<h2>distance = 523 cm</h2>
Explanation:
( a ) The rotational speed of the ladybug = 25 r.p.m = 25/60 r.p.s
= 5/12 rev/sec
( b ) The definition of frequency is the number of rotations per second .
Here the number of rotations per second is 5/12 . Thus frequency = 5/12 Hz
( c ) The tangential speed is v = angular velocity x radius of rotation
The angular velocity ω = 2π x n , where n is the number of rotations per second
Thus angular velocity = 2π x 5/12 = 5π/6 rad/sec
The linear velocity = angular velocity x distance from center of record
Thus tangential speed = 5π/6 x 10 = 25π/3 cm/sec
Angular displacement in 20 sec = ω x t = 5π/6 x 20 = 50π/3 rad
Linear displacement = angular displacement x distance from center of record
= 50π/3 x 10 = 500π/3 = 523 cm
Answer:
1. the pencil would have the momentum and would keep going until it hits the windshield. 2. when the car suddenly accelerates, the pencil would be inert and it would move toward the back of the car until a constant speed from the car is reached.
<u>Communications:</u> Laura develops programs that flag suspicious network activity.
<h3>What is
communication?</h3>
Communication involves the transfer of information from one person (sender) to another (recipient), especially through the use of semiotics, symbols, signs and network devices.
<h3>How to match the scenarios?</h3>
- <u>Communications:</u> Laura develops programs that flag suspicious network activity.
- <u>Healthcare:</u> Greg calibrates electronic scales.
- <u>Manufacturing:</u> Tim ensures the output of a process remains in a defined range.
Read more on communication here: brainly.com/question/26152499
#SPJ1
.Answer:
The value of the work done is 
.
Explanation:
When a charged particle having charge 
is moving through an electric field 
, the net force (
) on the charge is
and the work done (
) by the particle is
Given, 
.
Substitute the value of electric field in equation (1) and then substitute the result in equation (2).
![W &=& \int\limits^7_0 {q\dfrac{A_{0}}{x^{1/2}}} \, dx \\&=& qA_{0} \int\limits^7_0 {x^{-1/2}} \, dx \\&=& 2qA_{0}[x^{1/2}]_{0}^{7}\\&=& 5.29 qA_{0}](https://tex.z-dn.net/?f=W%20%26%3D%26%20%5Cint%5Climits%5E7_0%20%7Bq%5Cdfrac%7BA_%7B0%7D%7D%7Bx%5E%7B1%2F2%7D%7D%7D%20%5C%2C%20dx%20%5C%5C%26%3D%26%20qA_%7B0%7D%20%5Cint%5Climits%5E7_0%20%7Bx%5E%7B-1%2F2%7D%7D%20%5C%2C%20dx%20%5C%5C%26%3D%26%202qA_%7B0%7D%5Bx%5E%7B1%2F2%7D%5D_%7B0%7D%5E%7B7%7D%5C%5C%26%3D%26%205.29%20qA_%7B0%7D)
As long as the graph is a straight line they both will give same answer. Edit: In the case of errors,I think the most fitting slope line is more accurate method,because taking average is same as drawing a line from initial to final point of the line and taking its slope.
Hope it helps
