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nalin [4]
3 years ago
5

If 51.9 C of charge are transferred in a specific lightning strike and the potential difference is ∆V = 10.93 MV, how much energ

y is released in this lightning strike? (in J
Physics
1 answer:
MA_775_DIABLO [31]3 years ago
8 0

Quantity of charge unloaded = 51.9 Coulombs

Potential difference = 10.93 megavolts = 1.093 x 10⁷ volts

1 volt = 1 joule per coulomb

1.093 x 10⁷ volts = 1.093 x 10⁷ joules per coulomb

Energy = (1.093 x 10⁷ J/C) x (51.9 C)

Energy = 567.267 megaJoules

That's <em>5.67 x 10⁸ Joules</em> .

==> My wife's blow-dryer is marked 1260 watts.

If the energy in this lightning strike could be collected, bottled, stored, and used as needed, it could run my wife's blow-dryer for 125 hours.  That would save us more than $30 on our electric bill !

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A 57 g ice cube can slide without friction up and down a 33 ∘ slope. The ice cube is pressed against a spring at the bottom of t
beks73 [17]
The springs stored energy is transferred to the cube as kinetic energy and then by the slop the KE is converted to height energy. 

<span>0.5 . k . x^2 = 0.5 . m . v^2 = m . g . ∆h </span>

<span>0.5 . 50 . (0.1^2) = 0.05 . 9.8 . ∆h </span>

<span>∆h = 0.51 m = 51 cm </span>

<span>This is the height gained </span>
<span>Distance along the slope = ∆h / sin 60 = 0.589 = 59 cm </span>

<span>In the second case, the stored spring energy is converted into height energy AND frictional heat energy. </span>

<span>The height energy is m . g . d sin 60 where d is the distance the cube moves along the slope. </span>

<span>The Frictional energy converted is F . d </span>

<span>F ( the frictional force ) = µ . N </span>

<span>N ( the reaction to the component of the gravity force perpendicular to the surface of the slope ) = m . g . cos60 </span>

<span>Total energy converted </span>

<span>0.5 . k . x^2 = (m . g . dsin60) + (µ . m . g . cos60 . d ) </span>

<span>Solve for d </span>

<span>d = 0.528 = 53 cm</span>
5 0
3 years ago
At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction
bogdanovich [222]

Answer:

magnitude of force on charge 2Q  = \frac{KQ^{2} }{I^{2} }

Direction of force on charge = 61 ⁰

Explanation:

The magnitude on the force on the charge can be evaluated by finding the net force acting on the charge 2Q  i.e x-component of the net force and the y-component of the net force

║F║ = \sqrt{f_{x}^{2} + f_{y}^{2}    }  =  after considering the forces coming from Q, 3Q and 4Q AND APPLYING COULOMBS LAW

magnitude of force acting on 2Q = \frac{KQ^{2} }{I^{2} }

The direction of the force on charge 2Q is calculated as

tan ∅ = \frac{f_{y} }{f_{x} } = 1.8284

therefore ∅ = tan^{-1}  1.8284

= 61⁰

3 0
3 years ago
Why do planets rings surround them and not the sun
Shalnov [3]

Well, the rings surrounding a planet are made out of rock. A ring surrounding the sun would be impossible since the sun can reach more than 27 million degrees Fahrenheit (15 million degrees Celsius.)

Hope this helped.

7 0
3 years ago
Compared to energy-flow in ecosystems, the flow of matter ________.
disa [49]
Compared to energy-flow in ecosystems, the flow of matter <span>reflects conservation and recycling.
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3 0
3 years ago
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statuscvo [17]

Explanation:

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hope this helps...

6 0
3 years ago
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