As per bernoulli's principle
here
= pressure upwards
= pressure downwards
= velocity of air upwards
= velocity of air downwards
now from this equation we can say that the pressure difference will be
now the force due to this pressure difference will be
so this is the above force which is given above
Because,
In left image pin is not touch to the wire.
In right image pin is touch to the wire.
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Every planet/moon has global wind that are mostly determined by the way the planet/moon rotates and how evenly the Sun illuminates it. On the Earth the equator gets much more Sun than the poles. resulting in warmer air at the equator than the poles and creating circulation cells (or "Hadley Cells") which consist of warm air rising over the equator and then moving North and South from it and back round.
The Earth is also rotating. When any solid body rotates, bits of it that are nearer its axis move slower than those which are further away. As you move north (or south) from the equator, you are moving closer to the axis of the Earth and so the air which started at the equator and moved north (or south) will be moving faster than the ground it is over (it has the rotation speed of the ground at the equator, not the ground which is is now over). This results in winds which always move from the west to the east in the mid latitudes.
Answer: friction reduces the speed during motion
Explanation:
The more the friction, the lesser the speed during motion
Answer:
Explanation:
Given that,
A point charge is placed between two charges
Q1 = 4 μC
Q2 = -1 μC
Distance between the two charges is 1m
We want to find the point when the electric field will be zero.
Electric field can be calculated using
E = kQ/r²
Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.
Then, the magnitude of the electric at point x is zero.
E = kQ1 / r² + kQ2 / r²
0 = kQ1 / x² - kQ2 / (1-x)²
kQ1 / x² = kQ2 / (1-x)²
Divide through by k
Q1 / x² = Q2 / (1-x)²
4μ / x² = 1μ / (1 - x)²
Divide through by μ
4 / x² = 1 / (1-x)²
Cross multiply
4(1-x)² = x²
4(1-2x+x²) = x²
4 - 8x + 4x² = x²
4x² - 8x + 4 - x² = 0
3x² - 8x + 4 = 0
Check attachment for solution of quadratic equation
We found that,
x = 2m or x = ⅔m
So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.
