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Aliun [14]
3 years ago
13

A copper wire (density equal to 8900 kg / m3) 0.3 mm in diameter "floats" due to the force of the earth's magnetic field, which

is horizontal, perpendicular to the wire, and 5.5x10-5 T in magnitude. What current should the wire carry?
Physics
1 answer:
Andrews [41]3 years ago
5 0

Answer:I=112.094\ A

Explanation:

Given

density \rho =8900\ kg/m^3

diameter d=0.3\ mm

Magnetic field B=5.5\times 10^{-5}\ T

Force on the current carrying conductor placed in a magnetic field

F=BIL\sin \theta

where L=length of conductor

\theta=angle between magnetic field and current

If the wire is floating then weight must be balanced by weight of wire

Weight=mg=\rho ALg

Therefore

\rho ALg=BIL\times \sin 90

8900\times \frac{\pi}{4}(0.3\times 10^{-3})^2L\times 9.8=5.5\times 10^{-5}\times I\times L

I=\frac{8900\times \pi\times 9\times 10^{-8}\times 9.8}{4\times 5.5\times 10^{-5}}

I=112.094\ A

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Molecular formula of water molecule is H₂O.
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8 0
3 years ago
Dave and Alex push on opposite ends of a car that has a mass of 875 kg. Dave pushes the car to the right with a force of 250 N,
White raven [17]

The net force acting on the car is 65 N to the left

The net force acting on an object is simply defined as the resultant force acting on the object.

From the question given, we obtained the following data:

  • Force applied to the right (Fᵣ) = 250 N
  • Force applied to the left (Fₗ) = 315 N
  • Net force (Fₙ) =?

The net force acting on the car can be obtained as follow:

Fₙ = Fₗ – Fᵣ

Fₙ = 315 – 250

<h3>Fₙ = 65 N to the left </h3>

Therefore, the net force acting on the car is 65 N to the left

Learn more on net force: brainly.com/question/19549734

8 0
2 years ago
Read 2 more answers
A skier is accelerating down a 30.0-degree hill at 3.80 m/s^2.
Bond [772]

Answer:

ax = -3.29[m/s²]

ay = -1.9[m/s²]

Explanation:

We must remember that acceleration is a vector and therefore has magnitude and direction.

In this case, it is accelerating downwards, therefore for a greater understanding we will make a diagram of said vector, this diagram is attached.

a_{x}=-3.8*cos(30) = -3.29 [m/s^{2}]\\ a_{y}=-3.8*sin(30) = -1.9 [m/s^{2}]

3 0
3 years ago
Three point charges are arranged along the x axis. Charge q1=-4.00nC is located at x= .250 m and q2= 2.40 nC is at the x= -.300m
Umnica [9.8K]

Answer:

q₃=5.3nC

Explanation:

First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:

F= k\frac{|q_aq_b|}{r^{2} } \\\\\\F_{13}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(-4.00*10^{-9}C)q_3|}{(.250m)^{2} } =576q_3N/C\\\\F_{23}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(2.40*10^{-9}C)q_3|}{(.300m)^{2} } =240q_3N/C\\

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

F_{Net}=F_{13}+F_{23}\\\\4.40*10^{-9} N=576q_3N/C+240q_3N/C\\\\4.40*10^{-6} N=816q_3N/C\\\\\implies q_3=5.3*10^{-9}C=5.3nC

In words, the value of q₃ must be 5.3nC.

7 0
3 years ago
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