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3 years ago

The gases cast off by a dying star are called

Physics

2 answers:

Nady [450]3 years ago

7 0

Answer:

Planetary Nebula

Explanation:

A star has its own life cycle very similar to a human life cycle where it takes birth, grows up, becomes old and then dies. The life cycle of stars and human differ in the duration of cycle and end result.

When a star dies it can end up differently basis its mass. Here one of the ends of a star is mentioned where it casts off all the gases. Such type of situation will create a planetary nebula. One famous example of such nebula is Crab Nebula.

lozanna [386]3 years ago

4 0

Planetary Nebulae hope that helps

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A submarine is at an ocean depth of 3.00 km.

Alexxx [7]

Answer:

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3 years ago

Would water molecules in Venus’ atmosphere, whose temperature is 740 K, escape into outer space? A water molecule has a mass tha

Akimi4 [234]

Answer:

The water molecule cannot escape, since the average velocity of the water molecules is less than one sixth of the escape velocity of venus.

Explanation:

The average speed of gas molecules is given by:

$v_{rms}=\sqrt{\frac{3RT}{M}}$

R is the gas constant, T is the temperature and M the molar mass of the gas.

We know that a water molecule has a mass that is 18 times that of a hydrogen atom:

$M_H=1.01*10^{-3}\frac{kg}{mol}\\M_{H2O}=18M_H=0.02\frac{kg}{mol}$

So, we have:

$v_{rms}=\sqrt{\frac{3(8.314\frac{J}{mol \cdot K})740K}{0.02\frac{kg}{mol}}}\\v_{rms}=960.65\frac{m}{s}*\frac{1km}{1000m}=0.96\frac{km}{s}$

The water molecule cannot escape, since the average velocity of the water molecules is less than one sixth of the escape velocity of venus:

$10\frac{km}{s}*\frac{1}{6}=1.6\frac{km}{s}\\0.96\frac{km}{s}$

4 0

3 years ago

A mass on a spring A oscillates at twice the frequency of the same mass on spring B. Which statement is correct?A.The spring con

Nataliya [291]

Answer:

A.The spring constant for B is one quarter of the spring constant for A.

Explanation:

If spring A oscillates at twice the frequency of spring B, and period is frequency inverted. It means spring B has a period twice of spring A's.

$T_B = 2T_A$

As $T = 2\pi\sqrt{\frac{m}{k}}$ , and the 2 springs have the same mass

$2\pi\sqrt{\frac{m}{k_B}} = 2\pi\sqrt{\frac{m}{k_A}}$

$\sqrt{k_A} = 2\sqrt{B}$

$k_A = 4k_B$

$k_B = k_A/4$

So A.The spring constant for B is one quarter of the spring constant for A. is the correct answer.

3 0

3 years ago

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

adell [148]

Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

3 0

3 years ago

You place a book on top of a spring and push down, compressing the spring by 10 cm. When you let go of the book, it is pushed up

enot [183]

At First, there is chemical Energy( in your muscels) which is Used to Push down the spring. This Energy becomes the Energy of the spring, which increases until you stop pushing. If you Put your hand away, the Energy of the spring will become kinetic energ. This Energy is at the highest Level the Moment the book ist Leaving the spring. Afterwards, the kinetic Energy decreases while the Gravitational Potential Energy increases.

4 0

3 years ago

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