Without a bulb energy cant go through and it would be an open circuit blocking the energy from coming out.
Answer:
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Answer:
Explanation:
Let the equilibrium position of third charge be x distance from q₁.
Force on third charge due to q₁
= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²
Force on third charge due to q₂
= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
Both the force will act in opposite direction and for balancing , they should be equal.
9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
5 / x² = 2 / ( .4 - x )²
Taking square root on both sides
2.236 / x = 1.414 / .4 - x
2.236 ( .4 - x ) = 1.414 x
.8944 - 2.236 x = 1.414 x
.8944 = 3.65 x
x = .245 m
24.5 cm
So the third charge should be at a distance of 24.5 cm from q₁ .
Answer:
we encounter ocassion where one or more object move in the which is not stationary with respect to another example a boat is cross a river that is flowing at some rate of aeroplane encountring wind durning it motion
The solution to the problem is as follows:
<span>First, I'd convert 188 mi/hr to ft/s. You should end up with about ~275.7 ft/s.
So now write down all the values you know:
Vfinal = 275.7 ft/s
Vinitial = 0 ft/s
distance = 299ft
</span>
<span>Now just plug in Vf, Vi and d to solve
</span>
<span>Vf^2 = Vi^2 + 2 a d
</span><span>BTW: That will give you the acceleration in ft/s^2. You can convert that to "g"s by dividing it by 32 since 1 g is 32 ft/s^2.</span>