How much force is required to drag a 90 lb. box up this "frictionless" inclined plane? 109 lb. 10 lb. 81 lb. 9 lb.
2 answers:
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Answer:
The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.
Explanation:
Let suppose that spaceship is accelerated uniformly. A yard equals 0.914 meters. A feet equals 0.304 meters. If air viscosity and friction can be neglected, then acceleration (), measured in meters per square second, is estimated by this kinematic formula:
(1)
Where:
- Travelled distance, measured in meters.
,
- Initial and final speeds of the spaceship, measured in meters.
If we know that ,
and
, then the acceleration experimented by the spaceship is:
The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.
Answer:
SKID
Explanation:
In general, airplane tracks are flat, they do not have cant, consequently the friction force is what keeps the bicycle in the circle.
Let's use Newton's second law, let's set a reference frame with the horizontal x-axis and the vertical y-axis.
Y axis y
N- W = 0
N = W
X axis (radial)
fr = m a
the acceleration in the curve is centripetal
a =
the friction force has the expression
fr = μ N
we substitute
μ mg = m v²/r
v =
we calculate
v =
v = 1,715 m / s
to compare with the cyclist's speed let's reduce to the SI system
v₀ = 18 km / h (1000 m / 1 km) (1 h / 3600 s) = 5 m / s
We can see that the speed that the cyclist is carrying is greater than the speed that the curve can take, therefore the cyclist will SKID