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Crazy boy
3 months ago
11

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

atter in megagrams per microliter (that is, Mg/µL)?
Physics
2 answers:
Sav [3.1K]3 months ago
8 0

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

kicyunya [3.2K]3 months ago
3 0

Answer: The density of nuclear matter will be 10^{-12}Mg/\mu L

Explanation:

Density is defined as the mass per unit volume.

Density=\frac{mass}{Volume}

Given:

The density of nuclear matter = 1018kg/m^3

1 ml = 1cm^3

1 kg = 0.001 Mg

Thus 1018kg=\frac{0.001}{1}\times 1018=1.018Mg

Additionally 1m^3=10^9\mu L

Substituting the values yields:

Density=\frac{0.001Mg}{10^9\mu L}=10^{-12}Mg/\mu L

Consequently, the density of nuclear matter will be 10^{-12}Mg/\mu L

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Response:

The speed at which the distance from the helicopter to you is changing (in ft/s) after 5 seconds is \sqrt{725} ft/ sec

Clarification:

Provided:

h(t) = 25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec. 5 = 125 ft

x(5) = 10 ft/sec. 5 = 50 ft

At this point, we can determine the distance between the individual and the helicopter utilizing the Pythagorean theorem

D(t) = \sqrt{h^2 + x^2}

Now, let's calculate the derivative of distance in relation to time

\frac{dD}{dt} (t) = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}

By plugging in the values for h(t) and x(t) and simplifying, we arrive at,

\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}

\frac{dh}{dt} = 25ft/sec

\frac{dx}{dt} = 10 ft/sec

\frac{Dd}{dt} (t) = \frac{1250t +200t}{2\sqrt{725}t} = \frac{725}{\sqrt{725}} = \sqrt{725} ft / sec

5 0
1 month ago
a straight, 2.5-m wire carries a typical household current of 1.5 a (in one direction) at location where the earth's magnetic fi
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a) The force's magnitude is F=2.1*10^{-4}N

, and it acts in the direction from south to north, meaning upward. b) The force's magnitude is F=2.1*10^{-4}N

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As computer structures get smaller and smaller, quantum rules start to create difficulties. Suppose electrons move through a cha
Softa [3030]

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

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