I think you're saying that once you start pushing on the cars, you want to be able to stop each one in the same time.
This is sneaky. At first, I thought it must be both 'c' and 'd'. But it's not
kinetic energy, for reasons I'm not ambitious enough to go into.
(And besides, there's no great honor awarded around here for explaining
why any given choice is NOT the answer.)
The answer is momentum.
Momentum is (mass x speed). Change in momentum is (force x time).
No matter the weight (mass) or speed of the car, the one with the greater
momentum is always the one that will require the greater (force x time)
to stop it. If the time is the same for any car, then more momentum
will always require more force.
Answer:
35 m
0.56 m/s west
Explanation:
A) Total distance is the length of the path taken.
30 m + 5 m = 35 m
B) Velocity is displacement over time. Displacement is the difference between the final position and the initial position.
If west is -x, and east is +x, then:
Δx = -30 m + 5 m
Δx = -25 m
v = Δx / t
v = -25 m / 45 s
v = -0.56 m/s
v = 0.56 m/s west
Answer:
40 N
Explanation:
We are given that
Speed of system is constant
Therefore, acceleration=a=0
Tension applied on block B=T=50 N
Friction force=f=10 N
We have to find the friction force acting on block A.
Let T' be the tension in string connecting block A and block B and friction force on block A be f'.
For Block B
Where 
=Mass of block B
Substitute the values
For block A
Where 
Mass of block A
Substitute the values
Hence, the friction force acting on block A=40 N
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.
