The periodic table is in increasing atomic mass. Hope this helped.
<span>v=u+at</span>
<span>
0=5.0283−9.8×t</span><span>
t=0.5130sec</span><span>
Total time = 2t = 1.0261 sec</span>
Answer
given,
v = (6 t - 3 t²) m/s
we know,
position of the particle
integrating both side
x = 3 t² - t³
Position of the particle at t= 3 s
x = 3 x 3² - 3³
x = 0 m
now, particle’s deceleration
a = 6 - 6 t
at t= 3 s
a = 6 - 6 x 3
a = -12 m/s²
distance traveled by the particle
x = 3 t² - t³
at t = 0 x = 0
t = 1 s , x = 3 (1)² - 1³ = 2 m
t = 2 s , x = 3(2)² - 2³ = 4 m
t = 3 s , x = 0 m
total distance traveled by the particle
D = distance in 0-1 s + distance in 1 -2 s + distance in 2 -3 s
D = 2 + 4 + 2 = 8 m
average speed of the particle
Answer:
It will be more than deta t
Explanation:
Because
deta t' = န deta t
But န= 1/√ (1 - v²/c²
So the observers in all the initial frames will be more than deta t
By definition, acceleration is the change in velocity per change of time. As time passes by, the time increases in value. So, when the acceleration is decreasing while the time is increasing, then that means that the change of velocity is also decreasing with time. So, optimally, the initial velocity and the velocity at any time are very relatively close to each other,
