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anygoal [31]
3 years ago
14

If 2.0 x 10^-4 C charge passes a point in 5.0 x 10^-5 s, what is the rate of current flow?

Physics
1 answer:
insens350 [35]3 years ago
3 0
Current is defined as the rate of charge flowing a point every second. Having a current of 1 Ampere signifies 1 Coulomb is flowing in a circuit every second. It is measured by the use of an ammeter which is positioned in series to the component to be measured. The current in the problem is calculated as follows:

I = 2.0 x 10^-4 C / 5.0 x 10^-5 s
<span>I = 4 A or 4.0 x 10^0 A</span>
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A 4.0 µC point charge and a 3.0 µC point charge are a distance L apart. Where should a third point charge be placed so that the
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Answer:

Q must be placed at 0.53 L

Explanation:

Given  data:

q_1 = 4.0 μC , q_2 = 3.0μC

Distance between charge is L

third charge q be placed at  distance x cm from q1

The force by charge q_1 due to q is

F1 = \frac{k q q_1}{x^2}

F1 = \frac{k q ( 4.0 μC )}{ x^2}                  ----1

The force by charge q_2 due to q is

F2 =  \frac{k q q_2}{(L-x)^2}

F2 = \frac{kq (3.0 μC)}{(L-x)^2}                   --2

we know that net electric force is equal to zero

F_1 = F_2

\frac{k q ( 4.0 μC )}{x^2}   =\frac{k q ( 3.0 μC )}{(l-x)^2}

\frac{4}{3}*(L-x)^2 = x^2

x = \sqrt{\frac{4}{3}*(L - x)

L-x = \frac{x}{1.15}

L = x + \frac{x}{1.15} = 1.86 x

x = 0.53 L

Q must be placed at 0.53 L

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Explanation:

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7. A mother pushes her 9.5 kg baby in her 5kg baby carriage over the grass with a force of 110N @ an angle
jasenka [17]

Weight of the carriage =(m+M)g =142.1\ N

Normal force =Fsin(\theta) + W = 197.1\ N

Frictional force =\mu N=27.59\ N

Acceleration =4.66\ m\ s^{-2}

Explanation:

We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

To calculate Weight we know that both the mass of the baby and the carriage will be added.

  • So Weight(W) =(m+M)\times g =(9.5+5)\ kg \times 9.8 =142.1\ Newton\ (N)

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with F_x, force of 110\ N acting vertically downward.Both are downward and Normal is upward so Normal force =Summation\ of\ both\ forces

  • Normal force (N) = Fsin(\theta)+W=110sin(30) + 142.1 =197.1\ N
  • Frictional force (f) =\mu N=0.14\times 197.1 =27.59\ N

To calculate acceleration we will use Newtons second law.

That is Force is product of mass and acceleration.

We can see in the diagram that F_y=Horizontal and F_x=Vertical component of forces.

So Fnet = Fy(Horizontal) - f(friction) = m\times a

  • Acceleration (a) =\frac{Fcos(\theta)-\mu N}{mass(m)} =\frac{(95.26-27.59)}{14.5}= 4.66\ m\ s{^2 }

So we have the weight of the carriage, normal force,frictional force and acceleration.

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2 years ago
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