Consider a railroad bridge over a highway. A train passing over the bridge dislodges a loose bolt from the bridge, which proceed
1 answer:
Answer:
12.35m
Explanation:
Hello! To solve this problem we must consider the following:
1. The car moves with constant speed, which means that the distance traveled is equal to the multiplication of time by speed.
X = VT
we solve the equation for time
2. The bolt moves with constant acceleration, with acceleration of 9.81m / s ^ 2, so we could apply the following equation.
note=remember that "a uniformly accelerated motion", means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
where
Vo = Initial speed
=0
T = time
=1.588s
g =gravity=9.8m/s^2
Y = bridge height
solving
You might be interested in
Answer:
The speed decreases 75%.
Explanation:
- Since no friction present, assuming no external forces acting during the three collisions, total momentum must be conserved.
- For the first collission, only mass 1 is moving before it, so we can write the following equation:
- Since both masses are identical, and they stick together after the collision, we can express the final momentum as follows:
- From (1) and (2) we get:
- v₁ = v₀/2 (3)
- Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1 and 2 combined together before colliding with mass 3 is just v₁, so the initial momentum prior the second collision (p₁) can be expressed as follows:
- Since after the collision the three masses stick together, we can express this final momentum (p₂) as follows:
- From (4) and (5) we get:
- v₂ = v₀/3 (6)
- Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1, 2 and 3 combined together before colliding with mass 4 is just v₂, so the initial momentum prior the third collision (p₂) can be expressed as follows:
- Since after the collision the four masses stick together, we can express this final momentum (p₃) as follows:
- From (7) and (8) we get:
- v₃ = v₀/4
- This means that after the last collision, the speed will have been reduced to a 25% of the initial value, so it will have been reduced in a 75% regarding the initial value of v₀.
= Heat released to cold reservoir
= Heat released to hot reservoir
= maximum amount of work
= temperature of cold reservoir
= temperature of hot reservoir
we know that
eq-1
maximum work is given as
=
-
using eq-1
=
-