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Annette [7]
2 years ago
12

If a 1-megaton nuclear weapon is exploded

Physics
1 answer:
Elenna [48]2 years ago
4 0

Answer:

1,780,000 N

Explanation:

0.2 atm × (1.013×10⁵ Pa/atm) = 20,260 Pa

Force = pressure × area

F = 20,260 Pa × (3.89 m × 22.6 m)

F = 1,780,000 N

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Flammable materials, like alcohol, should never be dispensed or used near
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A circular loop in the plane of a paper lies in a 0.45 T magnetic field pointing into the paper. The loop's diameter changes fro
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Answer:

(A). The direction of the induced current will be clockwise.

(B). The magnitude of the average induced emf 16.87 mV.

(C). The induced current is 6.75 mA.

Explanation:

Given that,

Magnetic field = 0.45 T

The loop's diameter changes from 17.0 cm to 6.0 cm .

Time = 0.53 sec

(A). We need to find the direction of the induced current.

Using Lenz law

If the direction of magnetic field shows into the paper then the direction of the induced current will be clockwise.

(B). We need to calculate the magnetic flux

Using formula of flux

\phi_{1}=BA\cos\theta

Put the value into the formula

\phi_{1}=0.45\times(\pi\times(8.5\times10^{-2})^2)\cos0

\phi_{1}=0.01021\ Wb

We need to calculate the magnetic flux

Using formula of flux

\phi_{2}=BA\cos\theta

Put the value into the formula

\phi_{2}=0.45\times(\pi\times(3\times10^{-2})^2)\cos0

\phi_{2}=0.00127\ Wb

We need to calculate the magnitude of the average induced emf

Using formula of emf

\epsilon=-N(\dfrac{\Delta \phi}{\Delta t})

Put the value into t5he formula

\epsilon=-1\times(\dfrac{0.00127-0.01021}{0.53})

\epsilon=0.016867\ V

\epsilon=16.87\ mV

(C). If the coil resistance is 2.5 Ω.

We need to calculate the induced current

Using formula of current

I=\dfrac{\epsilon}{R}

Put the value into the formula

I=\dfrac{0.016867}{2.5}

I=0.00675\ A

I=6.75\ mA

Hence, (A). The direction of the induced current will be clockwise.

(B). The magnitude of the average induced emf 16.87 mV.

(C). The induced current is 6.75 mA.

5 0
3 years ago
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