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3 years ago

(121 m/s, or 43

Physics

1 answer:

mart [117]3 years ago

3 0

Answer:

(a) r = 328 m

(b) F = 6271.8 N

Explanation:

speed, v = 540 km/h = 150 m/s

Acceleration, a = 7 g

(a) Let the radius is r.

$a =\frac{v^{2}}{r}\\7\times 9.8 = \frac{150\times 150}{r}\\r = 328 m$

(b) The force on 80 kg pilot at the lowest point is

$F =m \times g + m\frac{v^{2}}{r}\\F = 80\times9.8 +80\times \frac{150\times 150}{328}\\\\F =784 + 5487.8 = 6271.8 N$

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where m is the mass of the block, k is the spring constant, A is the amplitude (in this case A = 14 cm) and v is the velocity of the block

Hence

$T = \sqrt{\frac{2 kg}{50 N/m}} = 0.2 s$

and by reeplacing it in ( 2 ):

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Answer:

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An extremely long wire laying parallel to the x -axis and passing through the origin carries a current of 250A running in the po

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Answer:

$1.232\times 10^{-5}\ T.$

Explanation:

<u>Given:</u>

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$\oint \vec B \cdot d\vec l=\mu_o I.$

In case of a circular loop, the directions of magnetic field and the line element $d\vec l$ , both are along the tangent of the loop at that point, therefore, $\vec B\cdot d\vec l = B\ dl$ .

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$\oint dl$ is the circumference of the Amperian loop = $2\pi r$

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It is the magnetic field due to a current carrying wire at a distance r from it.

For the first wire, passing through the origin:

Consider an Amperian loop of radius 3.510 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, $r_1 = 3.510\ m.$

The magnetic field at the given point due to this wire is given by:

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For the first wire, passing through the y-axis:

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The magnetic field at the given point due to this wire is given by:

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Thus, the net magnetic field at $r_r=-3.510\ m$ is given by