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3 years ago

15

The hydrogen stored inside a large weather balloon has a mass of 13.558 g. What is the volume of this balloon if the density of

hydrogen is 0.089 g/L?

2 answers:

sergij07 [2.7K]3 years ago

8 0

Answer:

152.337 L

Explanation:

To calculate density using volume we use the common equation

$D = \frac{m}{V} \\$

where

D is the density

m is the mass

V is the volume

This equation can be manipulated to work out volume to:

$V = \frac{m}{D} \\\\V = \frac{13.558g}{0.089g/L} \\\\V= 152.337 L$

sergejj [24]3 years ago

6 0

(13.558 gm) · (1 L / 0.089 gm) = 152.34 L (rounded)

(fraction equal to ' 1 ') ^

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O que é multifásicos?

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Answer:

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3 years ago

The sun's radiation has different wavelengths. the shorter is the radiation's wavelength, the weaker is the radiation's energy.

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It is false. Because the amount of energy carried in the wave is inversely related to the length of the waves wavelength. To correct the statement it should be that the shorter the radiation's wavelength the stronger is the radiation's energy.

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3 years ago

Whenever two Apollo astronauts were on the surface of the Moon, a third astronaut orbited the Moon. Assume the orbit to be circu

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Answer:

(a) Determine the astronaut's orbital speed

The velocity of the astronaut is 1436m/s.

(b) Determine the period of the orbit.

The period of the orbit is 10413 seconds.

Explanation:

<em>(a) Determine the astronaut's orbital speed</em>

The orbital speed of astronaut can be found by means of the Universal law of gravity:

$F = G\frac{M \cdot m}{r^{2}}$ (1)

Then, replacing Newton's second law in equation 3 it is gotten:

$m\cdot a = G\frac{M \cdot m}{r^{2}}$ (2)

However, a is the centripetal acceleration since the astronaut describes a circular motion around the Moon:

$a = \frac{v^{2}}{r}$ (3)

Replacing equation 3 in equation 2 it is gotten:

$m\frac{v^{2}}{r} = G\frac{M \cdot m}{r^{2}}$

$m \cdot v^{2} = G \frac{M \cdot m}{r^{2}}r$

$m \cdot v^{2} = G \frac{M \cdot m}{r}$

$v^{2} = G \frac{M \cdot m}{rm}$

$v^{2} = G \frac{M}{r}$

$v = \sqrt{\frac{G M}{r}}$ (4)

Where v is the orbital speed, G is the gravitational constant, M is the mass of the Moon, and r is the orbital radius.

Notice that the orbital radius will be given by the sum of the radius of the Moon and the height of the astronaut above the surface.

But it is necessary to express the height of the astronaut above the surface in units of meters before it can be used.

$r = 680km \cdot \frac{1000m}{1km}$ ⇒ $680000m$

$r = 1.70x10^{6}m+680000m$

$r = 2380000m$

However it is neccesary to find the mass of Moon in order to use equation 4.

Then Newton's second law ( $F = ma$ ) will be replaced in equation (1):

$ma = G\frac{Mm}{r^{2}}$

Then, M will be isolated

$M = \frac{r^{2}a}{G}$ (5)

Where r is the orbital radius of the astronaut and a is the acceleration due to gravity.

$M = \frac{(2380000m)^{2}(0.867 m/s^{2})}{6.67x10^{-11}N.m^{2}/kg^{2}}$ (5)

$M = 7.36x10^{22}Kg$

$v = \sqrt{\frac{(6.67x10^{-11}N.m^{2}/kg^{2})(7.36x10^{22}Kg)}{2380000m}}$

$v = 1436m/s$

Hence, the velocity of the astronaut is 1436m/s.

<em>(b) Determine the period of the orbit. </em>

The period of the orbit can be determined by the next equation:

$v = \frac{2\pi r}{T}$ (6)

Where v is the orbital velocity, r is the orbital radius and T is the period of the orbit.

Then, T can be isolated from equation 6.

$T = \frac{2\pi r}{v}$ (7)

$T = \frac{2\pi (2380000m)}{1436m/s}$

$T = 10413s$

Hence, the period of the orbit is 10413 seconds.

6 0

3 years ago

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Elden [556K]

Answer:

E₀ = 5.97 10⁸ N / C

Explanation:

Coulomb's law in a material medium is written

$F = \frac{1}{4\pi \epsilon } \frac{q_{1} q_{2} }{r^{2} }$

ε is the electrical permittivity of the material

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F = q E

in dielectric materials such as mica there is an electric field inside the material that decreases the external field

E = E₀ /ε

where ε is the electrical permittivity of the material; in general it is tabulated in the form

$\epsilon _{r} =\frac{\epsilon}{\epsilon_{o}}$

$\epsilon = \epsilon_{r} \ \epsilon_{o}$

we substitute

F = q E₀ / \epsilon_{r} \epsilon_{o}

$E_{o} = \frac{F \ \epsilon_{r} \ \epsilon_{o} }{q}$

for the case of mica it is equal to er = 5.4 and suppose a test charge equal to the charge of the electron

we calculate

E₀ = 2 5.4 8.85 10⁻¹² / 1.6 10⁻¹⁹

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3 years ago

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lukranit [14]

Answer:

v=10.75m/s

Explanation:

It is given that,

Pete drives 215 Meters in 20 seconds. He is driving down 7th floor. We need to find his speed.

Speed = distance/time

⇒

$v=\dfrac{215\ m}{20\ s}\\\\v=10.75\ m/s$

Hence, the speed of Pete is 10.75 m/s.

8 0

3 years ago