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Ivahew [28]
4 years ago
5

Which cruising altitude is appropriate for vfr flight on a magnetic course of 135°

Physics
1 answer:
Nata [24]4 years ago
5 0

Answer:odd thousands plus 500 feet.

Explanation:

On a magnetic course of zero through 179 degrees, select an odd thousand foot cruising altitude plus 500 feet, such as 3,500, 5,500, up to and including 17,500. Even and odd thousands are reserved for those aircraft on an active instrument flight plan. Even thousands plus 500 feet are for aircraft flying between 180 and 359 degrees.

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Two forces, one four times as large as the other, pull in the same direction on a 10kg mass and impart to it an acceleration of
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Answer:

The acceleration of the mass is 2 meters per square second.

Explanation:

By Newton's second law, we know that force (F), measured in newtons, is the product of mass (m), measured in kilograms, and net acceleration (a), measured in meters per square second. That is:

F = m\cdot a (1)

The initial force applied in the mass is:

F = (10\,kg)\cdot \left(2.5\,\frac{m}{s^{2}} \right)

F = 25\,N

In addition, we know that force is directly proportional to acceleration. If the smaller force is removed, then the initial force is reduced to \frac{4}{5} of the initial force. The acceleration of the mass is:

\frac{25\,N}{20\,N} = \frac{2.5\,\frac{m}{s^{2}} }{a}

a = 2\,\frac{m}{s^{2}}

The acceleration of the mass is 2 meters per square second.

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A coil of wire containing N turns is in an external magnetic field that is perpendicular to the plane of the coil and it steadil
krok68 [10]

Answer:

The Resultant Induced Emf in coil is 4∈.

Explanation:

Given that,

A coil of wire containing having N turns in an External magnetic Field that is perpendicular to the plane of the coil which is steadily changing. An Emf (∈) is induced in the coil.

To find :-

find the induced Emf if rate of change of the magnetic field and the number of turns in the coil are Doubled (but nothing else changes).

So,

   Emf induced in the coil represented by formula

                          ∈  =   -N\frac{d\phi}{dt}                                  ...................(1)

                                          Where:

                                                    .   \phi = BAcos\theta     { B is magnetic field }

                                                                                 {A is cross-sectional area}

                                                    .  N = No. of turns in coil.

                                                    .  \frac{d\phi}{dt} = Rate change of induced Emf.

Here,

Considering the case :-

                                    N1 = 2N  &      \frac{d\phi1}{dt} = 2\frac{d\phi}{dt}

Putting these value in the equation (1) and finding the  new emf induced (∈1)

                           

                                      ∈1 =-N1\times\frac{d\phi1}{dt}

                                      ∈1 =-2N\times2\frac{d\phi}{dt}

                                       ∈1 =4 [-N\times\frac{d\phi}{dt}]

                                        ∈1 = 4∈             ...............{from Equation (1)}      

Hence,

The Resultant Induced Emf in coil is 4∈.        

                           

8 0
3 years ago
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