The velocity of the ball when it reaches the ground is equal to B. 68.6 m/s. This value was obtained from the formula Vf = Vi + at. Vf is the final velocity. Vi is the initial velocity. The acceleration is "a", while the time of travel is "t". The solution is:
<span>Vf = Vi + at
</span>Vf = 0 + (-9.8 m/s^2) (7 s)
Vf = -68.6 m/s
The negative sign denotes the direction of the ball.
Answer:

Explanation:
The speed increased from 2.0 * 10^7 m/s to 4.0 * 10^7 m/s over a 1.2 cm distance.
Let us find the acceleration:


Electric force is given as the product of charge and electric field strength:
F = qE
where q = electric charge
E = Electric field strength
Force is generally given as:
F = ma
where m = mass
a = acceleration
Equating both:
ma = qE
E = ma / q
For an electron:
m = 9.11 × 10^{-31} kg
q = 1.602 × 10^{-19} C
Therefore, the electric field strength of the electron is:

Answer:
T = 676 N
Explanation:
Given that: f = 65 Hz, L = 2.0 m, and ρ = 5.0 g
= 0.005 kg
A stationary wave that is set up in the string has a frequency of;
f = 

⇒ T = 4
M
Where: t is the tension in the wire, L is the length of the wire, f is the frequency of the waves produced by the wire and M is the mass per unit length of the wire.
But M = L × ρ = (2 × 0.005) = 0.01 kg/m
T = 4 ×
×
× 0.01
= 4 × 4 ×4225 × 0.01
= 676 N
Tension of the wire is 676 N.