Answer: 0.0180701 s
Explanation:
Given the following :
Length of string (L) = 10 m
Weight of string (W) = 0.32 N
Weight attached to lower end = 1kN = 1×10^3
Using the relation:
Time (t) = √ (weight of string * Length) / weight attached to lower end * acceleration due to gravity
g = acceleration due to gravity = 9.8m/s^2
Weight of string = 0.32N
Time(t) = √ (0.32 * 10) / [(1*10^3) * (9.8)]
Time = √3.2 / 9800
= √0.0003265
= 0.0180701s
The term "latency" describes the 27 seconds that pass after someone has finished texting or conversing on the phone before they can concentrate again on driving. The given statement is true.
<h3>What is latency?</h3>
A measure of delay is latency. The amount of time it takes for some data to travel across a network is known as latency.
The amount of time it takes for information to travel from its source to its destination and back is often quantified as a round trip delay.
The 27 seconds that pass after someone has done texting or talking on the phone before they can focus on driving again are referred to as "latency." The assertion is accurate.
Hence, the given statement is true.
To learn more about the latency refer to;
brainly.com/question/14264521
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Answer: 0.5 m
Explanation:
Given
Mass of the person is 
Trampoline launches the person into the air up to height of 
Force experience by springs is 
Here, the work done on displacing the springs is equivalent to the Potential energy acquired by the person i.e.
![\Rightarrow F\cdot x=mgh\quad [\text{x=displacement of the trampoline}]\\\\\text{Insert the values}\\\\\Rightarrow x=\dfrac{50\times 9.8\times 2}{1960}\\\\\Rightarrow x=\dfrac{980}{1960}\\\\\Rightarrow x=0.5\ m](https://tex.z-dn.net/?f=%5CRightarrow%20F%5Ccdot%20x%3Dmgh%5Cquad%20%5B%5Ctext%7Bx%3Ddisplacement%20of%20the%20trampoline%7D%5D%5C%5C%5C%5C%5Ctext%7BInsert%20the%20values%7D%5C%5C%5C%5C%5CRightarrow%20x%3D%5Cdfrac%7B50%5Ctimes%209.8%5Ctimes%202%7D%7B1960%7D%5C%5C%5C%5C%5CRightarrow%20x%3D%5Cdfrac%7B980%7D%7B1960%7D%5C%5C%5C%5C%5CRightarrow%20x%3D0.5%5C%20m)
Answer:
There is no atmspheric pressure
