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1 year ago

What two forces keep stars in equilibrium?

Physics

1 answer:

natima [27]1 year ago

3 0

According to the given information self gravity two forces keep stars in equilibrium.

The correct option is B.

<h3>What you mean by forces?</h3>

The push or pull on a mass-containing item alters its velocity. An external force is an agent that has the power to alter the rest or flowing condition of a body. It has a direction and a magnitude. Like we discovered in a previous session, forces are significant because they create motion changes.

<h3>What is effect of force?</h3>

An object can change its dimensions or shape, begin moving, stop moving, accelerate, or decelerate as a consequence of the a force acting on it. When two things contact, they exert pressures on each other that are identical in size but directed in the opposite direction.

To know more about Force visit:

brainly.com/question/13191643

#SPJ1

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An object moving at 38m/s takes 6s to come to a stop. What is the object's acceleration? _______m/s2

bearhunter [10]

Final velocity = 0m/s
Initial velocity = 38m/s

Acceleration = 0-38/6 = -38/6 = -6.33m/s2

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4 years ago

What is an isotope?

NikAS [45]

Answer:

Explanation:

they are atoms of the same element that have the same number of protons but different number of neutrons.

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3 years ago

A constant electric field with magnitude 1.50 ✕ 103 N/C is pointing in the positive x-direction. An electron is fired from x = −

romanna [79]

Answer:

The speed of electron is $1.5\times10^{7}\ m/s$

Explanation:

Given that,

Electric field $E=1.50\times10^{3}\ N/C$

Distance = -0.0200

The electron's speed has fallen by half when it reaches x = 0.190 m.

Potential energy $P.E=5.04\times10^{-17}\ J$

Change in potential energy $\Delta P.E=-9.60\times10^{-17}\ J$ as it goes x = 0.190 m to x = -0.210 m

We need to calculate the work done by the electric field

Using formula of work done

$W=-eE\Delta x$

Put the value into the formula

$W=-1.6\times10^{-19}\times1.50\times10^{3}\times(0.190-(-0.0200))$

$W=-5.04\times10^{-17}\ J$

We need to calculate the initial velocity

Using change in kinetic energy,

$\Delta K.E = \dfrac{1}{2}m(\dfrac{v}{2})^2+\dfrac{1}{2}mv^2$

$\Delta K.E=\dfrac{-3mv^2}{8}$

Now, using work energy theorem

$\Delta K.E=W$

$\Delta K.E=\Delta U$

So, $\Delta U=W$

Put the value in the equation

$\dfrac{-3mv^2}{8}=-5.04\times10^{-17}$

$v^2=\dfrac{8\times(5.04\times10^{-17})}{3m}$

Put the value of m

$v=\sqrt{\dfrac{8\times(5.04\times10^{-17})}{3\times9.1\times10^{-31}}}$

$v=1.21\times10^{7}\ m/s$

We need to calculate the change in potential energy

Using given potential energy

$\Delta U=-9.60\times10^{-17}-(-5.04\times10^{-17})$

$\Delta U=-4.56\times10^{-17}\ J$

We need to calculate the speed of electron

Using change in energy

$\Delta U=-W=-\Delta K.E$

$\Delta K.E=\Delta U$

$\dfrac{1}{2}m(v_{f}^2-v_{i}^2)=4.56\times10^{-17}$

Put the value into the formula

$v_{f}=\sqrt{\dfrac{2\times4.56\times10^{-17}}{9.1\times10^{-31}}+(1.21\times10^{7})^2}$

$v_{f}=1.5\times10^{7}\ m/s$

Hence, The speed of electron is $1.5\times10^{7}\ m/s$

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3 years ago

How much power would be needed to lift a 1.0x10^3N crate 1.5m in 5 seconds?

lina2011 [118]

Answer:power =300watts

Explanation:

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