The table is:
t(s) vx(m/s)
0 0
10 23
20 46
30 69
a) from the data in the table, we observe that the acceleration is constant (because the rate of change in velocity is the same for each time interval of 10 seconds), so we can choose just one interval and calculate the acceleration as the ratio between the change in velocity and the change in time. Taking the first interval, we find

b) To find the jet's acceleration in g's, we just need to divide the acceleration in m/s^2 by the value of g, the acceleration of gravity (9.81 m/s^2), so we find

c) the wheels leave the ground when the jet reaches its take-off velocity, which is 82 m/s.
At t=0s, the velocity of the jet is 0. We know that the acceleration is constant (a=2.3 m/s^2), so we can find the time t at which the jet reaches a velocity vf=82 m/s by using the equation

Re-arranging and substituting numbers, we find

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Answer:

0.3619sec
Explanation:
Given that
Mass,m=148 g
Length,L=13 cm
Velocity,u'(0)=10 cm/s
We have to find the position u of the mass at any time t
We know that

Where 

u(0)=0
Substitute the value

Substitute u'(0)=10


Substitute the values

Period =T = 2π/8.68
After half period
π/8.68 it returns to equilibruim
π/8.68 = 0.3619sec
Answer:
d = 105 m
Explanation:
Speed of a car, v = 21 m/s
We need to find the distance traveled by the dar during those 5 s before it stops. Let the distance is d. It can be calculated as :
d = v × t
d = 21 m/s × 5 s
d = 105 m
So, it will cover 105 m before it stops.