Answer:
5.571 sec
Explanation:
angular frequency = √ (k/m) = √ (49.3 / 5) = 3.14 rad/s
Period To = 2π / angular frequency
Period To = 2π/3.14 = 2 × 3.14 / 3.142 = 2.00 sec which you got
T measured by the observer = To / (√ (1 - (v²/c²))) = 2 / √( 1 - 0.871111) = 2 / 0.35901 = 5.571 sec
t=2.00/(1-√((2.80*10^8)^2/(3.00*10^8)^2))= should have been ( To / (√ (1 - (v²/c²))). where To = 2.00 sec
Answer: height of building = 18.8m
Explanation: The question is a projectile motion, a two dimensional motion with a vertical constant acceleration (g = - 9.8m/s²) and a constant horizontal velocity (thus making horizontal component of acceleration zero).
From the question, distance between bottom of building and where the object lands = 64m, initial velocity for throwing the object = 19.6m/s
The horizontal range formulae is given as
d= vt
Where d= horizontal range = 64m, v = initial velocity of throw.
64 = 19.6 × t
t = 64/ 19.6
t = 3.265 s.
Height (h) of the building is gotten by using the formulae
h =vt - 1/2gt²
h = (19.6×3.265) - 1/2×9.8×(3.265)²
h = 71.05 - (104.47/2)
h = 71.05 - 52.235
h = 18.8m
Answer:
32.9166667 m / s^2
Explanation:
s = 4.25km (1000m / 1km)
= 4250m
u = 20m/s
delta T = 20min (60sec / 1min)
= 1200s
Use formula s = ut + (1/2)at^2
4250m = 20m/s * 1200s + (1/2)a*1200s^2
Rearrange it to find a
a = (s-ut) / (1/2 * t^2)
a = (4250m - 20m/s*1200s) / (1/2 * 1200s^2)
a = -32.9166667 m / s^2
Kinetic energy =(1/2) (mass) (speed²)
First object: KE = (1/2) (2 kg) (2m/s)² = 4 joules during the lift.
Second object: KE = (1/2) (4kg) (3 m/s)² = 18 joules during the lift.
The second object has more kinetic energy while it's being lifted
than the first object has while it's being lifted. Once they reach their
final heights and stop, neither object has any kinetic energy.