You place a 3.0-m-long board symmetrically across a 0.5-m-wide chair to seat three physics students at a party at your house. If
60-kg Dan sits on the left end of the board and 50-kg Tahreen on the right end of the board, where should 54-kg Komila sit to keep the board stable? Ignore the mass of the board and treat each student as point-like objects The positive r-direction is to the right, and the origin is at the center of the chair
Part A Determine the coordinate of the leftmost point on the board where Komila can sit to keep the board stable
Part B Determine the coordinate of the rightmost point on the board where Komila can sit to keep the board stable
Part A Determine the coordinate of the leftmost point on the board where Komila can sit to keep the board stable
Part B Determine the coordinate of the rightmost point on the board where Komila can sit to keep the board stable
1 answer:
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Answer:0.318 revolutions
Explanation:
Given
Initially Propeller is at rest i.e.
after
using
Revolutions turned in 2 s
To get revolution
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Answer:
The answer to your question is: total energy = 30100.4 J
Explanation:
Kinetic energy (KE) is the energy due to the movement of and object, its units are joules (J)
Data
mass = 1280 kg
speed = 4.92 m/s
Force = 509 N
distance = 28.7 m
Formula
Work = Fd
Process
- Calculate Kinetic energy
- Calculate work
- Add both results
KE =
KE = 15492.1 J
Work = (509)(28.7)
Work = 14608.3 J
Total = 15492.1 + 14608.3
Total energy = 30100.4 J
Explanation:
It is given that,
Velocity of the particle moving in straight line is :
We need to find the distance (x) traveled by the particle during the first t seconds. It is given by :
Using by parts integration, we get the value of x as :
Hence, this is the required solution.