Answer:
45 s .
Explanation:
The accelerator will first accelerate , then move with uniform velocity and at last it will decelerate to rest .
displacement s = ?
acceleration a = 1 m /s²
Final speed v = 5 m/s
initial speed u = 0
v² = u² + 2as
5² = 0 + 2 x 1 x s
s = 12.5 m
B) Let time of acceleration or deceleration be t
v = u + a t
5 = 0 + 1 t
t = 5 s
Similarly displacement during deceleration = 12.5 m
Total distance during uniform motion = 200 - ( 12.5 + 12.5 ) = 175 m .
velocity of uniform motion = 5 m /s
time during which there was uniform velocity = 175 / 5 = 35 s
Total time = 5 + 35 + 5 = 45 s .
Answer:
I believe D
Explanation:
You need to have a more accurate reading and you want to test it multiple times throughout the week though to get a base resting rate.
I hope this is correct good luck!
This problems a perfect application for this acceleration formula:
Distance = (1/2) (acceleration) (time)² .
During the speeding-up half: 1,600 meters = (1/2) (1.3 m/s²) T²
During the slowing-down half: 1,600 meters = (1/2) (1.3 m/s²) T²
Pick either half, and divide each side by 0.65 m/s²:
T² = (1600 m) / (0.65 m/s²)
T = square root of (1600 / 0.65) seconds
Time for the total trip between the stations is double that time.
T = 2 √(1600/0.65) = <em>99.2 seconds</em> (rounded)
Answer:
dont know because I am a student lol
Answer:
Each of the joints represents a degree of freedom in the manipulator system and allows translation and rotary motion :) Hope this helps
