<span>The medium in which it travels through</span>
Answer:
The force of static friction acting on the luggage is, Fₓ = 180.32 N
Explanation:
Given data,
The mass of the luggage, m = 23 kg
You pulled the luggage with a force of, F = 77 N
The coefficient of static friction of luggage and floor, μₓ = 0.8
The formula for static frictional force is,
Fₓ = μₓ · η
Where,
η - normal force acting on the luggage 'mg'
Substituting the values in the above equation,
Fₓ = 0.8 x 23 x 9.8
= 180.32 N
Hence, the minimum force require to pull the luggage is, Fₓ = 180.32 N
Answer:
The speed with which the man flies forward is 5.5 m/s
Explanation:
The mass of the man = 100 kg
The mass of the scooter = 10 kg
The speed with which the man was traveling on the scooter = 5 m/s
The speed of the scooter after it hits the rock = 0 m/s
Let v represent the speed with which the man flies forward
The formula for momentum, P, is P = Mass × Velocity
The conservation of linear momentum principle is, the total initial momentum = The total final momentum, therefore, we have;
The total initial momentum = (100 kg + 10 kg) × 5 m/s = 550 kg·m/s
The total final momentum = 100 kg × v + 10 kg × 0 m/s = 100 kg × v
When the momentum is conserved, we have;
550 kg·m/s = 100 kg × v
∴ v = 550 kg·m/s/(100 kg) = 5.5 m/s.
The speed with which the man flies forward = v = 5.5 m/s
Explanation:
Draw a free body diagram for each disc.
Disc A has three forces acting on it: 86.5 N up, T₁ down, and Wa down.
∑F = ma
86.5 N − T₁ − Wa = 0
Wa = 86.5 N − T₁
ma × 9.8 m/s² = 86.5 N − 55.6 N
ma = 3.2 kg
Disc B has three forces acting on it: T₁ up, T₂ down, and Wb down.
∑F = ma
T₁ − T₂ − Wb = 0
Wb = T₁ − T₂
mb × 9.8 m/s² = 55.6 N − 36.5 N
mb = 1.9 kg
Disc C has three forces acting on it: T₂ up, T₃ down, and Wc down.
∑F = ma
T₂ − T₃ − Wc = 0
Wc = T₂ − T₃
mc × 9.8 m/s² = 36.5 N − 9.6 N
mc = 2.7 kg
Disc D has two forces acting on it: T₃ up and Wd down.
∑F = ma
T₃ − Wd = 0
Wd = T₃
md × 9.8 m/s² = 9.6 N
md = 0.98 kg
R1 + R4 = 1430 + 1350 = 2780 = R14 series combination of R1 & R4
R2 + R5 = 1350 + 1150 = 2500 = R25
The circuit has been reduced to 3 resistors in parallel
R314 = 2780 * 1100 / (2780 + 1100) = 788 this is the resistance of the parallel combination of R14 and R3
R31425 = 2500 * 788 / (2500 + 788) = 599 which is the equivalent of the circuit - you can also use the formula for 3 resistors in parallel but this seems simpler
