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Bess [88]
2 years ago
11

A small cork with an excess charge of +6.0µC is placed 0.12 m from another cork, which carries a charge of -4.3µC.

Physics
1 answer:
Volgvan2 years ago
6 0

A) 16.1 N

The magnitude of the electric force between the corks is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q_1 = 6.0 \mu C=6.0 \cdot 10^{-6} C is the magnitude of the charge on the first cork

q_2 = 4.3 \mu C = 4.3 \cdot 10^{-6}C is the magnitude of the charge of the second cork

r = 0.12 m is the separation between the two corks

Substituting numbers into the formula, we find

F=(9\cdot 10^9 N m^2 C^{-2} )\frac{(6.0\cdot 10^{-6}C)(4.3\cdot 10^{-6} C)}{(0.12 m)^2}=16.1 N

B) Attractive

According to Coulomb's law, the direction of the electric force between two charged objects depends on the sign of the charge of the two objects.

In particular, we have:

- if the two objects have charges with same sign (e.g. positive-positive or negative-negative), the force is repulsive

- if the two objects have charges with opposite sign (e.g. positive-negative), the force is attractive

In this problem, we have

Cork 1 has a positive charge

Cork 2 has a negative charge

So, the force between them is attractive.

C) 2.69\cdot 10^{13}

The net charge of the negative cork is

q_2 = -4.3 \cdot 10^{-6}C

We know that the charge of a single electron is

e=-1.6\cdot 10^{-19}C

The net charge on the negative cork is due to the presence of N excess electrons, so we can write

q_2 = Ne

and solving for N, we find the number of excess electrons:

N=\frac{q_2}{e}=\frac{-4.3\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=2.69\cdot 10^{13}

D) 3.75\cdot 10^{13}

The net charge on the positive cork is

q_1 = +6.0\cdot 10^{-6}C

We know that the charge of a single electron is

e=-1.6\cdot 10^{-19}C

The net charge on the positive cork is due to the "absence" of N excess electrons, so we can write

q_1 = -Ne

and solving for N, we find the number of electrons lost by the cork:

N=-\frac{q_1}{e}=-\frac{+6.0\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=3.75\cdot 10^{13}

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A shopper pushes a grocery cart 41.9 m on level ground, against a 44.5 N frictional force. The cart has a mass of 16.3 kg. He pu
Rashid [163]

Answer:

Fp = 26.59[N]

Explanation:

This problem can be solved using the principle of work and energy conservation, i.e. the final kinetic energy of a body will be equal to the sum of the forces that do work on the body plus the initial kinetic energy.

We need to identify the initial data:

d = distance = 41.9[m]

Ff = friction force = 44.5 [N]

m = mass = 16.3 [kg]

v1 = 1.9 [m/s]

v2 = 12.6 [m/s]

The kinetic energy at the beginning can be calculated as follows:

E_{k1}= \frac{1}{2}*m*v_{1}^2 \\E_{k1}= \frac{1}{2}*16.3*(1.9)_{1}^2\\E_{k1}= 29.42[J]

And the final kinetic energy.

E_{k2}= \frac{1}{2}*m*v_{2}^2 \\E_{k2}= \frac{1}{2}*16.3*(12.6)^2\\E_{k2}= 1294[J]

The work is performed by two forces, the friction force and the pushing force, it is important to clarify that these forces are opposite in direction.

The weight of the cart also performs a work in the direction of movement since the plane is tilted down, this component of the weight of the cart must be parallel to the surface of the inclined plane.

W_{1-2}=-(44.5*41.9)+(16.3*9.81*sin(17.5)*41.9)+(F_{p}*41.9) \\therefore:\\E_{k1}+W_{1-2}=E_{k2}\\29.42+150.16+(F_{p}*41.9)=1294\\F_{p}=1114.42/41.9\\F_{p}=26.59[N]

5 0
2 years ago
If the Force exerted by the intern is doubled and the distance is halved, does the done by the intern increase, decrease, or rem
Jlenok [28]

Remain the same

Explanation:

If the force exerted by the intern is doubled and the distance is halved, the work done by the intern remains the same.

 Work done is the force applied to move a body through a distance.

Work done = F x d

where F is the applied force

            d is the distance moved

Now;

if:

    f = 2f

    d = \frac{1}{2}d

Input the parameter:

 Work done = fxd = 2f x \frac{1}{2}d  = fd

The work done will still remain the same

learn more:

Work done brainly.com/question/9100769

#learnwithBrainly

3 0
2 years ago
The surface charge density on an infinite charged plane is - 2.10 ×10−6C/m2. A proton is shot straight away from the plane at 2.
inn [45]

Explanation:

Formula to calculate electric field because of the plate is as follows.

         E = \frac{\sigma}{2 \times \epsilon_{o}}

            = \frac{2.10 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}

           = 1.18 \times 10^{5} N/C

Now, we will consider that equilibrium of forces are present there. So,

                   ma = qE

       a = \frac{1.6 \times 10^{-19} \times 1.18 \times 10^{5}}{1.67 \times 10^{-27}}

          = 1.13 \times 10^{13} m/s^2

According to the third equation of motion,

         v^{2} = 2 \times a \times d

or,      d = \frac{v^{2}}{2d}

             = \frac{(2.4 \times 10^{6})^{2}}{2 \times 1.13 \times 10^{13}}

             = 0.254 m

Thus, we can conclude that the proton will travel 0.254 m before reaching its turning point.

7 0
2 years ago
A basketball player standing up with the hoop launches the ball straight up with an initial velocity of v_o = 3.75 m/s from 2.5
denis23 [38]

Answer:

a) The maximum height the ball will achieve above the launch point is 0.2 m.

b) The minimum velocity with which the ball must be launched is 4.43 m/s or 0.174 in/ms.

Explanation:

a)

For the height reached, we use 3rd equation of motion:

2gh = Vf² - Vo²

Here,

Vo = 3.75 m/s

Vf =  0m/s, since ball stops at the highest point

g = -9.8 m/s² (negative sign for upward motion)

h = maximum height reached by ball

therefore, eqn becomes:

2(-9.8m/s²)(h) = (0 m/s)² - (3.75 m/s²)²

<u>h = 0.2 m</u>

b)

To find out the initial speed to reach the hoop at height of 3.5 m, we again use 3rd eqn. of motion with h= 3.5 m - 2.5m = 1 m (taking launch point as reference), and Vo as unknown:

2(-9.8m/s²)(1 m) = (0 m/s)² - (Vo)²

(Vo)² = 19.6 m²/s²

Vo = √19.6 m²/s²

<u>Vo = 4.43 m/s</u>

Vo = (4.43 m/s)(1 s/1000 ms)(39.37 in/1 m)

<u>Vo = 0.174 in/ms</u>

<u />

6 0
2 years ago
What is the speed of light (in m/s) in air? (Enter your answer to at least four significant figures. Assume the speed of light i
jenyasd209 [6]

Answer:

The speed of light in air is 2.996x10⁸ m/s, and polystyrene is 1.873x10⁸ m/s.

Explanation:

To find the speed of light in air and in polystyrene we need to use the following equation:

c_{m} = \frac{c}{n}

Where:

c_{m}: is the speed of light in the medium

n: is the refractive index of the medium

In air:

c_{a} = \frac{c}{n_{a}} = \frac{2.997 \cdot 10^{8} m/s}{1.0003} = 2.996 \cdot 10^{8} m/s

In polystyrene:

c_{p} = \frac{c}{n_{p}} = \frac{2.997 \cdot 10^{8} m/s}{1.6} = 1.873 \cdot 10^{8} m/s  

Therefore, the speed of light in air is 2.996x10⁸ m/s, and polystyrene is 1.873x10⁸ m/s.

I hope it helps you!

8 0
2 years ago
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