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TEA [102]
3 years ago
9

Given that an electric field of 3×106V/m3×106V/m is required to produce an electrical spark within a volume of air, estimate the

length of a thundercloud lightning bolt.
Physics
1 answer:
Andre45 [30]3 years ago
5 0

Answer:

Length, l = 33.4 m

Explanation:

Given that,

Electrical field, E=3\times 10^6\ V/m

Let the electrical potential is, V=10^8\ V

We need to find the length of a thundercloud lightning bolt. The relation between electric field and the electric potential is given by :

V=E\times d\\\\d=\dfrac{V}{E}\\\\d=\dfrac{10^8}{3\times 10^6}\\\\d=33.4\ m

So, the length of a thundercloud lightning bolt is 33.4 meters. Hence, this is the required solution.

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Write down any 5 example of conservation of momentum?​
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Which of the following objects cannot make a shadow? Broken piece of glass from a window/wooden pane of the same window? Explain
Lorico [155]

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broken pieces of glass from a window can't form a shadow.

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4 0
2 years ago
Two boats leave the same port at the same time, with boat A traveling north at 15 knots (nautical miles per hour) and boat B tra
Mrrafil [7]

Answer:

The chance in distance is 25 knots

Explanation:

The distance between the two particles is given by:

s^2 = (x_A - x_B)^2+(y_A - y_B)^2  (1)

Since A is traveling north and B is traveling east we can say that their displacement vector are perpendicular and therefore (1) transformed as:

s^2 = x_B^2+y_A^2 (2)

Taking the differential with respect to time:

\displaystyle{2s\frac{ds}{dt}= 2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt}}  (3)

where \displaystyle{\frac{dx_B}{dt}}=v_B and \displaystyle{\frac{dx_A}{dt}}=v_A are the respective given velocities of the boats. To find s and x_B we make use of the given position for A, y_A=30, the Pythagoras theorem and the relation between distance and velocity for a movement with constant velocity.

\displaystyle{y_A = v_A\cdot t\rightarrow t = \frac{y_A}{v_A}=\frac{30}{15}=2 h

with this time, we know can now calculate the distance at which B is:

\displaystyle{x_B = v_B\cdot t= 20 \cdot 2 = 40\ nmi

and applying Pythagoras:

\displaystyle{s = \sqrt{x_B^2+y_A^2}=\sqrt{30^2 + 40^2}=\sqrt{2500}=50}

Now substituting all the values in (3) and solving for  \displaystyle{\frac{ds}{dt} } we get:

\displaystyle{\frac{ds}{dt} = \frac{1}{2s}(2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt})}\\\displaystyle{\frac{ds}{dt} = 25 \ knots}

4 0
3 years ago
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