Answer & Explanation:
a)
Lenz's law states that the direction of induced electric current is always such that, it opposes the change in magnetic flux.
In a drop ride, the hub on which we sit and are hung to is an electromagnet and there are many such magnets mounted on the columns of the support. what happens is these electromagnets (in support) generate a repulsive magnetic field with respect to the field generated by the hub solenoids. this results in lift generation till the top of ride. reaching the top, the bar solenoids are at their maximum repulsive force. Then the solenoids in column are set current less means electric supply is cut off. this makes you fall under the effect of gravity. by the time you are half way down, column solenoids are turned on again. As the hub solenoid approaches every single electromagnet in supporting columns. Due to change in magnetic field (with respect to lenz's law) an opposing current induces further providing resistance to the fall, this continues until the ride comes to rest completely. This is how it works.
c) In addition, highly compressive springs, dampers, viscous dampers, etc. could be used in its place.
but the above listed cannot provide a differential braking,
have a limited lifecycle,
will provide resistance during lift also,
require higher maintenance
Answer:
Explanation:
The only thing I can figure you need here is the accleration of the sled. The equation we need to find this is Newton's Second Law that says that sum of the forces acting on an object is equal to the object's mass times its acceleration. For us, that looks like this because of the friction working against the sled:
F - f = ma but of course it's much more involved than that simple equation! We have the F value as 230 N, and we have the mass as 105, but we do not have the frictional force, f, and we need it to solve for a in the above equation. We know that
f = μ
where μ is the coefficient of friction, and is the normal force, aka weight of the object. We will use the coefficient of friction and find the weight in order to fill in for f:
so
so the weight of the sled is
1.0 × 10³ with the correct number of sig dig there. Now to find f:
f = (.025)(1.0 × 10³) so
f = 25 to the correct number of sig fig. Now on to our "real" equation:
F - f = ma and
230 - 25 = 105a. We have to do the subtraction first, round, and then divide since the rules for addition and subtraction are different from the rules for dividing and multiplying.
230 - 25 will round to the tens place giving us 210. Then
210 = 105a. 210 has 2 sig figs in it while 105 has 3, so we will divide and round to 2 sig fig:
a = 2.0 m/sec²
Answer:
none
Explanation:
it's to high up to be affected by the gravity
Answer:
T = 295.57 s
Explanation:
given,
mass of the rocket = 200 Kg
mass of the fuel = 100 Kg
acceleration = 35 m/s²
time, t = 35 s
time taken by the rocket to hit the ground, = ?
Final speed of the rocket when fuel is empty
using equation of motion
v = u + a t
v = 0 + 35 x 35
v = 1225 m/s
height of the rocket where fuel is empty
v² = u² + 2 a s
1225² = 0 + 2 x 35 x h₁
h₁ = 21437.5 m
After 35 s the rocket will be moving upward till the final velocity becomes zero.
Now, using equation of motion to find the height after 35 s
v² = u² + 2 g h₂
0² = 1225² + 2 x (-9.8) h₂
h₂ = 76562.5 m
total height = h₁ + h₂
= 76562.5 m + 21437.5 m = 98000 m
now, time taken by before the rocket hit the ground
using equation of motion
negative sign is used because the distance travel by the rocket is downward.
4.9 t² - 1225 t - 13500 = 0
t = 260.57 s
neglecting the negative sign
total time the rocket was in air
T = t₁ + t₂
T = 35 + 260.57
T = 295.57 s
Time for which rocket was in air is equal to 295.57 s.
