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:


Explanation:
<u>Horizontal Launch</u>
When an object is thrown horizontally with a speed v from a height h, it describes a curved path ruled by gravity until it eventually hits the ground.
The horizontal component of the velocity is always constant because no acceleration acts in that direction, thus:
vx=v
The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:

The horizontal component of the velocity is always the same:

The vertical component at t=5.5 s is:


The boiling point of water, or any liquid, varies according to the surrounding atmospheric pressure. A liquid boils, or begins turning to vapor, when its internal vapor pressure equals the atmospheric pressure.
Answer:
u = - 20 cm
m =
Given:
Radius of curvature, R = 10 cm
image distance, v = 4 cm
Solution:
Focal length of the convex mirror, f:
f = 
Using Lens' maker formula:

Substitute the given values in the above formula:


u = - 20 cm
where
u = object distance
Now, magnification is the ratio of image distance to the object distance:
magnification, m =
magnification, m =
m =
m =