90for each walk because 6• 15 = 90
Complete Question
A 95 kg clock initially at rest on a horizontal floor requires a 650 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 560 N keeps it moving with a constant velocity. Find the coefficient of static friction and the coefficient of kinetic friction.
Answer:
The value for static friction is ![\mu_s = 0.60](https://tex.z-dn.net/?f=%5Cmu_s%20%3D%20%200.60)
The value for static friction is ![\mu_k = 0.70](https://tex.z-dn.net/?f=%5Cmu_k%20%3D%20%200.70)
Explanation:
From the question we are told that
The mass of the clock is ![m = 95 \ kg](https://tex.z-dn.net/?f=m%20%20%3D%20%2095%20%5C%20%20kg)
The first horizontal force is ![F _s = 560 \ N](https://tex.z-dn.net/?f=F%20_s%20%20%3D%20%20560%20%5C%20%20N)
The second horizontal force is ![F _k = 650 \ N](https://tex.z-dn.net/?f=F%20_k%20%20%3D%20%20650%20%20%5C%20%20N)
Generally the static frictional force is equal to the first horizontal force
So
![F _s = m * g * \mu_s](https://tex.z-dn.net/?f=F%20_s%20%20%3D%20%20m%20%20%2A%20%20g%20%20%2A%20%20%5Cmu_s)
=> ![560 = 95 * 9.8 * \mu_s](https://tex.z-dn.net/?f=560%20%20%3D%20%2095%20%20%2A%20%209.8%20%20%2A%20%20%5Cmu_s)
=> ![\mu_s = 0.60](https://tex.z-dn.net/?f=%5Cmu_s%20%3D%20%200.60)
Generally the kinetic frictional force is equal to the second horizontal force
So
![F _k = m * g * \mu_k](https://tex.z-dn.net/?f=F%20_k%20%20%3D%20%20m%20%20%2A%20%20g%20%20%2A%20%20%5Cmu_k)
![650 = 95 * 9.8 * \mu_k](https://tex.z-dn.net/?f=650%20%3D%20%2095%20%20%2A%20%209.8%20%20%2A%20%20%5Cmu_k)
![\mu_k = 0.70](https://tex.z-dn.net/?f=%5Cmu_k%20%3D%20%200.70)
Answer:
e. design programming
Explanation:
The planning techniques are responsible for structuring the tasks to be performed within the project, defining the duration and the order of execution of the same, while the programming techniques try to organize the activities so that the logical temporal relationships between them, determining the calendar or the moments of time in which each one must be realized. The programming must be consistent with the objectives pursued and respect existing restrictions (resources, costs, workloads).
The programming therefore consists in setting, in an approximate way, the moments of beginning and termination of each activity. Some activities may have slack and others are critical activities (fixed over time).
STEPS:
Build a time diagram (moments of beginning and slack of activities).
Establish the times of each activity.
Analyze project costs and adjust clearances (minimum cost project).
longitude and latitude<span />
Answer:
Winner wins by 0.969 s
Explanation:
For the Porche:
Given:
Displacement of Porsche s = 400 m
Acceleration of Porsche a = 3.4 m/s^2
From Newton's second equation of motion,
(u = 0 as the car was initially at rest)
Substituting the values into the equation, we have
![t^2 = (2 * 400) / 3.4](https://tex.z-dn.net/?f=%20t%5E2%20%3D%20%282%20%2A%20400%29%20%2F%203.4%20)
= 235.29 / 3.4
t = 15.33 s
For the Honda:
Displacement of Honda = 310 m
Acceleration of Honda = 3 m/s^2
Applying Newton's second equation of motion
(u = 0 for same reason)
Substituting the values into the equation, we obtain
![t^2 = (2 * 310) / 3](https://tex.z-dn.net/?f=t%5E2%20%3D%20%282%20%2A%20310%29%20%2F%203%20)
= 620 / 3
t = 14.37 s
Hence
The winner (honda) wins by a time interval of = 15.33 - 14.37
=0.969 s