D)LT^-1 speed=distance(L)/time(T)——>L/T
Answer:
The potential energy of the rock = 10.5 kN
Explanation:
Mass of rock = 25 kg
Acceleration due to gravity = 10 m/s²
Height = 42 m
Potential energy, PE = mgh, where m is the mass, g is acceleration due to gravity and h is the height.
PE = 25 x 10 x 42 = 10500 N = 10.5 kN
The potential energy of the rock = 10.5 kN
Answer:
here as we increase the distance the intensity will decrease and hence the amplitude of the electric field will decrease and vice-versa
Explanation:
As wee know that the amplitude of the wave will decide the energy of the wave
Here we know that energy density of electromagnetic wave is given as
now we have
so here we can say that intensity of the wave at the given distance from the source is given by formula
so here as we increase the distance the intensity will decrease and hence the amplitude of the electric field will decrease and vice-versa.
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 .
