Like a lot of other things, (gravity, sound, electrostatic force), brightness also decreases as the square of the distance.
When the source moves to a new position that's 4 times as far away, its apparent brightness becomes (1/4^2) its original value.
That's 1/16 .
Answer:
v = √k/m x
Explanation:
We can solve this exercise using the energy conservation relationships
starting point. Fully compressed spring
Em₀ =
= ½ k x²
final point. Cart after leaving the spring
= K = ½ m v²
Em₀ = Em_{f}
½ k x² = ½ m v²
v = √k/m x
Answer: 6700m/s^2 E
Explanation:
Given the following :
Velocity of ball before being hit by a racquet = initial velocity(u) = 55m/s W
Velocity of the ball after being hit = final Velocity (v) = 45m/s E
Time of contact between ball and racquet (t) = 1.5 × 10^-2
The acceleration of a body is the change in Velocity of a body with time. It given by:
Acceleration (a) = [final Velocity(v) - initial velocity(u)] / time(t)
The westward direction is towards the left and thus Velocity takes a negative value, similarly, if it's towards the right, Velocity takes a positive value.
Therefore,
a = [45 - (-55)] / 1.5×10^-2
a = [45 + 55] / 0.015
a = 100 / 0.015
a = 6666.6666 m/s^2
a = 6700m/s^2 E
Since the value of acceleration is positive, the direction is towards the East (acceleration is in the direction of the ball's final Velocity).
Answer:

Explanation:
The equation of charge on the capacitor given its capacitance and a voltage applied to it is Q=CV.
Our capacitance is expressed in <em>microfarads</em>, where micro stands for
, so in S.I. we sill have:

Where the results has been expresed with <em>two significant figures</em>.