<u>Answer:</u>
The final velocity of the two railroad cars is 1.09 m/s
<u>Explanation:</u>
Since we are given that the two cars lock together it shows that the collision is inelastic in nature. The final velocity due to inelastic collision is given by

where
V= Final velocity
M1= mass of the first object in kgs = 12000
M2= mas of the second object in kgs = 10000
V1= initial velocity of the first object in m/s = 2m/s
V2= initial velocity of the second object in m/s = 0 (given at rest)
Substituting the given values in the formula we get
V = 2×12000 + 0x100012000 + 10000= 2400022000= 1.09 m/s

Which is the final velocity of the two railroad cars
Explanation:
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Answer:
Part a)

Part b)

Since the distance of other building is 15 m so YES it can make it to other building
Part c)

direction of velocity is given as
![[tex]\theta = 26.35 degree](https://tex.z-dn.net/?f=%5Btex%5D%5Ctheta%20%3D%2026.35%20degree)
Explanation:
Part a)
acceleration due to gravity on this planet is 3/4 times the gravity on earth
So the acceleration due to gravity on this new planet is given as


now the vertical displacement covered by the canister is given as

now by kinematics we have



Part b)
Horizontal speed of the canister is given as

now the distance moved by it



Since the distance of other building is 15 m so YES it can make it to other building
Part c)
Final velocity in X direction will remains the same

final velocity in Y direction



now magnitude of velocity is given as



direction of velocity is given as


![[tex]\theta = 26.35 degree](https://tex.z-dn.net/?f=%5Btex%5D%5Ctheta%20%3D%2026.35%20degree)
Answer:
The answer is True
Explanation:
Statistical Multiplexing is considered an example of communication link sharing which makes it comparable to DBA (Dynamic Bandwidth Allocation). Here, communication channels are broken down into data streams to optimize the communication process.
In Statistical Time-division Multiplexing, time slots are allocated to data streams for communication optimization. This method makes sure that no time slot or bandwidth is wasted.
Hence, the sum of combined circuits must not be equal to the capacity of the circuit to work effectively.