A mechanic pushes a 3.60 ✕ 103-kg car from rest to a speed of v, doing 5,130 j of work in the process. during this time, the c
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Answer:
The time taken by the wave to travel along the combination of two wires is 458 ms.
Explanation:
Given that,
Length of steel wire= 27.0 m
Length of copper wire = 48.0 m
Tension = 145 N
Radius of both wires = 0.450 mm
Density of steel wire
Density of copper wire
We need to calculate the linear density of steel wire
Using formula of linear density
Put the value into the formula
We need to calculate the linear density of copper wire
Using formula of linear density
Put the value into the formula
We need to calculate the velocity of the wave along the steel wire
Using formula of velocity
We need to calculate the velocity of the wave along the steel wire
Using formula of velocity
We need to calculate the time taken by the wave to travel along the combination of two wires
Put the value into the formula
Hence, The time taken by the wave to travel along the combination of two wires is 458 ms.
Answer:
a) F₁ = 267.3 N, N₁ = 1300 N, b) μ = 0.324
Explanation:
For this exercise we use the rotational equilibrium condition, we have a reference system is the floor and the anticlockwise rotations as positive, in the adjoint we can see a diagram of the forces
let's use subscript 1 for the ladder and 2 for the firefighter
∑ τ = 0
-W₁ x₁ - W₂ x₂ + N₁ y = 0
N₁ = (1)
the center of mass of the ladder is at its geometric center,
d = L / 2 = 15/2 = 7.5 m
cos 60 = x₁ / d₁
x₁ = d₁ cos 60
x₁ = 7.5 cos 60
x₁ = 3.75 m
for the firefighter d₂ = 4 m
cos 60 = x₂ / d₂
x₂ = d₂ cos 60
x₂ = 4 cos 60 = 2 m
for the fulcrum d₃ = 15 m
sin 60 = y / d₃
y = d₃ sin 60
y = 15 sin 60
y = 13 m
we look for the Normal by substituting in equation 1
N₂ =
N₂ = 267.3 N
now let's use the translational equilibrium relations
X axis
F₁ - N₂ = 0
F₁ = N₂
F₁ = 267.3 N
Axis y
N₁ - W₁ -W₂ = 0
N₁ = W₁ + W₂
N₁ = 500 + 800
N₁ = 1300 N
b) for this case change the firefighter's distance d₂ = 9 m
x₂ = 9 cos 60
x₂ = 4.5 m
we substitute in 1
N₂ = \frac{500 \ 3.75 \ + 800 \ 4.5}{13}
N₂ = 421.15 N
of the translational equilibrium equation on the x-axis
fr = F₁ = N₂
fr = 421.15 N
friction force has the expression
fr = μ N
in this case the reaction of the Earth to the support of the ladder is N1 = 1300N
μ = fr / N₁
μ = 421.15 / 1300
μ = 0.324
