Answer:
a) F = 3.2 10⁻¹⁰ N
, b) v = 9.9 10⁷ m / s
Explanation:
a) The electric force is
F = q E
The electric field is related to the potential reference
V = E d
E = V / d
Let's replace
F = e V / d
Let's calculate
F = 1.6 10⁻¹⁹ 28 10³ / 1.4 10⁻²
F = 3.2 10⁻¹⁰ N
b) For this part we can use kinematics
v² = v₀ + 2 a d
v = √ 2 ad
Acceleration can be found with Newton's second law
e V / d = m a
a = e / m V / d
a = 1.6 10⁻¹⁹ / 9.1 10⁻³¹ 28 10³ / 1.4 10⁻²
a = 3,516 10⁻¹⁷ m / s²
Let's calculate the speed
v = √ (2 3,516 10¹⁷ 1.4 10⁻²)
v = √ (98,448 10¹⁴)
v = 9.9 10⁷ m / s
Refer to the diagram shown below.
Assume that air resistance is ignored.
Note:
The distance, h, of a falling object with initial vertical velocity of zero at time t is
h = (1/2)gt²
where
g = 9.8 m/s²
The initial vertical velocity of the supplies is 0 m/s.
It the time taken for the supplies to reach the ground is t, then
(50 m) = (1/2)*(9.8 m/s²)*(t s)²
Hence obtain
t² = 50/4.9 = 10.2041
t = 3.1944 s
The horizontal distance traveled at a speed of 100 m/s is
d = (100 m/s)*(3.1944 s) = 319.44 m
Answer: 319.4 m (nearest tenth)
Energy that transfers through the medium