It mimics the movement of the waves
Answer and explanation:
When you are changing a car tire, the most important thing is to keep the total diameter as equal as possible.
The total car tire diameter can be calculated as:

The profile of this tire is 75 (the higher/taller relation), therefore a 5 percent lower profile would be:
pr=0.95·75=71.25
The problem is that the profiles are normalized and the nearest profile available is 70.
If we take a theorical tire with a profile of 71.25:

The theorical tire size should be 205/71 R15.
If we look for a real tire size, we should look for a tire with a diameter nearest to 26.5'' and a profile of 70.
The best option for real tire size is: Tire 225/70 R14 (wheel diameter of 26.4'') or 205/70 R15 (wheel diameter of 26.3'').
Answer:51.44 units
Explanation:
Given
x component of vector is 
y component of vector is 
so position vector is

Magnitude of vector is


|r|=51.44 units
Direction

vector is in 2nd quadrant thus


Answer:
150.6 km
Explanation:
One mile is about 1.61 km so multiply 93.6 by 1.6 which gives you above 150.6
Answer: 13.2 seconds.
Explanation: using equation of motion; S= ut +1/2at² where u = initial velocity=0
S= distance travelled
a = acceleration due gravity
t= time.
1 foot = 0.305m so,
S= 2860 feet =872.3m
S= ut+1/2 at²
872.3 = 0×t + 1/2×10 × t²
872.3 =0 + 5t²
T²= 872.3/5
T²= 174.46
Take the square root of T we then have;
t = 13.2 seconds to one decimal place.