Answer:
<em>The end of the ramp is 38.416 m high</em>
Explanation:
<u>Horizontal Motion
</u>
When an object is thrown horizontally with an initial speed v and from a height h, it follows a curved path ruled by gravity.
The maximum horizontal distance traveled by the object can be calculated as follows:
If the maximum horizontal distance is known, we can solve the above equation for h:
The skier initiates the horizontal motion at v=25 m/s and lands at a distance d=70 m from the base of the ramp. The height is now calculated:
h= 38.416 m
The end of the ramp is 38.416 m high
4% of 110 is 4.4. So the possible range of speeds is the interval from 110-4.4 till 110+4.4.
105.6 till 114.4
Answer:
0.333 m/s
Explanation:
avg speed= (total distance)/(total time)
10m/30s
0.333 m/s
1. Based on Scenario A, multiple frames will minimize re-transmission overhead and should be preferred in the encapsulation of packets.
2. Based on Scenario B, the encapsulation of packets should be in a single frame because of the high level of network reliability and accuracy.
<u>Justification:</u>
There will not be further need to re-transmit the packets in a highly reliable and accurate network environment, unlike in an environment that is very prone to errors. The reliable and accurate network environment makes a single frame economically better.
Encapsulation involves the process of wrapping code and data together within a class so that data is protected and access to code is restricted.
With encapsulation, each layer:
- provides a service to the layer above it
- communicates with a corresponding receiving node
Thus, in a reliable and accurate network environment, single frames should be used to enhance transmission and minimize re-transmission overhead. This is unlike in an environment that is very prone to errors, where multiple frames should rather be used to minimize re-transmission overhead.
Learn more about encapsulation of packets here: brainly.com/question/22471914
