Answer:
C?
Explanation:
My best guess would be C as it's the only answer that gives a reason behind the statement.
Answer:
It will require a force of 1/5, answer A.
Explanation:
In the attached image we can see an example of an array of pulleys that will lift a 100 kg-f load.
If we analyze pulleys A,B, and C as in the image we can check a force in the cable of 20kg-f.
In the pulley D we have three forces of 20 kg-f each and those forces plus the forces in the pulley B, sum a total of 100 kg-f (60+40). This matches the mechanical advantage (100/5) = 20 kg-f
(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:
4987N
Explanation:
Step 1:
Data obtained from the question include:
Mass (m) = 0.140 kg
Initial velocity (U) = 28.9 m/s
Time (t) = 1.85 ms = 1.85x10^-3s
Final velocity (V) = 37.0 m/s
Force (F) =?
Step 2:
Determination of the magnitude of the horizontal force applied. This can be obtained by applying the formula:
F = m(V + U) /t
F = 0.140(37+ 28.9) /1.85x10^-3
F = 9.226/1.85x10^-3
F = 4987N
Therefore, the magnitude of the horizontal force applied is 4987N
At time t = 0 the velocity is v1. Therefore C1 = v1 and C2 = x1. Equations (1), (2), (3), and (4) fully describe the motion of particles, or bodies experiencing rectilinear (straight-line) motion, where acceleration a is constant.