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tamaranim1 [39]
4 years ago
11

What nuclear reaction is the energy source of a white dwarf?

Physics
1 answer:
sammy [17]4 years ago
7 0

Answer:

Multiple choice answer would be "None"

Explanation:

White dwarfs are radiating stored heat from earlier reactions.  

Technically, it would be the last fusion stage the star went through  

BEFORE it became a white dwarf, but that's nit-picking.

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What would it mean if neither the red or blue litmus paper changes colors?
Anna35 [415]

if the color changes, it is neutral but if it stays the same, it is an acid.

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3 years ago
How does the electric force between two charged particles change if particle's charge is reduced by a factor of 3?
MAVERICK [17]

The force between two charges is proportional to the product of the charges.

If only one of the charges is reduced by a factor of 3, then the force is reduced by a factor of 3.

If both charges are reduced by a factor of 3, then the force is reduced by a factor of 9.

3 0
4 years ago
Read 2 more answers
In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 78.5 m/s. Th
Andrews [41]

Answer:

FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

Explanation:

First you have to consider that the Ford Thunderbird (FT) follows a rectilinear motion with varying acceleration, while Mercedez Benz (MB) has a constant velocity (no acceleration). So if you finde the time spent by FT in each section, and the distance, then you will find the distance for MB.

1) Vf² = Vi² + 2ad, where Vf: final velocity, Vi: ionitial velocity, a: acceleration and d: distance.

For the first portion  (0 m/s)² = (78.5 m/s)² + 2a(250 m) ⇒

-(78.5 m/s)² / 2(250m) = a ⇒ a = -12.3 m/s².

Now, you can find the corresponding time for this section with the following formule: Vf = Vi + at ⇒ 0 m/s = 78.5 m/s + (-12.3 m/s²) t

⇒ t= (-78.5 m/s)/ (-12.3 m/s²) ⇒ t= 6.4 seconds.

2) Then FT spent 5 seconds in the pit.

3) The the FT accelerates until reach 78.5 m/s again in a distance of 370 m.

Vf² = Vi² + 2ad ⇒ (78.5 m/s)² = (0 m/s)² + 2a(370 m)

⇒ (78.5 m/s)²/ 2(370 m) = a ⇒ a = 8.3 m/s²

Then, Vf = Vi + at ⇒ 78.5 m/s = 0 m/2 + (8.3 m/s²) t

⇒ (78.5 m/s)/(8.3 m/s²) = t ⇒ t = 9.5 seconds.

4) Summarizing, the FT moves 620 meters (250 + 370 mts) in 20.9 seconds ( 6.4 s + 5 s + 9.5 s).

5) During this time, MB moves

Velocity = distance/ time ⇒ Velocity x time = Distance

⇒ Distance = (78.5 m/s) x  (20.9 seconds) ⇒ Distance = 1640.6 meters

6) Finally, the FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

3 0
3 years ago
How do scientist learn about stars and galaxies even tough humans have never been there?
sergiy2304 [10]
Through satellites they launch into space.
5 0
3 years ago
A 5.17 kg block free to move on a horizontal, frictionless surface is attached to one end of a light horizontal spring. The othe
Andreyy89

Answer:

v=0.816 m/s

Explanation:

The force of the spring and the motion of the block are in equilibrium so without any force of friction the motion is

E_{s}=E_{k}

\frac{1}{2}*k*d_{s}^2=\frac{1}{2}*m*v^2

First determinate the constant of the spring that produce the kinetic energy of the bloc

k=\frac{m*v^2}{d_{s}^2}

k=\frac{5.17kg*(1.30\frac{m}{s})^2}{0.102^2m}

k=839.8 \frac{kg}{s^2}

Now the motion with the force of friction in the kinetic

E_{s}=E_{k}-W_{k}

\frac{1}{2}*k*d_{s}^2=\frac{1}{2}*m*v^2-u*m*g

Resolve to v

v^2=\frac{(k*d_{s}^2)+(2*g*u)}{m}

v=\sqrt{\frac{839.8*(0.102m)^2-2*9.8*0.270}{5.17}}

v=0.81 \frac{m}{s}

3 0
3 years ago
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