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ladessa [460]
2 years ago
15

Compare the events in the life of the Sun with those of a star that starts with less

Physics
1 answer:
GarryVolchara [31]2 years ago
6 0

A star with greater mass will die out faster than the Sun.

<h3>What factors star is dependent on?</h3>

A star's future relies upon its mass. For the most part, the more huge the star, the quicker it consumes its fuel supply, and the more limited its life. The most huge stars can wear out and detonate in a cosmic explosion after two or three million years of combination.

Our Sun is a typical estimated star: there are more modest stars and bigger stars, even up to multiple times bigger. Numerous other planetary groups have different suns, while our own simply has one. The Sun is made for the most part out of hydrogen and helium gas.

In this manner, one correlation in the occasions in the existence of the Sun with those of a star that beginnings with a mass multiple times more prominent than the Sun's is a star that has a more noteworthy mass will vanish quicker.

Learn more about Star.

brainly.com/question/21458024

#SPJ1

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How much energy in Joules (J) would an electric heater that draws 9.5 A when connected to a 120 V supply use if the heater were
leonid [27]
<h2>Energy used by heater is 8.21 x 10⁶ J</h2>

Explanation:

Energy = Power x Time

Power = Voltage x Current

Voltage = 120 V

Current = 9.5 A

Power = Voltage x Current

Power = 120 x 9.5 = 1140 W

Time = 2 hours = 2 x 60 x 60 = 7200 s

Energy = Power x Time

Energy = 1140 x 7200

Energy = 8208000 J

Energy used by heater is 8.21 x 10⁶ J

7 0
3 years ago
"Two uniform identical solid spherical balls each of mass M and radius R" and moment of inertia about its center 2/5 MR2 are rel
adelina 88 [10]

Answer:

he sphere that uses less time is sphere A

Explanation:

Let's start with ball A, for this let's use the kinematics relations

        v² = v₀² - 2g (y-y₀)

indicate that the sphere is released therefore its initial velocity is zero and when it reaches the floor its height is zero y = 0

         v² = 0 - 2 g (0- y₀)

         v = \sqrt{2g y_o}

         v = \sqrt{2 \ 9.8\ H}

         v = 4.427 √H

Now let's work the sphere B, in this case it rolls down a ramp, let's use the conservation of energy

starting point. At the highest point, before you start to move

         Em₀ = U = m g y

final point. At the bottom of the ramp

         Em_f = K = ½ m v² + ½ I w²

notice that we include the kinetic energy of translation and rotation

energy is conserved

          Em₀ = Em_f

          mg H = ½ m v² + ½ I w²

angular and linear velocity are related

          v = w r

          w = v / r

the momentorot of inertia indicates that it is worth

          I = \frac{2}{5} m r²

we substitute

           m g H = ½ m v² + ½ (\frac{2}{5}  m r²) (\frac{v}{r} )²

           gH = \frac{1}{2}  v² + \frac{1}{5}  v² = \frac{7}{10}  v²

           v = \sqrt{\frac{10}{7} \ g H}

           v = \sqrt{ \frac{10}{7}  \ 9.8 \ H}

           v=3.742 √H

Taking the final speeds of the sphere, let's analyze the distance traveled, sphere A falls into the air, so the distance traveled is H.  The ball B rolls in a plane, so the distance (L) traveled can be found with trigonometry

           sin θ = H / L

           L = H /sin θ

we can see that L> H

In summary, ball A arrives with more speed and travels a shorter distance, therefore it must use a shorter time

Consequently the sphere that uses less time is sphere A

5 0
2 years ago
Two equal, but oppositely charged particles are attracted to each other electrically. The size of the force of attraction is 48.
galina1969 [7]

Given:

The force of attraction is F = 48.1 N

The separation between the charges is

\begin{gathered} r=\text{ 60.9 cm} \\ =\text{ 60.9}\times10^{-2}\text{ m} \end{gathered}

Also, the magnitude of charge q1 = q2 = q.

To find the magnitude of charge.

Explanation:

The magnitude of charge can be calculated by the formula

\begin{gathered} F=\frac{k(2q)}{r^2} \\ q=\frac{Fr^2}{2k} \end{gathered}

Here, k is the Coulomb's constant whose value is

k\text{ = 9}\times10^9\text{ N m}^2\text{ /C}^2

On substituting the values, the magnitude of charge will be

\begin{gathered} q=\frac{48.1\times(60.9^\times10^{-2})^2}{2\times9\times10^{^9}} \\ =9.91\times10^{-10}\text{ C} \\ =9.91\times10^{-4}\text{ }\mu C \end{gathered}

Thus, the magnitude of each charge is 9.91 x 10^(-4) micro Coulombs.

6 0
1 year ago
Differentiate the following functions with respect to x <br>xsin x​
gtnhenbr [62]

Answer:

Here is your answer

Hope it helps

6 0
3 years ago
Read 2 more answers
If a 70-kg swimmer pushed off a pool wall with a force of 250N at what rate will the swimmer accelerate from the wall
Vlad [161]
F = ma
250 = 70 x a
a = 250/70
a = 3.57
5 0
3 years ago
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