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Eva8 [605]
3 years ago
8

A cylindrical coil of radius 6 cm is placed in magnetic field, which is changing in time with the rate 0.5 T/s. The magnetic fie

ld direction is parallel to the cylinder axis. What should be the number of turns in the coil to induce the emf
Physics
1 answer:
Jlenok [28]3 years ago
3 0

induced emf value is missing..

please correct the question

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How was mount Everest made? Please i am handing out 100 points
Simora [160]

Answer:

Mount Everest formed from a tectonic smashup between the Indian and Eurasian tectonic plates tens of millions of years ago.

Explanation:

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2 years ago
What part of a circuit has the main job of conducting the electricity?
Vesnalui [34]
The main job of conducting electricity is the power source.
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which observation about the Mid-Atlantic Ridge region provides the best evidence of the seafloor has been spreading over for mil
Kitty [74]

Magnetic stripes on the sea floor

Explanation:

The observation of magnetic stripes patterns at the region of the Mid-Atlantic Ridge galvanized the evidence that sea floor spreading has been on for several millions of years.

  • Geomagnetic field of the earth is similar to that of bar magnet having the north and south pole.
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Sea floor spreading brainly.com/question/9912731

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8 0
3 years ago
What is 64 nanometers to m?
defon
\sf Hello!

\sf We\: know \:that,
\sf 1\: meter = \sf 10^{9} nm

\sf Then,

\sf Distance\: in \:m = \sf Distance\: in\: nm × \dfrac{\sf 1}{\sf 10^{9}}\: \sf m

⇒ \sf Distance\: in \:m = \sf 64 × 10^{-9} \:m

⇒ \sf Distance\: in\: m = \sf 6.4 × 10^{-8} \:m

~ \sf iCarl
3 0
3 years ago
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A projectile is launched from ground level with an initial speed of 53.7 m/s. Find the launch angle (the angle the initial veloc
Simora [160]

Answer:\theta =75.52^{\circ}

Explanation:

Given

initial speed of Launch(u)=53.7 m/s

Range of Projectile =Maximum height of Projectile

Range is given by R=\frac{u^2\sin 2\theta }{g}

Maximum height is given by H_{max}=\frac{u^2\sin ^2\theta }{2g}

R=H_{max}

\frac{u^2\sin 2\theta }{g}=\frac{u^2\sin ^2\theta }{2g}

\frac{u^2\sin 2\theta }{g}-\frac{u^2\sin ^2\theta }{2g}=0

\frac{u^2\sin \theta }{g}\cdot \left [ 2\cos \theta -\frac{1}{2}\right ]=0

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6 0
3 years ago
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