Explanation:
It is given that,
Initial speed of sprinter, u = 0
Final speed of sprinter, v = 10 m/s
Time taken, t = 1.28 s
a. We need to find the acceleration of sprinter. It can be calculated using first equation of motion as :



b. Final speed of the sprinter, v = 36 km/h
Time, t = 0.000355 h
Acceleration, 

Hence, this is the required solution.
<span>I think they were also too skeptic to believe the continent did move or pull apart, even today do you believe that the
continents broke from one big flat plate, and that they pulled apart?
They also wonder what large force would be responsible for the movement.
It
was much later that evidences from plant and animal features that had
similarity from two different planets came up that scientists began
accepting the idea of continental drift.
And similar rock strata from two different opposite continents, showed similar rock strata.
All these evidences came up much later after Alfred Wengener.
So Alfred Wengener was honored Posthumously</span>
Answer:
xf = 5.68 × 10³ m
yf = 8.57 × 10³ m
Explanation:
given data
vi = 290 m/s
θ = 57.0°
t = 36.0 s
solution
firsa we get here origin (0,0) to where the shell is launched
xi = 0 yi = 0
xf = ? yf = ?
vxi = vicosθ vyi = visinθ
ax = 0 ay = −9.8 m/s
now we solve x motion: that is
xf = xi + vxi × t + 0.5 × ax × t² ............1
simplfy it we get
xf = 0 + vicosθ × t + 0
put here value and we get
xf = 0 + (290 m/s) cos(57) (36.0 s)
xf = 5.68 × 10³ m
and
now we solve for y motion: that is
yf = yi + vyi × t + 0.5 × ay × t
² ............2
put here value and we get
yf = 0 + (290 m/s) × sin(57) × (36.0 s) + 0.5 × (−9.8 m/s2) × (36.0 s) ²
yf = 8.57 × 10³ m
Answer:
Explanation:
Using the equation of motion v² = u²+2as
v is the final velocity = 40m/s
u is the iniyail velocity 0m/s
a is the acceleration
s is the displacement
Substituting in the formula;
40² = 0²+2a(50)
1600 = 100a
Divide both sides by 100
100a/100 = 1600/100
a = 16
Hence the car acceleration is 16m/s²