(a) Equating centripetal force to friction force, one finds the relation
v² = kar
for car speed v, coefficient of friction k, radius of curvature r, and downward acceleration a.
There is already downward acceleration due to gravity. The additional accceleration due to the wing is
a = F/m = 10600 N/(805 kg) ≈ 13.1677 m/s²
We presume this is added to the 9.80 m/s² gravity provides, so the coefficient of friction is
k = v²/(ar) = (54 m/s)²/((13.1677 m/s² +9.80 m/s²)·(155 m))
k ≈ 0.8191
(b) The maximum speed is proportional to the square root of the downward acceleration. Changing that by a factor of 9.80/(9.80+13.17) changes the maximum speed by the square root of this factor.
max speed with no wing effect = (54 m/s)√(9.8/22.97) ≈ 35.27 m/s
Answer:
The distance close to the sidewalk can the flower pot fall is
x = 14.83 m
Explanation:
Given
Δt = 0.300 s
d = 19.6 m
h = 1.79 m
Knowing as the velocity of the sound as a 330 m/s
t = (19.6 - 1.79)m / 330 m/s
t = 0.0539 s
Total time
tₙ = 0.3 + 0.0539 = 0.3539 s
Time for flower-pot
s = ¹/₂ * g * t²
tₐ = √[(2 * 17.81m)/9.81m/s²]
tₐ = 1.34 s, t' = 0.3539
1.34 - 0.3539 = 0.9861 s
19.6 m - x = ¹/₂* g * t ²
x = 19.6 - ¹/₂ * (9.81) * (0.9861)²
x = 14.83 m
Answer:
The bus moved after our girl Sandy passed out = S = 905.04 m
Explanation:
At t = 0 sec Vi = 80mph = 35.76 m/s
t = 30 sec Vf = 55mph = 24.58 m/s
According to the first equation of motion
<u>Vf = Vi + at</u>
24.587 = 35.76 + a(30)
a = - 0.373 m/s2 (negative sign shows deceleration)
Now for distance travelled in 30 sec we have
second equation of motion
<u> S=Vit+1/2at2</u>
S = 35.76 x 30 + 1/2 x -0.373 x 30 x 30
S = 905.04 m
Answer:
b) True. the force of air drag on him is equal to his weight.
Explanation:
Let us propose the solution of the problem in order to analyze the given statements.
The problem must be solved with Newton's second law.
When he jumps off the plane
fr - w = ma
Where the friction force has some form of type.
fr = G v + H v²
Let's replace
(G v + H v²) - mg = m dv / dt
We can see that the friction force increases as the speed increases
At the equilibrium point
fr - w = 0
fr = mg
(G v + H v2) = mg
For low speeds the quadratic depended is not important, so we can reduce the equation to
G v = mg
v = mg / G
This is the terminal speed.
Now let's analyze the claims
a) False is g between the friction force constant
b) True.
c) False. It is equal to the weight
d) False. In the terminal speed the acceleration is zero
e) False. The friction force is equal to the weight
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