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Talja [164]
3 years ago
15

Suppose that scientists observe violent gas eruptions on a planet with an acceleration due to gravity of 3.9 m/s2. The jets thro

w sand and dust about 75 m above the surface. What is the speed i of the material just as it leaves the surface?
Physics
1 answer:
Lina20 [59]3 years ago
4 0

Answer:

The velocity with which the sand throw is 24.2 m/s.

Explanation:

acceleration due to gravity, a =  3.9 m/s2

height, h = 75 m

final velocity, v = 0

Let the initial  velocity at the time of throw is u.

Use third equation of motion

v^{2} = u^{2} - 2 g h \\\\0 = u^{2} -2\times 3.9\times 75\\\\u=24.2m/s

The velocity with which the sand throw is 24.2 m/s.

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El espectro visible en el aire está comprendido entre la longitud de onda 450 nm del color azul, Determina la velocidad de propa
TiliK225 [7]

Answer:

v = 2,99913 10⁸ m / s

Explanation:

The velocity of propagation of a wave is

         v = λ f

in the case of an electromagnetic wave in a vacuum the speed that speed of light

         v = c

When the wave reaches a material medium, it is transmitted through a resonant type process, whereby the molecules of the medium vibrate at the same frequency as the wave, as the speed of the wave decreases the only way that they remain the relationship is that the donut length changes in the material medium

         λ = λ₀ / n

where n is the index of refraction of the material medium.

Therefore the expression is

           v = \frac{\lambda_o}{n} f

Let's look for the frequency of blue light in a vacuum

           f =\frac{c }{\lambda_o}  

           f = \frac{3 \ 10^8}{450 \ 10^{-9}}

           f = 6.667 10¹⁴ Hz

the refractive index of air is tabulated

           n = 1,00029

let's calculate

           v = \frac{450 \ 10^{-9} }{1.00029}  \ 6.667 \ 10^{14}450 10-9 / 1,00029 6,667 1014

            v = 2,99913 10⁸ m / s

we can see that the decrease in speed is very small

8 0
3 years ago
As part of the simulation, you will manipulate magnet polarity. This is the variable. The induced current will vary based on how
Bogdan [553]

Answer: strength.

Explanation: The simulation made the strength an independent variable and dependent variable.

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3 years ago
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Answer:

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Delia has a genetic disease that causes her to be weak and tired because her blood cells are not able to carry enough oxygen to
ddd [48]

Answer:

idk i’m not smart i think it’s A her blood cells have to much insulin. but again idk...

Explanation:

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4 years ago
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A 0.500-mol sample of an ideal monatomic gas at 400 kPa and 300 K, expands quasi-statically until the pressure decreases to 160
Marianna [84]

Answer:

(a). The final volume is 2.5 times the initial volume.

The work done is 1143 J.

(b). The final temperature is 207.9 K

The work done is 574 J.

Explanation:

Given that,

Sample of an ideal gas = 0.500 mol

Initial pressure = 400 kPa

Final pressure = 160 kPa

Temperature = 300 K

(a) for isothermal,

Temperature will be same.

We need to calculate the volume of gas

P_{f}V_{f}=P_{i}V_{i}

V_{f}=(\dfrac{P_{i}}{P_{f}})V_{i}

Put the value into the fomrula

V_{f}=(\dfrac{400}{160})V_{i}

V_{f}=2.5 V_{i}

We need to calculate the work done

Using equation of energy

dQ=dW

dQ=nRTln(\dfrac{P_{in}}{P_{f}})

dQ=0.500\times8.314\times300\times ln(\dfrac{400}{160})

dQ=1143\ J

(b). For adiabatic,

No transfer of heat between system and surroundings

We need to calculate the final temperature

Using formula of gas

P_{f}^{1-\gamma}T_{f}^{\gamma}=P_{i}^{1-\gamma}T^{\gamma}

T_{f}=(\dfrac{P_{i}}{P_{f}})^{\frac{1-\gamma}{\gamma}}T_{i}

Put the value into the formula

T_{f}=(\dfrac{400}{160})^{\frac{1-\frac{5}{3}}{\frac{5}{3}}}\times300

T_{f}=207.9\ K

We need to calculate the wok done in adiabatic

Using formula of work done

W=\dfrac{nR(T_{i}-T_{f})}{\gamma-1}

W=\dfrac{0.500\times8.314\times(300-207.9)}{\dfrac{5}{3}-1}

W=574\ J

Hence, (a). The final volume is 2.5 times the initial volume.

The work done is 1143 J.

(b). The final temperature is 207.9 K

The work done is 574 J.

6 0
3 years ago
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