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3 years ago

The atomic bomb dropped on Hiroshima converted about 7.00x10-4kg of mass to energy. How much energy did that bomb produce?

Physics

1 answer:

strojnjashka [21]3 years ago

6 0

Answer:

$\sf \: given \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \bf \: mass \: \: \: m \: = 7.00 \times {10}^{ - 4} \: kg \\ \\ \bf \: E=mc^2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = > E=7.00 \times {10}^{ - 4} \times ({3 \times {10}^{8} })^{2} \\ \\ = > \green{ \boxed{ E = 6.3 \times {10}^{13} \: J}}$

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Nuetrik [128]

Answer:

The velocity of the ship relative to the earth V = 9.05 $\frac{m}{s}$

Explanation:

The local ocean current is = 1.52 m/s

Direction $\theta$ = 40°

Velocity component in X - direction $V_{x}$ = 1.52 $\cos 40$ °

$V_{x}$ = 1.164 $\frac{m}{s}$

Velocity component in Y - direction $V_{y}$ = 8 + 1.52 $\sin 40$ °

$V_{y}$ = 8.97 $\frac{m}{s}$

The velocity of the ship relative to the earth

$V = \sqrt{V_{x}^{2} + V_{y} ^{2} }$

Put the values of $V_{x}$ and $V_{y}$ we get,

⇒ $V = \sqrt{1.164^{2} + 8.97 ^{2} }$

⇒ V = 9.05 $\frac{m}{s}$

This is the velocity of the ship relative to the earth.

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3 years ago

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kkurt [141]

F = ma

We have mass = 0.2kg

and acceleration = 20 m/s^2

So..

F = (0.2)(20)

F = 4 N

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3 years ago

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kenny6666 [7]

Answer:46.5

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3 years ago