Based on Archimedes' principle, the greatest buoyant force an object can experience in water is determined by which quantity?
1 answer:
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Answer:
Explanation:
Let the extension in the spring be x .
restoring force = weight of block
kx = mg
x =
= 23.84 cm
b )
When the elevator is going upwards
Restoring force = mg + ma
k x₁ = 10.9 ( 9.8 + 1.89 )
x₁ = 28.44 cm
( y coordinate will be - ( 28.44 - 23.84 ) = - 4.6 cm )
c ) When the cable snaps , both elevator and block undergo free fall . In this case apparent g = 0
Since the spring is stretched by 28.44 cm , a restoring force continues to act on the block which is equal to
.2844 x 448
= 127.41 N
So a net acceleration a will act on the block
a = 127.41 / 10.9
= 11.68 m / s²
The block will undergo SHM with amplitude equal to 28.44 cm .
a) The train collide after 22.5 seconds
b) The trains collide at the location x = 537.5 m
c) See graph in attachment
d) The freight train must have a head start of 500 m
e) The deceleration must be smaller (towards negative value) than
f) The two trains avoid collision if the acceleration of the freight train is at least
Explanation:
a)
We can describe the position of the passenger train at time t with the equation
where
is the initial velocity of the passenger train
is the deceleration of the train
On the other hand, the position of the freight train is given by
where
is the initial position of the freight train
is the constant velocity of the train
The collision occurs if the two trains meet, so
This is a second-order equation that has two solutions:
t = 22.5 s
t = 177.5 s
We are interested in the 1st solution, which is the first time at which the passenger train collides with the freight train, so t = 22.5 seconds.
b)
In order to find the location of the collision, we just need to substitute the time of the collision into one of the expression of the position of the trains.
The position of the freight train is
And substituting t = 22.5 s, we find:
We can verify that the passenger train is at the same position at the time of the collision:
So, the two trains collide at x = 537.5 m.
c)
In the graph in attachment, the position-time graph of each train is represented. We have:
- The freight train is moving at constant speed, therefore it is represented with a straight line with constant slope (the slope corresponds to its velocity, so 15.0 m/s)
- The passenger train has a uniformly accelerated motion, so it is a parabola: at the beginning, the slope (the velocity) is higher than that of the freight train, however later it decreases due to the fact that the train is decelerating
The two trains meet at t = 22.5 s, where the position is 537.5 m.
d)
In order to avoid the collision, the freight train must have a initial position of
such that the two trains never meet.
We said that the two trains meet if:
Re-arranging,
Substituting the values for the acceleration and the velocity,
The solution of this equation is given by the formula
The two trains never meet if the discrimant is negative (so that there are no solutions to the equation), therefore
Therefore, the freight train must have a head start of 500 m.
e)
In this case, we want to find the acceleration of the passenger train such that the two trains do not collide.
We solve the problem similarly to part d):
Re-arranging
Substituting,
The solution to this equation is
Again, the two trains never meet if the discriminant is negative, so
So, the deceleration must be smaller (towards negative value) than
f)
In this case, the motion of the freight train is also accelerated, so its position at time t is given by
where is the acceleration of the freight train.
Then we solve the problem similarly to the previous part: the two trains collide if their position is the same,
Re-arranging,
And the solution is
Again, the two trains avoid collision if the discriminant is negative, so
Learn more about accelerated motion:
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