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4 years ago

A nebula, or swirling cloud of interstellar dust and gases, is seen. What happens next in the star formation process?

Physics

2 answers:

AVprozaik [17]4 years ago

5 0

Atoms begin to gravitate together to form a center

geniusboy [140]4 years ago

5 0

Answer: A)Atoms begin to gravitate together to form a center.

Explanation:

A nebula is a stellar nursery where stars are born. A nebula is a cloud of dust and gases. The disk starts swirling and atoms begin to gravitate and come together. A center is formed. This center forms the core. When enough mass is gathered to generate accurate temperature and pressure, nuclear fusion reaction kicks in to produce heat and light and a star is born.

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I'm walking 1.6m/s to 7-11 and it started to rain so I sped up to 2.7m/s in 1.2

olga nikolaevna [1]

Answer:

Explanation:

$a = \frac{v_f-v_0}{t}$ which is the final velocity minus the initial velocity in the numerator, and the change in time in the denominator. For us:

$a=\frac{2.7-1.6}{1.2}$ so

a = .92 m/s/s (NOT negative because you're speeding up)

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3 years ago

The movement of a magnetic pole away from the actual pole

Lunna [17]

The North Magnetic Pole is the point on the surface of Earth's Northern Hemisphere at which the planet's magnetic field points vertically downwards (in other words, if a magnetic compass needle is allowed to rotate about a horizontal axis, it will point straight down). There is only one location where this occurs, near (but distinct from) the Geographic North Pole and the Geomagnetic North Pole.

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4 years ago

A person with mass mp = 76 kg stands on a spinning platform disk with a radius of R = 1.98 m and mass md = 191 kg. The disk is i

nalin [4]

<span>1.7 rad/s The key thing here is conservation of angular momentum. The system as a whole will retain the same angular momentum. The initial velocity is 1.7 rad/s. As the person walks closer to the center of the spinning disk, the speed will increase. But I'm not going to bother calculating by how much. Just remember the speed will increase. And then as the person walks back out to the rim to the same distance that the person originally started, the speed will decrease. But during the entire walk, the total angular momentum remained constant. And since the initial mass distribution matches the final mass distribution, the final angular speed will match the initial angular speed.</span>

3 0

4 years ago

what is the main reason that attitudes are more often revealed in spoken rather than written language

pishuonlain [190]

It's a combination of all those things. probably because we are taught from an early age to write in an academic fashion, giving balanced arguments and a conclusion. When speaking from the heart, there is no opposing argument nor is there a conclusion, just emotion.

5 0

3 years ago

A 10-kg package drops from chute into a 25-kg cart with a velocity of 3 m/s. The cart is initially at rest and can roll freely w

amid [387]

Answer:

(a) the final velocity of the cart is 0.857 m/s

(b) the impulse experienced by the package is 21.43 kg.m/s

Explanation:

Given;

mass of the package, m₁ = 10 kg

mass of the cart, m₂ = 25 kg

initial velocity of the package, u₁ = 3 m/s

initial velocity of the cart, u₂ = 0

let the final velocity of the cart = v

(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

10 x 3 + 25 x 0 = v(10 + 25)

30 = 35v

v = 30 / 35

v = 0.857 m/s

(b) the impulse experienced by the package;

The impulse = change in momentum of the package

J = ΔP = m₁v - m₁u₁

J = m₁(v - u₁)

J = 10(0.857 - 3)

J = -21.43 kg.m/s

the magnitude of the impulse experienced by the package = 21.43 kg.m/s

(c)

the initial kinetic energy of the package is calculated as;

$K.E_i = \frac{1}{2} mu_1^2\\\\K.E_i = \frac{1}{2} \times 10 \times (3)^2\\\\K.E_i = 45 \ J\\\\$

the final kinetic energy of the package;

$K.E_f = \frac{1}{2} (m_1 + m_2)v^2\\\\K.E_f = \frac{1}{2} \times (10 + 25) \times 0.857^2\\\\K.E_f = 12.85 \ J$

the fraction of the initial energy lost;

$= \frac{\Delta K.E}{K.E_i} = \frac{45 -12.85}{45} = 0.71$

7 0

3 years ago