Answer:
The block didn't slide due to balancing of gravitational force with friction force
Explanation:
When the block was given a flick the force provided an acceleration to it and it moved up the inclined plane. when the block reached top it was expected that it would slide back but it didn't this happened because of the frictional force acting on the bottom the block which was balancing the gravitational force component along the plane and this prevented sliding back of the block.
static friction was balancing mg*sin(theta)
fs = mg*sin(theta)
Answer:
Particles in a: gas are well separated with no regular arrangement. liquid are close together with no regular arrangement. solid are tightly packed, usually in a regular pattern.
i hope this helps your answer
Answer:
The force on q₁ due to q₂ is (0.00973i + 0.02798j) N
Explanation:
F₂₁ = 
Where;
F₂₁ is the vector force on q₁ due to q₂
K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²
r₂₁ is the unit vector
|r₂₁| is the magnitude of the unit vector
|q₁| is the absolute charge on point charge one
|q₂| is the absolute charge on point charge two
r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)
|r₂₁| = 
(|r₂₁|)² = 148.25
= 0.050938(0.19107i + 0.54933j) N
= (0.00973i + 0.02798j) N
Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N
Answer:
2.32 s
Explanation:
Using the equation of motion,
s = ut+g't²/2............................ Equation 1
Where s = distance, u = initial velocity, g' = acceleration due to gravity of the moon, t = time.
Note: Since Onur drops the basket ball from a height, u = 0 m/s
Then,
s = g't²/2
make t the subject of the equation,
t = √(2s/g')...................... Equation 2
Given: s = 10 m, g' = 3.7 m/s²
Substitute this value into equation 2
t = √(2×10/3.7)
t = √(20/3.7)
t = √(5.405)
t = 2.32 s.
