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Cerrena [4.2K]
3 years ago
7

If the equipotential surfaces due to some charge distribution are vertical planes, what can you say about the electric field dir

ection in this region: (a) it is vertically upward, (b) it is vertically downward, (c) it is horizontally to the left, (d) it is horizontally to the right, or (e)either (c) or (d) could be correct?
Physics
1 answer:
Elina [12.6K]3 years ago
7 0

Answer:

The correct option is

(e)either (c) or (d) could be correct.

Explanation:

The electric field of a charge radiates out in all directions and the intensity of the electric field strength given by E = F/q₀, diminishes as the lines of force moves further away from the source. The direction of F and E is in the line of potential motion of the source charge in the field.

Equipotential surfaces are locations in the radiated electric that have the same field strength or electric potential. The work done in moving within an equipotential surface is zero and as such since

Work = Force × distance = 0 where distance ≠ 0.

The force acting between two points on an equipotential surface is also zero or the component of the force within an equipotential surface is zero and since there is a force in the electric field, it is acting at right angles to the equipotential surface which could be horizontally to the left or right directions where the equipotential surfaces due to the charge distribution are in the vertical plane.

Therefore it is either horizontally to the left, or horizontally to the right.

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Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. S
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Answer:

The required angular speed the neutron star is 10992.32 rad/s

Explanation:

Given the data in the question;

mass of the sun M_S = 1.99 × 10³⁰ kg

Mass of the neutron star

M_N = 2( M_S )

M_N = 2( 1.99 × 10³⁰ kg )

M_N = ( 3.98 × 10³⁰ kg )

Radius of neutron star R_N = 13.0 km = 13 × 10³ m

Now, let mass of a small object on the neutron star be m

angular speed be ω_N.

During rotational motion, the gravitational force on the object supplies the necessary centripetal force.

GmM_N = / R_N² = mR_Nω_N²

ω_N² = GM_N = / R_N³

ω_N = √(GM_N = / R_N³)

we know that gravitational G = 6.67 × 10⁻¹¹ Nm²/kg²

we substitute

ω_N = √( (  6.67 × 10⁻¹¹ )( 3.98 × 10³⁰ ) ) / (13 × 10³ )³)

ω_N = √( 2.65466 × 10²⁰ / 2.197 × 10¹²

ω_N = √ 120831133.3636777

ω_N = 10992.32 rad/s

Therefore, The required angular speed the neutron star is 10992.32 rad/s

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3 years ago
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The famous Newton’s Third Law states that “For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object.”

By using this,

10grams or 0.01kg of bullet with speed 400 m/sec and 5kg gun recoil with speed suppose ‘v’.

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8 0
3 years ago
Two objects, m1 = 0.6 kg and m2 = 4.4 kg undergo a one-dimensional head-on collision
Sidana [21]

a) The velocity after the collision.is 11.456 m/s.

b) The kinetic energy lost due to the collision is 44.564 J.

<h3>What is conservation of momentum principle?</h3>

When two bodies of different masses move together each other and have head on collision, they travel to same or different direction after collision.

The external force is not acting here, so the initial momentum is equal to the final momentum. For inelastic collision, final velocity is the common velocity for both the bodies.

m₁u₁ +m₂u₂ =(m₁ +m₂) v

Given are the two objects, m1 = 0.6 kg and m2 = 4.4 kg undergo a one-dimensional head-on collision. Their initial velocities along the one-dimension path are vi1 = 32.4 m/s [right] and vi2 = 8.6 m/s [left].

(a) Substitute the values, then the final velocity will be

0.6 x32.4 +4.4 x 8.6 = (0.6+4.4)v

v = 11.456 m/s

Thus, the velocity after collision is 11.456 m/s.

(b) Kinetic energy lost due to collision will be the difference between the kinetic energy before and after collision.

= [1/2m₁u₁² +1/2m₂u₂² ] - [1/2(m₁ +m₂) v²]

Substitute the value, we have

= [1/2 x 0.6 x32.4² + 1/2 x4.4 x 8.6²] - [1/2 x(0.6+4.4)11.456²]

= 44.564 J

Thus, the kinetic energy lost due to the collision is 44.564 J.

Learn more about conservation of momentum principle

brainly.com/question/14033058

#SPJ2

4 0
2 years ago
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