Answer:
4 times around
Explanation:
The total number of teeth involved will be the same for each gear. If the front gear is connected to the pedal and it goes around twice, then 2·24 = 48 teeth will have passed the reference point.
If the rear gear is attached to the wheel, and 48 teeth pass the reference point, then it will have made ...
(48 teeth)/(12 teeth/turn) = 4 turns
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Answer:
938.7 milliseconds
Explanation:
Since the transmission rate is in bits, we will need to convert the packet size to Bits.
1 bytes = 8 bits
1 MiB = 2^20 bytes = 8 × 2^20 bits
5 MiB = 5 × 8 × 2^20 bits.
The formula for queueing delay of <em>n-th</em> packet is : (n - 1) × L/R
where L : packet size = 5 × 8 × 2^20 bits, n: packet number = 48 and R : transmission rate = 2.1 Gbps = 2.1 × 10^9 bits per second.
Therefore queueing delay for 48th packet = ( (48-1) ×5 × 8 × 2^20)/2.1 × 10^9
queueing delay for 48th packet = (47 ×40× 2^20)/2.1 × 10^9
queueing delay for 48th packet = 0.938725181 seconds
queueing delay for 48th packet = 938.725181 milliseconds = 938.7 milliseconds
Answer:
Tension in cable BE= 196.2 N
Reactions A and D both are 73.575 N
Explanation:
The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence
hence
Therefore, tension in the cable, 
Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then
Similarly,
Therefore, both reactions at A and D are 73.575 N
