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3 years ago

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At some point, not close to its ends, within a solenoid of arbitrary length, calculate the approximate magnetic field if the sol

enoid carries a current 10.0 A and has 110.0 turns per meter of the solenoid's length.

1 answer:

GarryVolchara [31]3 years ago

5 0

Answer: The approximate magnetic field is 1.38 × 10^-3T

Explanation: Please see the attachment below

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Zero or +2

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A block of mass m is placed on a smooth wedge of inclination θ . The whole system is accelerated horizontally so that the block

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Answer: mg/Cosθ

Explanation:

Taking horizontal acceleration of wedge as 'a'

FCosΘ = FsinΘ

F = mass(m) × acceleration(a) = ma

For horizontal resolution g = 0

Therefore,

Horizontal = Vertical

maCosΘ = mgSinΘ

aCosΘ = gSinΘ

a = gSinΘ/CosΘ

Recall from trigonometry :

SinΘ/Cosθ = tanΘ

Therefore,

a = gtanΘ

Normal force acing on the wedge:

mgCosΘ + maSinΘ - - - - (y)

Substitute a = gtanΘ into (y)

mgCosΘ + mgtanΘsinΘ

tanΘ = sinΘ/cosΘ

mgCosΘ + mgsinΘ/cosΘsinΘ

mgCosΘ + mgsin^2Θ/cosΘ

Factorizing

mg(Cosθ + sin^2Θ/cosΘ)

Taking the L. C. M

mg[(Cos^2θ + sin^2Θ) /Cosθ]

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mg[ 1 /Cosθ]

mg/Cosθ

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Which of the following processes transfers heat? A. radiation B. conduction C. convection D. all of these

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All three transfer heat in different ways!

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A mass of 10.0 kg is in a gravitational field of 3.50 N/kg. What force acts on the mass?

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Explanation:

Given data

mass m=10.0 kg

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Force

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From definition of gravitational field intensity.

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consider a charge of -15.0 mCmoving to the right a 2.00x10^6 m/s in a mganetic field of .0300 T pointing upwards. What is the ma

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Answer:

a = 1.13×10^-8 m/s²

Explanation:

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3 years ago