Answer:
a = 0.50 m/s²
v = 6.0 m/s
Explanation:
Step 1: Given data
- Initial velocity (u): 0 m/s (rest)
- Displacement of the marble (s): 36 m
Step 2: Calculate the acceleration
We will use the following suvat equation.
s = u × t + 1/2 × a × t²
s = 0 × 12.0 s + 1/2 × a × (12.0 s)²
36 m = a × 72.0 s²
a = 0.50 m/s²
Step 3: Calculate the final velocity
We will use the following suvat equation.
v = u + a × t
v = 0 + 0.50 m/s² × 12.0 s
v = 6.0 m/s
From a to b speed is 600+40 = 640
from b to a speed is 600-40 = 560
let t be the number of hours of flight. This would mean it would have traveled a distance of 640 miles and the distance yet to travel is 2400-640t
Time left will be (2400-640t)/640. But if they were to return to a it would fly 640t miles at 560mph which will take (640t/560) hrs
(2400-640t) / 640 = 640t / 560
560(2400 - 640t) = 640t x 640
t = 1.75hrs
Before you even look at the questions, look over the graph, so you know what kind of information is there.
The x-axis is "time". OK. You know that as the graph moves from left to right, it shows what's happening as time goes on.
The y-axis is "speed" of something. OK. When the graph is high, the thing is moving fast. When the graph is low, the thing is moving slow. When the graph slopes up, the thing is gaining speed. When the graph slopes down, the thing is slowing down. When the graph is flat, the speed isn't changing, so the thing is moving at a constant speed.
NOW you can look at the questions.
OMG ! It's only ONE question: What's happening from 'c' to 'd' ? Well I don't know. Perhaps we can figure it out if we LOOK AT THE GRAPH !
-- Between c and d, the graph is flat. The speed is not changing. It's the same speed at d as it was back at c .
What speed is it ?
-- Look back at the y-axis. The speed at the height of c and d is 'zero' .
-- The 2nd and 4th choices are both correct. From c to d, <em>the speed is constant</em>. The constant speed is zero. <em>The car is not moving</em>.
Answer:
0.208 N
Explanation:
We are given that


Distance,d=0.41 m
The magnitude of the net electrostatic force experienced by any charge at point 4
Net force,






Where 

