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sertanlavr [38]
3 years ago
10

A long, straight wire carries a current of 8.60

Physics
1 answer:
Kay [80]3 years ago
6 0

First of all, we need to calculate the magnetic field generated by the wire at a distance r=3.90 cm=0.039 m, which is given by:

B=\frac{\mu_0 I}{2 \pi r}

where I=8.60 A is the current. Substituting numbers in the equation, we find

B=\frac{(4 \pi \cdot 10^{-7})(8.6 A)}{2 \pi (0.039 m)}=4.4 \cdot 10^{-5} T

And now we can calculate the force exerted on the electron, which is given by:

F=qvB

where q=1.6 \cdot 10^{-19} C is the charge of the electron and v=5.0 \cdot 10^4 m/s is its speed. Substituting data in the formula, we find

F=(1.6 \cdot 10^{-19}C)(5.00 \cdot 10^4 m/s)(4.4 \cdot 10^{-5} T)=3.5 \cdot 10^{-19} N

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While rearranging a dorm room, a student does 310 J of work in moving a desk 2.9 m. What was the magnitude of the applied horizo
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Answer:

The horizontal force is 106.89 N.

Explanation:

Given that,

Work done = 310 J

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Where, \theta=0^{\circ}

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310=F\cdot 2.9

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Mass wasting is also commonly triggered by the vibrational waves associated with earthquakes. Explain how earthquakes can disrup
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The vibrations will put the particles into suspension reducing the frictional forces between them.

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Shearing of soil grains from S-waves rounds jagged corners reducing frictional resistance. Also, Shaking from the seismic waves increases the water content of the material.

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A finite line of charge with linear charge density λ = 3.35 × 10-6 C/m, and length L = 0.808 m is located along the x axis (from
Andrews [41]

Answer: magnitude = 169.66N/C

direction = 8.6859°

Explanation:

Given from the question, we have that;

the Length of line of charge L = 0.808 m

Linear charge density λ = 3.35 × 10⁻⁶ C/m

charge q = -7.32 × 10⁻⁷ C

Coulombs force constant K = 1/(4π ε0) = 8.99 × 10⁹ N·m²/C².

NB. The picture uploaded gives a diagrammatic description of the problem.

From Pythagoras theorem we have,

tan Θ = 3.75 / (10.7-1.56)  

Θ = 22.3076 °

recall that the Electric field at point P due to the finite wire is;

È = Kλ (L / b(L+b)) Î .............. (1)

where È rep the electric field.

from equation 1, we have that  

È = 8.99 × 10⁹

where È rep the electric field.

from equation 1, we have that  

È = 8.99 × 10⁹ × 3.35 × 10⁻⁶ (0.808/ 10.7(10.7 – 0.808))

È = 0.229905 × 10³ N/C Î

recall also that the Electric field at point -P due to -q is;

È  = (8.99 × 10⁹ × 7.32 × 10⁻⁷) / ((3.75)² + (10.75-1.56)²) = 0.6742 × 10² N/C

where E = -E₁cosθÎ  + E₁sinθĴ

E = - 0.62446 ×10²Î   + 0.2562 ×10²Ĵ

The Resultant Electric charge Er is given as;

Er = 1.6771 ×10²Î + 0.2562 ×10²

Er =  [√(1.6771)² + (0.2562)² ] × 10² = 169.66 N/C

∴ Magnitude = 169.66 N/C

Having gotten the magnitude, let us find the direction;

⇒ Direction = tan Ф = 0.25621/1.6771 = 8.6859°

Direction = 8.6859°

cheers i hope this helps

7 0
3 years ago
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