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3 years ago

Under what circumstances will a binary star produce a nova?

1 answer:

5 0

When a star uses up all of it's energy and begins to die, it swells up to become a red giant star. This causes its surface gravity to decrease, thereby allowing some of its mass to escape into space. A binary star is a pair of stars that orbit each other because of their gravitational attraction to each other. When one member of the binary pair uses up all of its energy and begins to die, it loses mass due to the reduction in surface gravity. But instead of escaping into space, this mass is attracted to the companion star because of its gravitational pull. That increases the mass of the companion star. In a process that takes thousands of years, enough matter is transfered that causes the temperature and pressure to increase sufficiently to result in nuclear fusion reactions on the companion star. When these nuclear reactions become extremely violent, the released nuclear energy increases the brightness of this companion star dramatically, thereby creating a nova. Therefore, it is the dying of one of the stars in a binary system along with a sufficient transfer of star mass to sustain nuclear reactions that results in a nova.

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A 1.90-m-long barbell has a 25.0 kg weight on its left end and a 37.0 kg weight on its right end. if you ignore the weight of th

gregori [183]

Answer: The center of gravity is 1.1338 m away from the left side of the barbell

Explanation:

Length of the barbell = 1.90 m

The distance center of gravity from left = x

Mass on the left side = 25 kg

The distance center of gravity from right = 1.90 - x

Mass on the right side = 37 kg

At the balance point: $m_1x_1=m_2x_2$

$25 kg\times x=37 kg\times (1.90-x)$

$x=1.1338 m$

The center of gravity is 1.1338 m away from the left side of the barbell

7 0

3 years ago

Read 2 more answers

A bottle lying on the windowsill falls off and takes 4.95 seconds to reach the ground. The distance from the windowsill to the g

Liula [17]

The distance an object falls from rest through gravity is
D = (1/2) (g) (t²)
   Distance = (1/2 acceleration of gravity) x (square of the falling time)

We want to see how the time will be affected
if ' D ' doesn't change but ' g ' does.
So I'm going to start by rearranging the equation
to solve for ' t '.                                                      D = (1/2) (g) (t²)

Multiply each side by 2 :         2 D =            g    t²

Divide each side by ' g ' :      2 D/g =                  t²

Square root each side:        t = √ (2D/g)

Looking at the equation now, we can see what happens to ' t ' when only ' g ' changes:

-- ' g ' is in the denominator; so bigger 'g' ==> shorter 't'

   and smaller 'g' ==> longer 't' .--

They don't change by the same factor, because 1/g is inside the square root. So 't' changes the same amount as √1/g does.

Gravity on the surface of the moon is roughly 1/6 the value of gravity on the surface of the Earth.

So we expect ' t ' to increase by √6 = 2.45 times.

It would take the same bottle (2.45 x 4.95) = 12.12 seconds to roll off the same window sill and fall 120 meters down to the surface of the Moon.

5 0

3 years ago

In Challenge Example 11.9 (p. 280), after the explosion, suppose that the m1 fragment shot directly north at 12 m/s and the m3 f

Inga [223]

The question is incomplete. The mass of the object is 10 gram and travelling at a speed of 2 m/s.

Solution:

It is given that mass of object before explosion is,m = 10 g

Speed of object before explosion, v = 2 m/s

Let $m_1, m_2 \text{ and}\ m_3$ be the masses of the three fragments.

Let $v_1, v_2 \text{ and}\ v_3$ be the velocities of the three fragments.

Therefore, according to the law of conservation of momentum,

$mv=m_1v_1 +m_2v_2+m_3v_3$

$10 \times 2 \hat i=3 \times 12 \hat{j} + 3(v_{2x} \hat{i}+v_{2y} \hat{j})-4 \times 9 \hat{j}$

So the x- component of the velocity of the m2 fragment after the explosion is,

$3v_{2x} = 20$

∴ $v_{2x} = 6.67 \ m/s$

6 0

3 years ago

A 0.0250-kg bullet is accelerated from rest to a speed of 550 m/s in a 3.00-kg rifle. The pain of the rifle’s kick is much worse

kondaur [170]

Answer:

a) 4.583 m/s

b) 31.505 J

c) 0.491 m/s

d) 3.375 J

p_player = (110 kg)(8 m/s) = 880 kg m/s

p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

Explanation:

HI!

We can calculate the recoil velocity by conservation of momentum, remember that p=mv.

The momentum of the bullet is:

p_b = (0.0250 kg)*(550 m/s )

The momentum of the rifle is:

p_r = (3 kg) * v

Since the total initial momentum is zero:

p_b = p_r

That is:

v = (550 m/s ) (0.0250 kg/ 3 kg ) = 4.583 m/s

The kinetic energy gained by the rifle is:

K = (1/2) m v^2 = (1/2) *(3 kg) *(4.583 m/s)^2 = 31.505 J

We use the same formula as in a), but with m=28kg instead of 3 kg

v = (550 m/s ) (0.0250 kg/ 28 kg ) = 0.491 m/s

Again, the same formula as b, but with m=28 and v=0.491 m/s

K = 3.375 J

p_player = (110 kg)(8 m/s) = 880 kg m/s

p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

I believe that the kinetic energy is more related to the problem than the momentum. The relation between these two quantities is:

K = p^2/(2m)

usiing this relation, we get:

K_player = 3520 J

K_ball = 128.125 J

Therefore the kinetic energy of the player is around 27 time larger than the kinetic energy of the ball, that being said, the pain of being tackled by that player is around 27 times greater that being hit by the ball!

4 0

3 years ago

A weightlifter lifts a 1250-N barbell 2 m in 3 s.How much power was used to lift the barbell?

Vlad [161]

The power is 833.3 W

Explanation:

First of all, we need to calculate the work done in lifting the barbell, which is equal to the change in gravitational potential energy of the barbell:

$W=(mg)h$