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Scilla [17]
3 years ago
10

Determine the magnitude of the force between two 42 m-long parallel wires separated by 0.03 m, both carrying 6.3 A in the same d

irection.
Physics
1 answer:
xz_007 [3.2K]3 years ago
6 0

Answer:

The magnitude of the force between the two parallel wires is 0.0111 N.

Explanation:

Given;

length of the two parallel wires, L = 42 m

distance between the two wires, r = 0.03 m

current in both wires, I₁, I₂ = 6.3 A

Therefore, the magnitude of the repulsive force between the two parallel wires is given by;

F = \frac{\mu_0 I_1I_2l}{2\pi r}\\\\where;\\\mu_0 \ is \ permeability \ of \ free \ space = 4\pi *10^{-7} \ T.m/A \\\\F = \frac{(4\pi *10^{-7})(6.3)^2(42)}{2\pi (0.03)}\\\\F =   0.0111 \ N

Therefore, the magnitude of the force between the two parallel wires is 0.0111 N.

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Christian made some pancakes she's so 3/5 of them in the morning and 1/4 of the remainder in the afternoon is she had 300 pancak
DiKsa [7]

Christian made 1000 pancakes.

Explanation:

Let us represent the total amount of Pancake made by Christian as = K

    From the problem;

 Christian ate \frac{3}{5} of the pancake in the morning =  \frac{3}{5}  * K =  \frac{3}{5} K

We know that Christian cannot eat her pancake and at the same time have it, the  remaining pancake will then be:

        total amount of cake - fraction eaten

Remainder = K -  \frac{3}{5} K=  \frac{2}{5} K

   

In the afternoon, we know that she ate 1/4 of the remaining cake:

        \frac{1}{5} K*  \frac{2}{5} K = \frac{1}{10} K

 The remaining cake in the afternoon will be:

    Total amount of cake remaining from morning - amount eaten in the afternoon

    =    \frac{2}{5} K -  \frac{1}{10} K

    =    \frac{3}{10} K

The fraction of the cake remaining in the afternoon is  \frac{3}{10} K

Since she had 300cakes left in the afternoon, then :

            \frac{3}{10} K= 300

                    K = 1000 pancakes

Therefore Christian made 1000 pancakes.

learn more:

Fractions brainly.com/question/1648978

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4 0
3 years ago
A 2 microcoulomb charge is placed at a distance of 0.25 m away from a 3.6 microcoulomb charge. Describe the type of electrostati
EleoNora [17]

Answer: 1.04N

Explanation:

Given

q1 = 2*10^-6C

q2 = 3.6*10^-6C

r = 0.25m

k = 9*10^9

Magnitude of electrostatic force can be calculated by using coulomb's law. Coulomb's law states that, "the magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them."

F =(kq1q2) / r²

F = (9*10^9 * 2*10^-6 * 3.6*10^-6) / 0.25²

F = 0.0648/0.0625

F = 1.04N

The type of electrostatic force between the charges is the repulsive force

7 0
3 years ago
Read 2 more answers
Would a vibrating proton produce an electromagnetic wave
Anon25 [30]

Answer:

No,

Explanation:

An electromagnetic wave is made of vibrating electric and magnetic fields that continually induce each other; matter is not needed for this to occur.

5 0
3 years ago
Solution A has a specific heat of 2.0 J/g◦C. Solution B has a specific heat of 3.8 J/g◦C. If equal masses of both solutions start
fgiga [73]

Answer: 2. Solution A attains a higher temperature.

Explanation: Specific heat simply means, that amount of heat which is when supplied to a unit mass of a substance will raise its temperature by 1°C.

In the given situation we have equal masses of two solutions A & B, out of which A has lower specific heat which means that a unit mass of solution A requires lesser energy to raise its temperature by 1°C than the solution B.

Since, the masses of both the solutions are same and equal heat is supplied to both, the proportional condition will follow.

<em>We have a formula for such condition,</em>

Q=m.c.\Delta T.....................................(1)

where:

  • \Delta T= temperature difference
  • Q= heat energy
  • m= mass of the body
  • c= specific heat of the body

<u>Proving mathematically:</u>

<em>According to the given conditions</em>

  • we have equal masses of two solutions A & B, i.e. m_A=m_B
  • equal heat is supplied to both the solutions, i.e. Q_A=Q_B
  • specific heat of solution A, c_{A}=2.0 J.g^{-1} .\degree C^{-1}
  • specific heat of solution B, c_{B}=3.8 J.g^{-1} .\degree C^{-1}
  • \Delta T_A & \Delta T_B are the change in temperatures of the respective solutions.

Now, putting the above values

Q_A=Q_B

m_A.c_A. \Delta T_A=m_B.c_B . \Delta T_B\\\\2.0\times \Delta T_A=3.8 \times \Delta T_B\\\\ \Delta T_A=\frac{3.8}{2.0}\times \Delta T_B\\\\\\\frac{\Delta T_{A}}{\Delta T_{B}} = \frac{3.8}{2.0}>1

Which proves that solution A attains a higher temperature than solution B.

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3 years ago
What is density?
il63 [147K]
<span>a. the amount of matter in a given volume </span>
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2 years ago
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