Answer:
Straight line parallel to time axis.
Explanation:
The slope of the position time graph gives the velocity.
As the man is still, that means the velocity is zero. So, the slope of the graph is zero. It is a straight line parallel to time axis.
Answer:
a. The station is rotating at
b. the rotation needed is
Explanation:
We know that the centripetal acceleration is
where is the rotational speed and r is the radius. As the centripetal acceleration is feel like an centrifugal acceleration in the rotating frame of reference (be careful, as the rotating frame of reference is <u>NOT INERTIAL,</u> the centrifugal force is a fictitious force, the real force is the centripetal).
<h3>a. </h3>
The rotational speed is :
Knowing that there are in a revolution and 60 seconds in a minute.
<h3>b. </h3>
The rotational speed needed is :
Knowing that there are in a revolution and 60 seconds in a minute.
Answer:
Explanation:
To solve this, we start by using one of the equations of motion. The very first one, in fact
1
V = U + at.
V = 0 + 0.8 * 3.4 = 2.72 m/s.
2.
V = 0 + 0.8 * 4.3 = 3.44 m/s.
3.
d = ½ * 0.8 * 4.3² + 3.44 * 12.9
d = 7.396 + 44.376
d = 51.77 m.
4.
d = 62 - 51.77 = 10.23 m. = Distance
traveled during deceleration.
a = (V² - Vo²) / 2d.
a = (0² - 3.44²) / 20.46
a = -11.8336 / 20.46 = -0.58 m/s²
5.
t = (V - Vo)/a =(0 - 3.44) / -0.58
t = -3.44/-.58 = 5.93 s
= Stop time.
T = 4.3 + 12.9 + 5.93 = 23.13 s. = Total
time the hare was moving.
6.
d = Vo * t + ½ * a * t² = 62 m.
0 + 0.5 * (23.13)² * a = 61
267.5a = 61
a = 61/267.5
a = 0.23 m/s²
S = ut + 0.5at^2
<span>10 = 0 + 0.5(9.81)t^2 {and if g = 10 then t^2 = 2 so t ~1.414} </span>
<span>t^2 ~ 2.04 </span>
<span>t ~ 1.43 seconds</span>
Answer:
Change in velocity, change in direction, change in both velocity and direction
Explanation: