Q: Two charges q1 and q2, that are distance d apart , repel each other with a force of 6.40 N. what would be the force between two charges q1'=2q1 and q2'=3q2 that that are distance d apart?
Answer:
The force = 38.4 N
Explanation:
From coulombs law,
F = kq₁q₂/r² ............................ Equation 1
Where F = Force of attraction or repulsion between the charges, q₁ and q₂ = first and second charge respectively, r = distance between the charges, k = constant of proportionality.
When, F = 6.4 N, r = d m.
6.4 = kq₁q₂/d²......................... Equation 1
When q₁' = 2q₁, q₂' = 3q₂, r = d cm
F = k(2q₁)(3q₂)/d²
F = 6kq₁q₂/d².......................... Equation 2
Dividing Equation 1 by equation 2
6.4/F = kq₁q₂/d²/(6kq₁q₂/d²)
6.4/F = 1/6
F = 6.4×6
F = 38.4 N.
Thus the force = 38.4 N
- We know, acceleration is the change of velocity by time.
- Velocity is the speed of an object which also indicates the direction.
- Hence, acceleration is both dependant upon the speed as well as the direction.
- So, if an object is moving at a constant speed in a changing direction, the acceleration will also change. It will not be zero.
- An example is that of uniform circular motion.
Answer:
if an object is moving at a constant speed in a changing direction, the acceleration of the object will not be zero.
Answer:
the question this morning
Explanation:
90+70 look than
Answer:
The increase in potential energy of the ball is 115.82 J
Explanation:
Conceptual analysis
Potential Energy (U) is the energy of a body located at a certain height (h) above the ground and is calculated as follows:
U = m × g × h
U: Potential Energy in Joules (J)
m: mass in kg
g: acceleration due to gravity in m/s²
h: height in m
Equivalences
1 kg = 1000 g
1 ft = 0.3048 m
1 N = 1 (kg×m)/s²
1 J = N × m
Known data
Problem development
ΔU: Potential energy change
ΔU = U₂ - U₁
U₂ - U₁ = mₓgₓh₂ - mₓgₓh₁
U₂ - U₁ = mₓg(h₂ - h₁)
The increase in potential energy of the ball is 115.82 J
I think it is c I'm only in 7th grade but I'm pretty sure that the answer is c
