Answer:17.08 s
Explanation:
Given
distance between First and second Runner is 45.6 m
speed of first runner
=3.1 m/s
speed of second runner
=4.65 m/s
Distance between first runner and finish line is 250 m
Second runner need to run a distance of 250+45.6=295.6 m
Time required by second runner 
time required by first runner to reach finish line
Thus second runner reach the finish line 80.64-63.56=17.08 s earlier
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.
Refer to the diagram shown below.
Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force
The applied torque is
T = F*r
= (260 N)*(0.33 m)
= 85.8 N-m
By definition,
T = I*α
Therefore,
I = T/α
= (85.8 N-m)/(0.81 rad/s²)
= 105.93 kg-m²
Answer: 105.93 kg-m²
Answer:
Friction is a force that holds back the movement of a sliding object.
Explanation:
The two types of friction: Static friction and Kinetic friction. Static friction operates between two surfaces that aren't moving relative to each other, while kinetic friction acts between objects in motion.
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