Answer: Increase in wave frequency
Explanation:
When we talk about acoustics we are dealing with sound waves, and one of their main components along with the velocity and wavelength is the <u>frequency.</u>
In this sense, the frequency of any wave refers to how fast (or slow) a wave oscillates. For example, in the especific case of sound waves when the oscillation is faster, the frequency is higher and the pitch gets higher as well.
Answer:

Explanation:
For the first ball, the moment of inertia and the kinetic energy is:


So, replacing, we get that:

At the same way, the moment of inertia and kinetic energy for second ball is:


So:

Then,
is equal to
, so:




Finally, solving for
, we get:

Answer:
v=5.86 m/s
Explanation:
Given that,
Length of the string, l = 0.8 m
Maximum tension tolerated by the string, F = 15 N
Mass of the ball, m = 0.35 kg
We need to find the maximum speed the ball can have at the top of the circle. The ball is moving under the action of the centripetal force. The length of the string will be the radius of the circular path. The centripetal force is given by the relation as follows :

v is the maximum speed

Hence, the maximum speed of the ball is 5.86 m/s.
Answer:
20 cm
Explanation:
Te electric potential enery U = kq₁q₂/r were q₁ = 5 nC = 5 × 10⁻⁹ C and q₂ = -2 nC = -2 × 10⁻⁹ C and r = √(x - 2)² + (0 - 0)² +(0 - 0)² = x - 2. U = -0.5 µJ = -0.5 × 10⁻⁶ J, k = 9 × 10⁹ Nm²/C².
So r = kq₁q₂/U
x - 2 = kq₁q₂/U
x = 0.02 + kq₁q₂/U m
x = 0.02 + 9 × 10⁹ Nm²/C² × 5 × 10⁻⁹ C × -2 × 10⁻⁹ C/-0.5 × 10⁻⁶ J
x = 0.02 - 90 × 10⁻⁹ Nm²/-0.5 × 10⁻⁶ J
x = 0.02 + 0.18 = 0.2 m = 20 cm