Distance is the total length of an object's path. Displacement is the overall change in position, ie. how far an object is from its initial position.
The court is 30 m long, so a path going back and forth once is 60 m long. Going along this path 6 times totals 360 m.
The end point is the same as the starting point, so the displacement is 0 m.
Mass of the block = 1.4 kg
Weight of the block = mg = 1.4 × 9.8 = 13.72 N
Normal force from the surface (N) = 13.72 N
Acceleration = 1.25 m/s^2
Let the coefficient of kinetic friction be μ
Friction force = μN
F(net) = ma
μmg = ma
μg = a
μ = 
μ = 
μ = 0.1275
Hence, the coefficient of kinetic friction is: μ = 0.1275
The solution for this problem:
Given:
f1 = 0.89 Hz
f2 = 0.63 Hz
Δm = m2 - m1 = 0.603 kg
The frequency of mass-spring oscillation is:
f = (1/2π)√(k/m)
k = m(2πf)²
Then we know that k is constant for both trials, we have:
k = k
m1(2πf1)² = m2(2πf2)²
m1 = m2(f2/f1)²
m1 = (m1+Δm)(f2/f1)²
m1 = Δm/((f1/f2)²-1)
m 1 = 0.603/
(0.89/0.63)^2 – 1
= 0.609 kg or 0.61kg or 610 g
Answer for the first one is 2. The answer for the 3rd one is 8
Answer:
F= 403429 kpa
Explanation:
Pressure is the product of force and area
Mathematically,
P=F*A -------where F is force and A is area.
A= 40 *0.1 = 4mm² -----convert to m²
A= 4e⁻⁶ m²
P= 4000000 pa
F= P/A = 4000000/4e⁻⁶
F= 403428793.493 pa
F= 403429 kpa