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lesya692 [45]
3 years ago
12

If we begin from star birth, the next key stage in the star-gas-star cycle occurs when a(n)_______ returns gas and new elements

into space. The expelled gas forms a(n)________ that is very hot and expands in size. As the gas cools and merges with other interstellar gas, it forms a(n)__________ . Further cooling allows the gas to become a(n)_________ , in which star formation occur.
Physics
2 answers:
Naily [24]3 years ago
8 0

Answer:

1. supernova

2. bubble

3. atomic hydrogen cloud

4. molecular cloud

Explanation:

Supernova involves the bright explosion of a star when this happens it returns gas and new elements in to space. The expelled hot gas from the explosion forms a bubble and starts expanding in size. As this bubble cools, it merges merges with other interstellar gas to form an atomic hydrogen cloud. The atomic hydrogen cloud cools further to form a molecular cloud where another star formation begins to occur.

Helga [31]3 years ago
6 0

Answer:

1. A Supernova

2. A Bubble

3. Atomic hydrogen cloud

4. Molecular cloud

Explanation:

If we begin from star birth, the next key stage in the star-gas-star cycle occurs when a Supernova returns gas and new elements into space. The expelled gas forms a Bubble that is very hot and expands in size. As the gas cools and merges with other interstellar gas, it forms an Atomic hydrogen cloud. Further cooling allows the gas to become a Molecular cloud, in which star formation occur.

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About once every 30 minutes, a geyser known as Old Faceful projects water 11.0 m straight up into the air. Use g = 9.80 m/s^2, a
Maurinko [17]

Answer:

The speed of the water is 14.68 m/s.

Explanation:

Given that,

Time = 30 minutes

Distance = 11.0 m

Pressure = 101.3 kPa

Density of water = 1000 kg/m³

We need to calculate the speed of the water

Using equation of motion

v^2=u^2+2gs

Where, u = speed of water

g = acceleration due to gravity

h = height

Put the value into the formula

0=u^2-2\times9.8\times11.0

u=\sqrt{2\times9.8\times11.0}

u=14.68\ m/s

Hence, The speed of the water is 14.68 m/s.

7 0
3 years ago
You have been called in by your neighborhood safety council to advise them about installing a barrier on a dangerous curve that
Alborosie

Answer: The softer barrier is the better option

Explanation:

1) When is a car is moving at a certain speed, it has a certain amount of momentum (p=mv). A collision against a barrier would cause its momentum to decrease to 0. A change in momentum is Impulse

2) The formula for Impulse: J = f * Δt

J is Impulse

f is the force applied during the time Δt

A tough barrier would produce a smaller Δt, which means more force is applied on the car. (J is always constant)

A softer barrier would apply less force on the car, which means Δt is large.

Answer: The softer barrier is the better option

4 0
3 years ago
You toss a rock up vertically at an initial speed of 39 feet per second and release it at an initial height of 6 feet. The rock
3241004551 [841]

Answer:

2.583 s, 29.77 ft and 1.219 s

Explanation:

Using equation of motion and taken the motion upward as positive, also a = g ( acceleration due to gravity) = - 32 fts⁻², V= 39 fts⁻¹ V₁ is final velocity, y is the distance in ft from the ground

H = 6 ft, the height from which it is tossed

V₁ = V + gt = V - gt

at maximum height the body came to rest momentarily V₁ = 0

0 = V - gt

-V = -gt

- 39 / -32 = t

t time to reach maximum height = 1.219 s

To Maximum height reached can be calculated with the formula

V₁² = V² + 2g( y - H) where H is the initial height reached by the tossed rock

where V₁ is the final velocity at maximum height which = 0

0 = V² - 2g(y-H) where y is the distance traveled from the ground

-V² = -2g(y-H)

₋V² / -2g = y-H

(V²/2g) + H = y in ft

(39² / (2 × 32)) + 6

y = 29.77 ft

The total time it will be in air can be calculated with the formula below

y = H + Vt - 0.5gt² from y-H = ut + 0.5at²

0.5gt² - Vt - H = 0 since the body returned to the ground ( y = 0)

0.5gt² - Vt - H = 0

using quadratic formula

- (-V)² ± √ ((-V²) - 4 × 0.5g × -H) / (2 × 0.5 × g)

(V ± √ (V² + 2gH)) ÷ g

substitute the values into the expression

t = (39 + √(39² + (2×-32× 6)))/ 32 or (39 - √ (39² + (2 × -32×6))/ 32

t = (39 + √(1521 +384))/32 = (39 + √1905) / 32  = 2.583 s

t = (39 - √1905) / 32 =  -0.15 s

The will remain in air (V ± √ (V² + 2gH)) / g seconds. It will reach a maximum height of (V²/2g) + H feet after V/g seconds

8 0
3 years ago
You are presently taking a weather observation. The sky is full of wispy cirrus clouds estimated to be about 10 km overhead. If
marshall27 [118]

Answer:

x = 2000 Km

Explanation:

Given

y = 10 km

Slope: 1 : 200

x = ?

We can apply the formula

y / x = 1 / 200   ⇒     x = 200*y = 200*10 Km

⇒     x = 2000 Km

7 0
3 years ago
Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i
s2008m [1.1K]

Answer:

The final speed for the mass 2m is v_{2y}=-1,51\ v_{i} and the final speed for the mass 9m is v_{1f} =0,85\ v_{i}.

The angle at which the particle 9m is scattered is \theta = -66,68^{o} with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

\vec{p}=m\vec{v}

p_{xi} =p_{xf}

In the x axis before the collision we have

p_{xi}=9m\ v_{i} - 2m\ v_{i}

and after the collision we have that

p_{xf} =9m\ v_{1x}

In the y axis before the collision p_{yi} =0

after the collision we have that

p_{yf} =9m\ v_{1y} - 2m\ v_{2y}

so

p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}

then

p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}

<u>Conservation of kinetic energy:</u>

\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}

so

\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}

Putting in one side of the equation each speed we get

\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the  speed of the 9m particle:

v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}

the magnitude of the final speed of the particle 9m is

v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }

v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}

The tangent that the speed of the particle 9m makes with the -y axis is

tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}

As a vector the speed of the particle 9m is:

\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}

As a vector the speed of the particle 2m is:

\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}

8 0
3 years ago
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