The sports car has a weight of 4900 lblb and center of gravity at GG. If it starts from rest it causes the rear wheels to slip a
s it accelerates. The coefficients of static and kinetic friction at the road are μsμs = 0.5 and μkμkmu_k = 0.35, respectively. Neglect the mass of the wheelDetermine how long it takes for it to reach a speed of 10 ft/s.What are the normal reactions at each of the rear wheels on the road?What are the normal reactions at each of the front wheels on the road?
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Answer:
2)
3)
Explanation:
1) Expressing the Division as the summation of the quotient and the remainder
for
118, knowing it is originally a decimal form:
118:2=59 +(0), 59/2 =29 + 1, 29/2=14+1, 14/2=7+0, 7/2=3+1, 3/2=1+1, 1/2=0+1
2)
Similarly, we'll start the process with the absolute value of -49 since we want the positive value of it. Then let's start the successive divisions till zero.
|-49|=49
49:2=24+1, 24:2=12+0,12:2=6+0,6:2=3+0,3:2=1+1,1:2=0+1
100011
3)
The first step on that is dividing by 16, and then dividing their quotient again by 16, so on and adding their remainders. Simply put: