Answer:
The horizontal force acting on m2 is F + 9.8m1
Explanation:
Given;
Block m1 on left of block m2
Make a sketch of this problem;
F →→→→→→→→→→→-------m1--------m2
Apply Newton's second law of motion;
F = ma
where;
m is the total mass of the body
a is the acceleration of the body
The horizontal force acting on block m2 is the force applied to block m1 and force due to weight of block m1
F₂ = F + W1
F₂ = F + m1g
F₂ = F + 9.8m1
Therefore, the horizontal force acting on m2 is F + 9.8m1
Answer:
The vehicle travels 56.25 metres in the interval during which body decelerates .
Explanation:
- Initial velocity of vehicle, u = 32 m/s
- Final velocity of vehicle, v = 22 m/s
- Rate of acceleration, a = - 4.8 m/
Let the distance travelled be s .
We have to determine the distance travelled by the vehicle during this time.
The equation of motion is given by
s =
<u>s = 56.25 metres</u>
The vehicle travels 56.25 metres in the interval during which body decelerates .
Answer:
The Work-in-Process limit at the Analyzing step of the Program Kanban is based on the overall availability of Product Management, other subject matter experts, and the available development capacity and the VA time percentage of each process.
Explanation:
The term work-in-progress (WIP) describes partially finished goods or raw materials, labor, and overhead costs at several stages of the production process to be completed in a production and supply-chain management . WIP (Work in Progress) limits restrict and balance, through kanban board columns,the amount of production to match demand to capacity so workflow is increased all along the line helping it to finish faster avoiding bottlenecks focusing only on current tasks without new work taken on.
A Kanban board requires knowledge to provide a clear view of the WIP limit reached point but it also requieres discipline and commitment from employees.
The answer to this question is d because prejudices is not based on experience or reasons
<h2>
Answer:</h2>
400N/m
<h2>
Explanation:</h2>
When n identical springs of stiffness k, are attached in series, the reciprocal of their equivalent stiffness (1 / m) is given by the sum of the reciprocal of their individual stiffnesses. i.e
= ∑ⁿ₁ [] -----------------------(i)
That is;
= + + + . . . + -------------------(ii)
If they have the same value of stiffness say s, then equation (ii) becomes;
= n x -----------------(iii)
Where;
n = number of springs
From the question,
There are 3 identical springs, each with stiffness of 1200N/m and they are attached in series. This implies that;
n = 3
s = 1200N/m
Now, to calculate the effective stiffness,m, (i.e the stiffness of a longer spring formed from the series combination of these springs), we substitute these values into equation (iii) above as follows;
= 3 x
=
=
Cross multiply;
m = 400N/m
Therefore, the stiffness of the longer spring is 400N/m