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4 years ago

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What two things are needed to create an emission nebula? A) interstellar gas and dust B) hydrogen fusion and helium ionization C

) cool stars and much interstellar dust D) hot stars and interstellar gas, particularly hydrogen E) hydrogen gas and carbon dust

1 answer:

wolverine [178]4 years ago

8 0

Answer:

D) hot stars and interstellar gas, particularly hydrogen

Explanation:

Emission nebula is the nebula which is formed of the ionized gases which emit light of different wavelengths. The common source of ionization of the emission nebula is high-energy photons mostly which are emitted from hot star present nearby to it.

<u>The emission nebulae include H II regions. An H II region is the region of the interstellar atomic hydrogen which is ionized. It is the cloud of the partially ionized gas in which star formation takes place.</u>

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What is the main determining factor in defining boundaries between layers of earth's atmosphere?

konstantin123 [22]

The main determining factor in defining boundaries between layers of earth's atmosphere would be temperature changes in these layers. Temperature is one essential property that varies in the atmosphere. Based from this variation, the atmosphere is divided into four major layers and further to three smaller layers - troposphere, tropopause, the stratosphere, stratopause, the mesosphere, mesopause, and the thermosphere.The troposphere is the layer that is nearest to the surface of the Earth. It is the part where humans, plants and animals survive. Also, it is the warmest layer of the atmosphere. And as we go higher the atmosphere, the temperature would drop making it much cooler.

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3 years ago

Read 2 more answers

A large balloon of mass 210 kg is filled with helium gas until its volume is 329 m3. Assume the density of air is 1.29 kg/m3 and

Nastasia [14]

(a) See figure in attachment (please note that the image should be rotated by 90 degrees clockwise)

There are only two forces acting on the balloon, if we neglect air resistance:

- The weight of the balloon, labelled with W, whose magnitude is

$W=mg$

where m is the mass of the balloon+the helium gas inside and g is the acceleration due to gravity, and whose direction is downward

- The Buoyant force, labelled with B, whose magnitude is

$B=\rho_a V g$

where $\rho_a$ is the air density, V is the volume of the balloon and g the acceleration due to gravity, and where the direction is upward

(b) 4159 N

The buoyant force is given by

$B=\rho_a V g$

where $\rho_a$ is the air density, V is the volume of the balloon and g the acceleration due to gravity.

In this case we have

$\rho_a = 1.29 kg/m^3$ is the air density

$V=329 m^3$ is the volume of the balloon

g = 9.8 m/s^2 is the acceleration due to gravity

So the buoyant force is

$B=(1.29 kg/m^3)(329 m^3)(9.8 m/s^2)=4159 N$

(c) 1524 N

The mass of the helium gas inside the balloon is

$m_h=\rho_h V=(0.179 kg/m^3)(329 m^3)=59 kg$

where $\rho_h$ is the helium density; so we the total mass of the balloon+helium gas inside is

$m=m_h+m_b=59 kg+210 kg=269 kg$

So now we can find the weight of the balloon:

$W=mg=(269 kg)(9.8 m/s^2)=2635 N$

And so, the net force on the balloon is

$F=B-W=4159 N-2635 N=1524 N$

(d) The balloon will rise

Explanation: we said that there are only two forces acting on the balloon: the buoyant force, upward, and the weight, downward. Since the magnitude of the buoyant force is larger than the magnitude of the weigth, this means that the net force on the balloon points upward, so according to Newton's second law, the balloon will have an acceleration pointing upward, so it will rise.

(e) 155 kg

The maximum additional mass that the balloon can support in equilibrium can be found by requiring that the buoyant force is equal to the new weight of the balloon:

$W'=(m'+m)g=B$

where m' is the additional mass. Re-arranging the equation for m', we find

$m'=\frac{B}{g}-m=\frac{4159 N}{9.8 m/s^2}-269 kg=155 kg$

(f) The balloon and its load will accelerate upward.

If the mass of the load is less than the value calculated in the previous part (155 kg), the balloon will accelerate upward, because the buoyant force will still be larger than the weight of the balloon, so the net force will still be pointing upward.

(g) The decrease in air density as the altitude increases

As the balloon rises and goes higher, the density of the air in the atmosphere decreases. As a result, the buoyant force that pushes the balloon upward will decrease, according to the formula

$B=\rho_a V g$

So, at a certain altitude h, the buoyant force will be no longer greater than the weight of the balloon, therefore the net force will become zero and the balloon will no longer rise.

4 0

3 years ago

If you cannot exert enough force to loosen a bolt with a wrench, which of the following should you do?

stepan [7]

Answer:

2.) Use a wrench with a longer handle.

Explanation:

It is suggested to use a wrench with longer handle if you cannot exert enough force to loosen a bolt with a wrench.

In doing this work, the torque which is the force that can turn the screw is not enough;

Torque = force x distance

If we increase the distance or the length of the handle, we can generate more torque to overcome the force needed.

6 0

3 years ago

A heavy rope 6.00 m long and weighing 29.4 N is attached at one end to a ceiling and hangs vertically. A 0.500-kg mass is suspen

aivan3 [116]

Answer:

a) $v=3.1252\ m.s^{-1}$

b) $v=39.0672\ m.s^{-1}$

c) $v=8.2685\ m.s^{-1}$

d) No,

No.

Explanation:

Given:

length of rope, $l=6\ m$

weight of the rope, $w=29.4\ N$

mass suspended at the lower end of the rope, $M=0.5\ kg$

<u>Now the mass of the rope:</u>

$m=\frac{w}{g}$

$m=\frac{29.4}{9.8}$

$m=3.01\ kg$

<u>So the linear mass density of rope:</u>

$\mu=\frac{m}{l}$

$\mu=\frac{3.01}{6}$

$\mu=0.5017\ kg.m^{-1}$

We know that the speed of wave in a tensed rope is given as:

$v=\sqrt{\frac{F_T}{\mu} }$

where:

$F_T=$ tension force in the rope

a)

At the bottom of the hanging rope we have an extra mass suspended. So the tension at the bottom of the rope:

$F_T=M\times g$

$F_T=0.5\times 9.8$

$F_T=4.9\ N$

Therefore the speed of the wave at the bottom point of the rope:

$v=\sqrt{\frac{4.9}{0.5017} }$

$v=3.1252\ m.s^{-1}$

b)

Tension at a point in the middle of the rope:

$F_T=M\times g+\frac{w}{2}$

$F_T=0.5\times 9.8+\frac{29.4}{2}$

$F_T=19.6\ N$

Now wave speed at this point:

$v=\sqrt{\frac{19.6}{0.5017} }$

$v=39.0672\ m.s^{-1}$

c)

Tension at a point in the top of the rope:

$F_T=M\times g+w$

$F_T=0.5\times 9.8+29.4$

$F_T=34.3\ N$

Now wave speed at this point:

$v=\sqrt{\frac{34.3}{0.5017} }$

$v=8.2685\ m.s^{-1}$

d)

Tension at the middle of the rope is not the average tension of tension at the top and bottom of the rope because we have an extra mass attached at the bottom end of the rope.

Also the wave speed at the mid of the rope is not the average f the speeds at the top and the bottom of the ropes because it depends upon the tension of the rope at the concerned points.

7 0

3 years ago

A mixture of 30.2 g of sand and 87.7 g of water has a temperature of 12.1 °C. What is the mass of water at 85.8° C must be added

marissa [1.9K]

Q out = Q in

Q mix = Q water

mcΔt sand + mcΔt water = mcΔt water

30.2 x 2.01 x (29.3-12.1) + 87.7 x 4.19 x (29.3-12.1)=m x 4.19 x (85.8-29.3)

1044.07+6320.36=m x 236.74

m = 31.11 g

6 0

3 years ago