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natima [27]
3 years ago
5

A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1 ° from the normal to the plane of the l

oop. The magnitude of the magnetic field is 4.35 T , and the radius of the wire loop is 0.280 m . Find the magnetic flux Φ through the loop.
Physics
1 answer:
umka2103 [35]3 years ago
8 0

Answer:

Magnetic flux through the loop is 1.03 T m²

Explanation:

Given:

Magnetic field, B = 4.35 T

Radius of the circular loop, r = 0.280 m

Angle between circular loop and magnetic field, θ = 15.1⁰

Magnetic flux is determine by the relation:

\Phi = BA\cos \theta   ....(1)

Here A represents area of the circular loop.

Area of circular loop, A = πr²

Hence, the equation (1) becomes:

\Phi=B\pi r^{2} \cos \theta

Substitute the suitable values in the above equation.

\Phi=4.35\times\pi (0.28)^{2} \cos 15.1

\Phi = 1.03 T m²

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Ostrovityanka [42]

The correct answer is:

Work is negative, the environment did work on the object, and the energy of the system decreases.

In fact, the work-energy theorem states that the work done by the system is equal to its variation of kinetic energy:

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The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static fricti
Colt1911 [192]

This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

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Answer:

- the magnitude of force at point A is 79.1033 lb

- since FA < FA_max; Ladder WILL NOT slip

Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

NB = W / 2tan∅ -------let this be equation 1

∑Fx = 0

⇒ FA - NB = 0

FA = NB

therefore from equation 1

FA = NB = W / 2tan∅

we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

NA - W = 0

NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

= √( (W)² + (W / 2tan∅)²)

we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

= √ 6257.3328

FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

Now maximum possible frictional force at A

FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

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Answer:

20 days, 480 hours

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A gold sphere of radius R=100 μm and density 19g/cm^3 falls through water. Given the viscosity of water is about 10^-3​ Pa s and
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The terminal velocity of gold sphere is 39.2 cm/s

<h3>What is terminal velocity?</h3>

Terminal velocity is the maximum velocity attainable for an object as it falls through a fluid.

<h3>How to calculate the terminal velocity of the gold sphere?</h3>

The terminal velocity of the gold sphere is given by v = 2gr²(ρ - σ)/9η where

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So, susbtituting the values of the variables into the equation, we have that

v = 2gr²(ρ - σ)/9η

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So, the terminal velocity is 39.2 cm/s

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