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Norma-Jean [14]
3 years ago
11

Object A and object B are initially uncharged and are separated by a distance of 1 meter. Suppose 10,000 electrons are removed f

rom object A and placed on object B, creating an attractive force between A and B. An additional 10,000 electrons are removed from A and placed on B and the objects are moved so that the distance between them increases to 2 meters. By what factor does the electric force between them change?
Physics
1 answer:
bija089 [108]3 years ago
6 0

Answer:

1

Explanation:

Let's call the magnitude of the charge of 10000 electrons q.

Because A loses q, it has a net charge +q (electrons are negative; losing them creates positive charges). Conversely, B has gained q, so it has a charge of -q. The distance between them is 1 m.

By Coulomb's law, the force between them is

F_1 = \dfrac{kq \times q}{1^2}

F_1 = kq^2

When A loses an additional 10000 electrons, the charge on it becomes +2q while that on B is -2q. The distance is now 2 m.

The force is now

F_2 = \dfrac{k\times2q \times 2q}{2^2}

F_2 = \dfrac{4kq^2}{4}

F_2 = kq^2 = F_1

\dfrac{F_2}{F_1}=1

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Lee skated 36 miles in 3 hours. What was the speed at which Lee
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12mph

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A student has a mass (including clothes and shoes) of 65.0 kg. She drinks a 12 oz. can of soda, with a nutritional energy conten
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She can climb 0.92 m without losing weight.

<u>Explanation</u>:

Gravitational potential energy is the energy consisting of the product of mass, gravity and height.

1 cal = 4184 J

140 cal = 585760 J

Energy = 585760 J,  m = 65.0 kg = 65000 g,   Efficiency = 20 %

                                 GPE = mgh

where m represents the mass

          g represents the gravity,

           h represents the height.

                             585760 = 65000 \times 9.8 \times h

                                        h = 0.92 m.

7 0
3 years ago
A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s
Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g=4.67\times 10^{-3} kg

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1,m_1=1177g=1.177kg

Mass of block 2,m_2=1626 g=1.626 kg

Velocity of block 1,v_1=0.681m/s

(a)

Let velocity of the second block  after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

mv=m_1v_1+(m+m_2)v_2

4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2

1.66719=0.801537+1.63067v_2

1.66719-0.801537=1.63067v_2

0.865653=1.63067v_2

v_2=\frac{0.865653}{1.63067}

v_2=0.531m/s

Hence, the  velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

K_i=\frac{1}{2}mv^2

k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)

k_i=297.59 J

Final kinetic energy after collision

K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2

K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2

K_f=0.5028 J

Now, he ratio of the total kinetic energy after the collision to that before the collision

=\frac{k_f}{k_i}=\frac{0.5028}{297.59}

=0.00169

5 0
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