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snow_lady [41]
4 years ago
13

Release an electron initially at rest in the presence of an electric field. The electron tends to go to the region of 1. same el

ectric potential. 2. indeterminate; sometimes higher, sometimes lower. 3. lower electric potential. 4. higher electric potential.
Physics
1 answer:
olga nikolaevna [1]4 years ago
4 0

Answer:

The electron tends to go to the region of 4. higher electric potential.

Explanation:

When a charged particle is immersed in an electric field, it experiences a force given by

F=qE

where

q is the charge of the particle

E is the electric field

The direction of the force depends on the sign of the charge. In particular:

- The force and the electric field have the same direction if the charge is positive

- The force and the electric field have opposite directions if the charge is negative

Therefore, an electron (negative charge) moves in the direction opposite to the electric field lines.

However, electric field lines go from points at higher potential to points at lower potential: so, electrons move from regions at lower potential to regions of higher potential.

Therefore, the correct answer is

The electron tends to go to the region of 4. higher electric potential.

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In an elastic collision momentum is conserved as is
zzz [600]

In an elastic collision momentum is conserved as well as the kinetic energy

Explanation:

In physics, there are two types of collisions:

  • Elastic collision: in an elastic collision, the total momentum of the system is  conserved, and the total kinetic energy of the system is conserved as well. This is because there are no internal frictions acting on the system, so the energy is conserved. An example of elastic collision is (approximately) that occurring between two billiard balls.
  • Inelastic collision: in an inelastic collision, the total momentum of the system is conserved, while the total kinetic energy is not. In fact, due to the presence of internal frictions, part of the total energy is converted into thermal energy and sound during the collision, and therefore "wasted", so the final total kinetic energy is less than the initial one. An example of inelastic collision is the collision between two cars. The maximum amount of kinetic energy is lost when the two objects stick together after the collision; in this case, we talk about perfectly inelastic collision.

Therefore, the complete sentence is

In an elastic collision momentum is conserved as well as the kinetic energy

Learn more about collisions:

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#LearnwithBrainly

7 0
3 years ago
How did unarmed defense techniques develop? Why were they needed? Why not just use a knife or a sword?
Arturiano [62]

Answer:

Buddhist monks traveled from india to china and because of their beliefs they weren't allowed to use weapons.

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6 0
3 years ago
A uniform thin rod of length 0.84 m and mass 4.6 kg can rotate in a horizontal plane about a vertical axis through its center. T
nasty-shy [4]

Answer:

The velocity of the bullet is  v_o=2.2886 *10^4 m/s

Explanation:

The free body diagram of this question is shown on the first uploaded image  

Since we are told that the bullet got lodged in the rod we are going to look at  the bullet and rod combined as a system

And this also means that the collision is inelastic since the both object stuck together

examining this question we can deduce that the pivot on the rotational axis also exerts an external force on the system (the rod + the bullet)

These eternal force does not cause rotation but hold the rod along the horizontal plane hence the these is no external which mean that the angular momentum is conserved

Generally angular momentum is mathematically represented as

                  L_b = +mvb

This positive because the bullet misses the axis to the right (Which means it would be moving in a counter clock- wise direction around the axis)

   Before the collision v =v_o

 And b is the minimum distance of approach as shown on the diagram

According to trigonometry

                         b = \frac{l}{2} sin \theta

Now the total angular momentum before collision is

                  L_i = L_b = m v_o\frac{l}{2} sin(60^o)

Generally moment of inertia is mathematically represented as

               I = mr^2

Where m is the mass and r is the distance to the axis of rotation

After collision since the bullet and the rod combined and became a system

           their moment of inertia would be the same and is mathematically given as

      Assuming that it was \frac{1}{3} \ of \ Mass \ of\  the \  rod \ that was\ hit\ by \  the \ bullet \ since \ they \\ stuck \ together

                  I = I_{rod} + mr^2 = \frac{1}{3} M(\frac{l}{2}) ^2 +m(\frac{l}{2} )^2 = (\frac{M}{12} +\frac{m}{4}  )l^2 = \frac{1}{12}(M + 3m)l^2

                 I = I_{rod} + mr^2 = \frac{1}{3 *4} Ml^2 +m(\frac{l}{2} )^2 = (\frac{M}{12} +\frac{m}{4}  )l^2 = \frac{1}{12}(M + 3m)l^2

                I = I_{rod} + mr^2 = \frac{1}{12} Ml^2 +m(\frac{l}{2} )^2 = (\frac{M}{12} +\frac{m}{4}  )l^2 = \frac{1}{12}(M + 3m)l^2

Now as the system rotates due the force of the bullet and the fact that the plane is fixed about the vertical axis at an angular speed of

                w_f = 12.0 rad/s

The final momentum is mathematically represented  as

                        L_f = Iw_f = \frac{1}{12}(M +3m)l^3w_f  

Since Angular momentum is conserved

                    L_i = L_f

               mv_o\frac{l}{2}sin(60^o) = \frac{1}{12} (M +3m)l^2w_f

Making v_o the subject of the formula gives

                      v_o = \frac{2}{mlsin(60^o)} \cdot  \frac{1}{12} (M+3m)l^2w_f =\frac{(M+3m)lw_f}{6msin(60^o)}  

                        = \frac{(4.6Kg+(3* 3.9 *\frac{1kg}{1000} ))(0.84m)(12 rad/s)}{6(3.9*\frac{1Kg}{1000} )(\frac{\sqrt{3} }{2} )}

                         v_o=2.2886 *10^4 m/s

                           

                               

                     

               

                 

8 0
3 years ago
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