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Andrews [41]
3 years ago
13

Which of the following is a likely life cycle of a star?

Physics
1 answer:
melomori [17]3 years ago
7 0

Answer:

D.

Explanation:

But this just happen for big stars, like more than 20x the Sun mass.

Shortly: A nebula is a cloud of gas and dust, the material starts to be acummuleted and became a protostar (is like a big planet, almost  a star). With enought mass this is a star, burn hydrogen and transform it in Helium.

This occurs in Main Sequence, is about almost all the life time of a star. Then starts the lack of hydrogen. Gravity compress everything, pressure goes up and heat all. Too much energy, Helium get burned and the star grews fast, became a Red Giant. Time pass and the fuel is over, no more making fusion, gravity compress the star, too much strenght, colapses, neutron star.

If it have pretty mass, ok. If have more than like 2x Sun mass, became a blackhole.

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Calculate the magnitude of the total impulse applied to the car to bring it to rest
n200080 [17]

Answer:

Total impulse = mv = Initial momentum of the car

Explanation:

Let the mass of the car be 'm' kg moving with a velocity 'v' m/s.

The final velocity of the car is 0 m/s as it is brought to rest.

Impulse is equal to the product of constant force applied to an object for a very small interval. Impulse is also calculated as the total change in the linear momentum of an object during the given time interval.

The magnitude of impulse is the absolute value of the change in momentum.

|J|=|p_f-p_i|

Momentum of an object is equal to the product of its mass and velocity.

So, the initial momentum of the car is given as:

p_i=mv

The final momentum of the car is given as:

p_f=m(0)=0

Therefore, the impulse is given as:

|J|=|p_f-p_i|=|0-mv|=|-mv|=mv

Hence, the magnitude of the impulse applied to the car to bring it to rest is equal to the initial momentum of the car.

5 0
3 years ago
When atoms are split, they release energy. this concept applies to
balu736 [363]
This applies to nuclear reactions, specifically nuclear fission.

This huge release of energy has been used in atomic bombs and in the nuclear reactors that generate electricity.
6 0
3 years ago
A piece of blue paper appears blue because the paper
vfiekz [6]

A piece of blue paper appears blue because the paper absorbs all colors of light except blue. <em> (b)</em>

So any light that bounces off of the paper and enters your eye must be blue light !

6 0
3 years ago
You throw a bouncy rubber ball and a wet lump of clay, both of mass m, at a wall. Both strike the wall at speed v, but while the
lana [24]

Answer:

<em>The fifth option is the correct answer: mv; 2 mv</em>

Explanation:

<u>Change of Momentum</u>

Assume an object has a momentum p1 and after some interaction it now has a momentum p2, the change of momentum is

\Delta p=p_2-p_1

The momentum is computed as

p=mv

Where m is the mass of the object and v its speed. Now let's analyze the situation of both the ball and the clay.

The clay has an initial speed v and a mass m, thus its initial momentum is

p_1=mv

When it hits the wall, it sticks, thus its final speed is 0 and

p_2=0

The change of momentum is

\Delta p=0-mv=-mv

The absolute change is mv

Now for the ball, the initial condition is the same as it was for the clay, but the ball hits back at the same speed, thus its final momentum is

p_2=-mv

The change of momentum is

\Delta p=-mv-mv=-2mv

The absolute change is 2mv

The fifth option is the correct answer: mv; 2 mv

3 0
3 years ago
A small lead ball, attached to a 1.70-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled
Otrada [13]

Answer:

H = 54.37

Explanation:

given,

lead ball attached at = 1.70 m

rate of revolution = 3 revolution/sec

height above the ground = 2 m

velocity = \dfrac{distance}{time}

circumference of the circle = 2 π r

                                             = 2 x π x 1.7

                                             = 10.68 m

velocity = \dfrac{3 \times 10.68}{1}

v = 32.04 m/s

using conservation of energy

\dfrac{1}{2}mv^2 + mgh = mgH

\dfrac{1}{2}v^2 + gh = gH

\dfrac{1}{2}32.04^2 + 9.8\times 2 = 9.8\times H

532.88 = 9.8\times H

H = 54.37

the maximum height reached by the ball is equal to H = 54.37

4 0
3 years ago
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