I would say the same thing as the first answer
Answer:
The First is an example of acceleration, the second is an example of velocity
Coulomb's law:
F = k×q₁×q₂/r² where k ≈ 9.00×10⁹NC⁻²m²
Given values:
q₁ = +1.0C
q₂ = -1.0C
F = 650N
Substitute the terms in Coulomb's law with our given values. We will have to use the absolute value of q₂ to so the algebra works out. Solve for r:
650 = 9.00×10⁹×1.0×1.0/r²
r = 3721m
Taking significant figures into account:
<h3>r = 3700m</h3>
Answer:
Average speed = 0.0075 m/s
Average velocity = 0.0025 m/s along forward direction
Explanation:
Speed is the ratio of distance and time and velocity is the ratio of displacement and time.
Distance traveled = 10 + 5 = 15 cm = 0.15 m
Displacement = 10 - 5 = 5 cm = 0.05 m
Time = 20 seconds
Average speed = 0.0075 m/s
Average velocity = 0.0025 m/s along forward direction
Answer:
a)Distance traveled during the first second = 4.905 m.
b)Final velocity at which the object hits the ground = 38.36 m/s
c)Distance traveled during the last second of motion before hitting the ground = 33.45 m
Explanation:
a) We have equation of motion
S = ut + 0.5at²
Here u = 0, and a = g
S = 0.5gt²
Distance traveled during the first second ( t =1 )
S = 0.5 x 9.81 x 1² = 4.905 m
Distance traveled during the first second = 4.905 m.
b) We have equation of motion
v² = u² + 2as
Here u = 0, s= 75 m and a = g
v² = 0² + 2 x g x 75 = 150 x 9.81
v = 38.36 m/s
Final velocity at which the object hits the ground = 38.36 m/s
c) We have S = 0.5gt²
75 = 0.5 x 9.81 x t²
t = 3.91 s
We need to find distance traveled last second
That is
S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m
Distance traveled during the last second of motion before hitting the ground = 33.45 m
