A 2498 kg car is moving at 17.1 m/s slams on its brakes and slows to 2.6 m/s. What is the magnitude (absolute value) of the impu
1 answer:
Answer:
<em>J=36221 Kg.m/s</em>
Explanation:
<u>Impulse-Momentum Theorem</u>
These two magnitudes are related in the following way. Suppose an object is moving at a certain speed and changes it to
. The impulse is numerically equivalent to the change of linear momentum. Let's recall the momentum is given by
The initial and final momentums are, respectively
The change of momentum is
It is numerically equal to the Impulse J
We are given
The impulse the car experiences during that time is
J=-36221 Kg.m/s
The magnitude of J is
J=36221 Kg.m/s
You might be interested in
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
