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3 years ago

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A careful photographic survey of Jupiter’s moon Io by the spacecraft Voyager 1 showed active volcanoes spewing liquid sulfur to

heights of 70 km above the surface of this moon. If the value of g on Io is 2.0 m/s2 , estimate the speed with which the liquid sulfur left the volcano.

1 answer:

Y_Kistochka [10]3 years ago

7 0

Answer:

529.15 m/s

Explanation:

h = Maximum height = 70000 m

g = Acceleration due to gravity = 2 m/s²

m = Mass of sulfur

As the potential and kinetic energies are conserved

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 2\times 70000}\\\Rightarrow v=529.15\ m/s$

The speed with which the liquid sulfur left the volcano is 529.15 m/s

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Which of the following types of light microscopy improves the resolution of thick specimens by illuminating one plane of the spe

Vilka [71]

Answer:

confocal microscopy

Explanation:

According to my research on different types of microscopes, I can say that based on the information provided within the question the tool being mentioned in this situation is a confocal microscopy. This is an extremely powerful microscope used to develop extremely sharp images of cells and tissues by viewing one plane of the specimen at a given time.

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3 years ago

Heat moves from hot to cold either through air or liquid <br>

KonstantinChe [14]

Answer:

both

Explanation:

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2 years ago

An insulating sphere is 8.00 cm in diameter and carries a 6.50 ÂµC charge uniformly distributed throughout its interior volume.

Kobotan [32]

Explanation:

(a) Formula to calculate the density is as follows.

$\rho = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}$

= $2.42 \times 10^{-2} C/m^{3}$

Now, calculate the charge as follows.

$q_{in} = \rho(\frac{4}{3} \pi r^{3})$

= $2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}$

= $10.106 \times 10^{-8}$ C

or, = 101.06 nC

(b) For r = 6.50 cm, the value of charge will be calculated as follows.

$q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}$

= 7.454 $\mu C$

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3 years ago

The kinetic energy of a rocket is increased by a factor of eight after its engines are fired, whereas its total mass is reduced

Rudik [331]

The momentum increases by a factor of 2

Explanation:

We can solve this problem by rewriting the momentum of the rocket in terms of the kinetic energy and the mass.

The kinetic energy of the rocket is:

$K=\frac{1}{2}mv^2$ (1)

where

m is the mass

v is the velocity

The momentum of the rocket is

$p=mv$ (2)

From eq.(1) we get

$v=\sqrt{\frac{2K}{m}}$

and substituting into (2),

$p=\sqrt{2mK}$

Now in this problem we have:

- The kinetic energy of the rocket is increased by a factor 8:

$K' = 8K$

- The mass is reduced by half:

$m'=\frac{m}{2}$

Substituting, we find the new momentum:

$p'=\sqrt{2(\frac{m}{2}(8K)}=\sqrt{4(2mK)}=2\sqrt{2mK}=2p$

So, the momentum increases by a factor of 2.

Learn more about momentum and kinetic energy:

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3 years ago

What is the volume V of a sample of 4.00 mol of copper? The atomic mass of copper (Cu) is 63.5 g/mol, and the density of copper

rusak2 [61]

Answer : The volume of a sample of 4.00 mol of copper is $28.5cm^3$

Explanation :

First we have to calculate the mass of copper.

$\text{ Mass of copper}=\text{ Moles of copper}\times \text{ Molar mass of copper}$

$\text{ Mass of copper}=(4.00moles)\times (63.5g/mole)=254g$

Now we have to calculate the volume of copper.

Formula used :

$Density=\frac{Mass}{Volume}$

Now put all the given values in this formula, we get:

$8.92\times 10^3kg/m^3=\frac{254g}{Volume}$

$Volume=\frac{254g}{8.92\times 10^3kg/m^3}=2.85\times 10^{-2}L=2.85\times 10^{-2}\times 10^3cm^3=28.5cm^3$

Conversion used :

$1kg/m^3=1g/L\\\\1L=10^3cm^3$

Therefore, the volume of a sample of 4.00 mol of copper is $28.5cm^3$

7 0

3 years ago

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