Calculate the acceleration of a car that begins at rest at a stop light and accelerates to a speed of 20m/s in a 10 sec time per
1 answer:
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No. of ways in the elements of n be permuted is n!/4
Number of elements = n
Number of face in which the element of n be permuted if 1 must proceed 2 and 3 is to precede 4.
n = 4
The possible permutation will
(1,2,3,4), (1,3,2,4),(1,3,4,2),(3,4,1,2),(3,1,2,4),(3,1,4,2)
There are 6 possible outcomes
For n= 5
With 1,2,3,4 we add 5 in the arrangement with above terms
The possible number of permutation
5×6 = 30
When n = 6
30×6 = 180.
So for an element
If 1 precede 2 and 3 is yo precede 4 is
<h2>= n! ×6/4!
= n!/4
No. of ways in the elements of n be permuted is n!/4
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Answer:
The minimum coefficient of static friction for the box on the floor of the trunk is 0.54.
Explanation:
Given that,
Radius of the unbanked curve, r = 48 m
The speed of the car on the curve, v = 16 m/s
We need to find the minimum coefficient of static friction for the box on the floor of the trunk. It can be calculated by balancing the centripetal force and the force of gravity as :
So, the minimum coefficient of static friction for the box on the floor of the trunk is 0.54. Hence, this is the required solution.