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natali 33 [55]
3 years ago
5

About how long will a 2-m¤ star live as a main-sequence star quizle

Physics
1 answer:
tia_tia [17]3 years ago
8 0
It's mass and its age.
About how long will a 2-M star live as a main-sequence star.
1 billion years.
<span>A low mass red giant star's energy comes from.</span>
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An astronomer discovers a massive galaxy which has four nuclei. What is a likely explanation for a galaxy having more than one n
statuscvo [17]

The universe is made of a lot of galaxies. The likely explanation for a galaxy having more than one nucleus is that the galaxy must have swallowed several smaller galaxies that were its neighbors.

  • There are some reason behind the theory that the eating up of a galaxy is by another galaxy. The galaxy is said to have a  powerful past due to the fact  that a lot of smaller galaxies were eaten up.

The universe is known to have a very largest galaxy called the giant elliptical galaxy that is made up of about a trillion stars.

Learn more from

brainly.com/question/22885678

6 0
2 years ago
If the action force is the swimmer pushing water in the leftward direction, what is the reaction force?
Sphinxa [80]
  • According to Newton's Third Law of Motion, to every action, there is an equal and opposite reaction; action and reaction act on different bodies.
  • Here, the action force is in the leftward direction, so the reaction will be in the opposite direction.
  • If the action force is the swimmer pushing water in the leftward direction, then the reaction force is in the rightward direction.
  • And the reaction force will be given by the water on the swimmer.

<u>Answer</u><u>:</u>

<u>The </u><u>reaction </u><u>force </u><u>is </u><u>the </u><u>water </u><u>pushing </u><u>the </u><u>swimmer </u><u>in </u><u>the </u><u>rightward </u><u>direction</u><u>.</u>

Hope you could get an idea from here.

Doubt clarification - use comment section.

3 0
2 years ago
: A 70 kg man and a 12 kg sled are on the frictionless ice of a frozen lake, 25 m apart but connected by a rope of negligible ma
e-lub [12.9K]

Answer:

x_1 = 3.74m

Explanation:

given,

mass of man = 70 kg

mass of sled = 12 kg

F = m a_s

a_s = \dfrac{F}{m}

a_s = \dfrac{8.2}{12}

a_s = 0.68\ m/s^2

F = m a_m

a_m = \dfrac{F}{m}

a_m = \dfrac{8.2}{70}

a_m = 0.12\ m/s^2

x_1+x_2 = 25

\dfrac{1}{2}a_ct^2+ \dfrac{1}{2}a_mt^2 = 25

(a_c+a_m)t^2=50

(0.12+0.68)t^2=50

t = \sqrt{\dfrac{50}{0.8}}

t = 7.90 s

x_1 = \dfrac{1}{2}a_ct^2

x_1 = 0.5\times 0.12 \times 7.90^2

x_1 = 3.74m

5 0
3 years ago
A 2. 0 μf and a 4. 0 μf capacitor are connected in series across an 8. 0-v dc source. what is the charge on the 2. 0 μf capacito
Nezavi [6.7K]

voltage across 2.0μf capacitor is 5.32v

Given:

C1=2.0μf

C2=4.0μf

since two capacitors are in series there equivalent capacitance will be

[tex] \frac{1}{c} = \frac{1}{c1} + \frac{1}{c2} [/tex]

c =  \frac{c1 \times c2}{c1 + c2}

=  \frac{2 \times 4}{2 + 4}

=1.33μf

As the capacitance of a capacitor is equal to the ratio of the stored charge to the potential difference across its plates, giving: C = Q/V, thus V = Q/C as Q is constant across all series connected capacitors, therefore the individual voltage drops across each capacitor is determined by its its capacitance value.

Q=CV

given,V=8v

= 1.33 \times 10 {}^{ - 6}  \times 8

= 10.64 \times 10 {}^{ - 6}

charge on 2.0μf capacitor is

\frac{Qeq}{2 \times 10 {}^{ - 6} }

=  \frac{10.64 \times 10 {}^{ - 6} }{2 \times 10 {}^{ - 6} }

=5.32v

learn more about series capacitance from here: brainly.com/question/28166078

#SPJ4

3 0
1 year ago
Two charged particles, with charges q1=q and q2=4q, are located at a distance d= 2.00cm apart on the x axis. A third charged par
Murrr4er [49]

Answer:

X₃₁ = 0.58 m  and  X₃₂ = -1.38 m

Explanation:

For this exercise we use Newton's second law where the force is the Coulomb force

        F₁₃ - F₂₃ = 0

        F₁₃ = F₂₃

Since all charges are of the same sign, forces are repulsive

        F₁₃ = k q₁ q₃ / r₁₃²

        F₂₃ = k q₂ q₃ / r₂₃²

Let's find the distances

         r₁₃ = x₃- 0

         r₂₃ = 2 –x₃

We substitute

      k q q / x₃² = k 4q q / (2-x₃)²

      q² (2 - x₃)² = 4 q² x₃²

        4- 4x₃ + x₃² = 4 x₃²

        5x₃² + 4 x₃ - 4 = 0

We solve the quadratic equation

        x₃ = [-4 ±√(16 - 4 5 (-4)) ] / 2  5

        x₃ = [-4 ± 9.80] 10

       X₃₁ = 0.58 m

       X₃₂ = -1.38 m

For this two distance it is given that the two forces are equal

7 0
3 years ago
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