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KiRa [710]
4 years ago
12

As material from the outer envelope of the larger star builds up on the surface of the white dwarf, its temperature and density

increase continuously. Under this scenario, what is the first event that will eventually happen?
Physics
1 answer:
stepan [7]4 years ago
7 0

Answer:

Hydrogen will undergo nuclear fusion ,

Explanation:

here, it is being explained:-

The first thing that will inevitably arise would be that nuclear fusion would arise with hydrogen.

When fuel from the larger star's outer shell piles up on the white dwarf's surface, the temperature and density continuously increase as a result of which hydrogen will undergo nuclear fusion.

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HELP!! WILL MARK BRAINLIEST!!!!!
krek1111 [17]

Answer:

Create

Broken

Explanation:

Bond formation or creation requires the use of energy. Energy is used during bond formation between chemical species. The energy is required for the reaction to occur.

  • When bonds are broken, energy is released
  • Bond breaking process is a procedure that releases energy.
  • This energy makes them able to react.
5 0
3 years ago
A slingshot fires a pebble from the top of a building at a speed of 14.7 m/s. The building is 36.0 m tall. Ignoring air resistan
monitta

Answer:

The final speed is <em>30.37 m/s</em> for the three directions.

Explanation:

Given the initial speed, v₀ = 14.7 m/s and the height of the building as 36.0 m (neglecting air resistance and height of slingshot), we apply the principle of Conversation of Mechanical Energy:

E = Kinetic Energy (k.E) + Potential Energy (P.E)

K.E = 0.5mv²

P.E = mgh

∴ <em>0.5mv₀² + mgh₀ = 0.5mv₁² + mgh₁</em>

Since the final height (h₁) is zero and <em>m</em> is a common term, then we can re-write the above equation as:

<em>0.5v₀² + gh₀ = 0.5v₁² + 0</em>

<em>0.5v₁² - 0.5v₀² = gh₀</em>

<em>0.5(v₁² - v₀²) = gh₀</em>

<em>v₁² = v₀² + 2gh₀</em>

Fitting in the given terms, we can calculate the final speed

v₁² = (14.7)² + (2)(9.81)(36.0)

   = 216.09 + 706.32 = 922.41

v₁ = √922.41 = 30.37 m/s.

The speed at which the pebble hits the ground remains the same because kinetic energy depends on the speed of the object and not the direction to which the object is thrown.

7 0
4 years ago
If a train travel from Addis Ababa to Dire Dawa at a constant velocity of 400Km/hr and the to Djibouti at a constant velocity of
madreJ [45]

The value of V is 640 km/hr.

<h3>What is Average velocity?</h3>

Average velocity is the change in position or displacement (∆x) divided by the time intervals (∆t) in which the displacement occurs.

To calculate the value of V from the question, we use the formula below.

  • V' = (V+U)/2............ Equation 1

Making V the subject of the equation

  • V = 2V'-U.............. Equation 2

From the question,

⇒ Given:

  • V' = 520 km/hr
  • U = 400 km/hr

Substitute these values into equation 2

  • V = (520×2)-400
  • V = 640 km/hr

Hence, the value of V is 640 km/hr.

Learn more about average velocity here: brainly.com/question/4931057

#SPJ1

4 0
2 years ago
The magnetic flux through a metal ring varies with time t according to ΦB = 10 at^3 − 5 bt^2, where Φ is in webers, a = 6.00 s^
mixas84 [53]

Answer bruh I really dont know im just trying to set this account

Explanation:

3 0
4 years ago
You have two small spheres, each with a mass of 2.40 grams, separated by a distance of 10.0 cm. You remove the same number of el
nordsb [41]

Answer:

q = 2.066* 10⁻¹³ C.

n = 1,291,250 electrons.

Explanation:

1)

  • If the gravitational attraction is equal to their electrical repulsion, we can write the following equation:

       F_{g} = F_{c} (1)

  • where Fg is the gravitational attraction, that can be written as follows        according Newton's Universal Law of Gravitation:

       F_{g} = G*\frac{m_{1}*m_{2}}{r_{12}^{2}} (2)

  • Fc, due to it is the electrical repulsion between both charged spheres, must obey Coulomb's Law (assuming  we can treat both spheres as point charges), as follows:

       F_{c} = k*\frac{q_{1}*q_{2}}{r_{12}^{2}} (3)

  • since m₁ = m₂ = 0.0024 kg, and  r₁₂ = 0.1m, G and k universal constants, and q₁ = q₂ = Q, we can replace the values in (2) and (3), so we can rewrite (1) as follows:

       G*\frac{(0.0024kg)^{2}}{r_{12}^{2}} = k*\frac{Q^{2}}{r_{12}^{2}} (4)

  • Since obviously the distance is the same on both sides, we can cancel them out, and solve (4) for Q² first, as follows:

       Q^{2} = \frac{6.67e-11*(0.0024kg)^{2}}{9e9Nm2/C2} = 4.27*e-26 C2 (5)

  • Since both charges are the same, the charge on each sphere is just the square root of (5):
  • Q = 2.066* 10⁻¹³ C.

2)

  • Assuming that both spheres were electrically neutral before being charged, the negative charge removed must be equal to the positive charge on the spheres.
  • Now, since each electron carries an elementary charge equal to -1.6*10⁻¹⁹ C, in order to get the number of electrons removed from each sphere, we need to divide the charge removed from each sphere (the outcome of part 1) with negative sign) by the elementary charge, as follows:
  • n_{e} =\frac{-2.066e-13C}{-1.6e-19C} = 1,291,250 electrons. (6)
4 0
3 years ago
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