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vfiekz [6]
4 years ago
9

After a great many contacts with the charged ball, how is any charge on the rod arranged (when the charged ball is far away)?The

re is positive charge on end B and negative charge on end A.There is negative charge spread evenly on both ends.There is negative charge on end A with end B remaining neutral.There is positive charge on end A with end B remaining neutral.The conducting rod is not grounded, so a negative charge accumulates on one end, but a charge cannot remain at a single end in a conductor, so it will spread to both ends:There is negative charge spread evenly on both ends
Physics
1 answer:
lozanna [386]4 years ago
8 0

Answer:

The correct option is that both the ends will remain neutral.

Explanation:

As the rod is grounded all of the prolonged charge will be converted to ground so the overall charge on the rod in absence of the charged ball will be equal to zero.Such that the both ends will not bear any charge.

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A small ball of mass 2.00 kilograms is moving at a velocity 1.50 meters/second. It hits a larger, stationary ball of mass 5.00 k
rewona [7]

The kinetic energy of the small ball before the collision is

                             KE  =  (1/2) (mass) (speed)²

                                     = (1/2) (2 kg) (1.5 m/s)

                                     =    (1 kg)  (2.25 m²/s²)

                                     =        2.25 joules.

Now is a good time to review the Law of Conservation of Energy:

                     Energy is never created or destroyed. 
                     If it seems that some energy disappeared,
                     it actually had to go somewhere.
                     And if it seems like some energy magically appeared,
                     it actually had to come from somewhere.

The small ball has 2.25 joules of kinetic energy before the collision.
If the small ball doesn't have a jet engine on it or a hamster inside,
and does not stop briefly to eat spinach, then there won't be any
more kinetic energy than that after the collision.  The large ball
and the small ball will just have to share the same 2.25 joules.

3 0
4 years ago
Block A has mass 1.00 kg and block B has mass 3.00 kg. The blocks collide and stick together on a level, frictionless surface. A
Pani-rosa [81]

Answer:

1/2mv²=0

1/2(4kg)(v²)=0

2=-v²

square root -2=v

v=1.414

5 0
3 years ago
True or False. Father of a multitude" obeyed God's command to kill His son because the Ten Commandments had not yet been given.
velikii [3]

Answer:

Explanation:

The father of the Multitude was Abraham. And he did precede the 10 commandments. But the story really has nothing to do with the 10 commandments.

He obeyed God's Command because it was God who issued it. Abraham had complete faith in God's ability to make things come out right. The sacrifice of Isaac was a test of faith.

I would advise you to answer true, but whoever wrote the question employed a twisted kind of logic. The sacrifice of Isaac had nothing to do with the 10 commandments. <u><em>Thou Shalt not murde</em></u>r is not really covered by this circumstance.

So true might not be the correct answer. The question is very complex and does not lend itself to a simple True/ False answer.

4 0
3 years ago
1. What is meant by half-life?
Bezzdna [24]
1. The time for a radioactive sample to reduce to half of its original mass.
4. 10% of 232 is 23.2 times 6 equals 139.2
68.9 x6 equals 413.4
5. 0.37kg
6. 2000 years
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8 0
4 years ago
One billiard ball is shot east at 2.2 m/s. A second, identical billiard ball is shot west at 0.80 m/s. The balls have a glancing
Leona [35]

Answer:

(a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

Explanation:

Given that,

Velocity of one ball u₁= 2.2i m/s

Velocity of second ball u₂=- 0.80i m/s

Final velocity of the second ball v₂= 1.36j m/s

The mass of the identical balls are

m = m_{1}=m_{2}

(a). We need to calculate the speed of the first ball after the collision

Using law of conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

Along X- axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}

v_{1}=u_{1}+u_{2}

Put the value into the formula

v_{1}=2.2i-0.80i

v_{1}=1.4i\ m/s

Along Y-axis

0=m_{1}v_{1}+m_{2}v_{2}

m_{1}v_{1}=-m_{2}v_{2}

v_{1}=-v_{2}

Put the value into the formula

v_{1}=-1.36j\ m/s

Then the final speed of the first ball

v_{1}=\sqrt{(1.4)^2+(1.36)^2}

v_{1}=1.95\ m/s

(b) We need to calculate the direction of the first ball after the collision

Using formula of direction

\tan\theta=\dfrac{v_{2}}{v_{1}}

\tan\theta=\dfrac{-1.36}{1.4}

\theta=\tan^{-1}\dfrac{-1.36}{1.4}

\theta=-44.16^{\circ}

Negative sign shows the direction of first ball .

Hence, (a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

7 0
3 years ago
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