density given as 12.6 g/ml and mass as 7.65 g, we find that volume equals 0.60714 ml; since 1 ml is equivalent to 1 cm³, volume also equals 0.60714 cm³. Finally, taking the cube root of the volume provides the side length of the cube in cm, which is 0.84677 cm.

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What happens to the particles of a liquid when energy is removed from them?

Softa [913]

Response:

D: The distance among the particles diminishes

Clarification:

Removing energy reduces the activity of molecules, similar to how one slows down in cold temperatures (I believe).

3 0

10 days ago

The density of nuclear matter is about 1018 kg/m3. Given that 1 mL is equal in volume to 1 cm3, what is the density of nuclear m

Sav [1095]

Answer:

The density comes out to be $10^{6}$ Mg/µL

Explanation:

Given data:

The density of nuclear matter is approximately $10^{18}$ kg/m³

1 ml corresponds to 1 cm³

To determine:

The density of nuclear matter in Mg/µL

Solution:

We recognize that:

1 Mg equals 1000 kg

Thus, 1 m³ is equal to $10^{6}$ cm³

Moreover, 1 cm³ is equivalent to 1 mL

Thus, we can conclude that 1 mL is equal to 10³ µL

With this, we convert the density as follows:

Density = $10^{18}$ kg/m³

Density = $10^{18}$ kg/m³ × $\frac{10^{-3} }{10^{6} }$ Mg/µL

Density = $10^{6}$ Mg/µL

8 0

2 days ago

Read 2 more answers

As computer structures get smaller and smaller, quantum rules start to create difficulties. Suppose electrons move through a cha

Softa [913]

Answer:

a) ∆x∆v = 5.78*10^-5

∆v = 1157.08 m/s

b) 4.32*10^{-11}

Explanation:

This problem can be addressed using Heisenberg's uncertainty principle, which is expressed as:

$\Delta x\Delta p \geq \frac{\hbar}{2}$

Where h represents Planck’s constant (6.62*10^-34 J s).

Assuming that the electron's mass remains the same, we proceed as follows:

$\Delta x \Delta (m_ev)=m_e\Delta x\Delta v \geq \frac{\hbar}{2}$

Utilizing the electron's mass (9.61*10^-31 kg) and the uncertainty in position (50 nm), we can compute ∆x∆v and ∆v:

$\Delta x \Delta v\geq\frac{\hbar}{2m_e}=\frac{(1.055*10^{-34}Js)}{2(9.1*10^{-31}kg)}=5.78*10^{-5}\ m^2/s$

$\Delta v\geq\frac{5.78*10^{-5}}{50*10^{-9}m}=1157.08\frac{m}{s}$

If we treat the electron like a classic particle, the time required to cross the channel is determined using the upper limit of the uncertainty in velocity:

$t=\frac{x}{v}=\frac{50*10^{-9}m}{1157.08m/s}=4.32*10^{-11}s$

6 0

6 days ago

Two charges of 15 pC and −40 pC are inside a cube with sides that are of 0.40-m length. Determine the net electric flux through

Keith_Richards [1021]

To address this issue, we will utilize the principles related to Gauss' law, which states that the electric flux across a surface corresponds to the object's charge divided by the permittivity of vacuum. In mathematical terms, this can be expressed as

$\phi = \frac{Q_{net}}{\epsilon_0}$