Answer:
Total time elapsed between the start and when he returns with the ball is 7.5s
Explanation:
From the question,
- The dog starts at the origin and runs forward at 6m/s for 1.5s. First, we will determine the distance covered while running forward.
From
Speed = Distance / Time
Distance = Speed × Time
Speed = 6m/s
Time = 1.5s
∴ Distance = 6m/s × 1.5s
Distance = 9m
That is, the dog covered a distance of 9m while running forward.
- The dog turns around and runs backward at 7m/s for 3s. Now, we will also determine the distance the dog covered backwards.
Distance = Speed × Time
Speed = 7m/s
Time = 3s
Distance = 7m/s × 3s
Distance = 21m
The dog's displacement from the origin is 21m - 9m = 12m
Now, to calculate how much time has elapsed between the start if the dog runs back to the origin at 4m/s, we will first determine the time the dog spent back to the origin and then add to the time spent for the first two distances.
To get back to the origin, the dog needs to cover 12m
From
Speed = Distance / Time
Time = Distance / Speed
Distance = 12m
Speed = 4m/s
∴ Time = (12m) / (4m/s)
Time = 3s
Therefore, the dog spent 3s to run back to the origin.
Hence, total time elapsed = 1.5s + 3s + 3s
Total time elapsed = 7.5s
Answer:
ΔT = 0.02412 s
Explanation:
We will simply calculate the time for both the waves to travel through rail distance.
FOR THE TRAVELING THROUGH RAIL:
FOR THE WAVE TRAVELING THROUGH AIR:
The separation in time between two pulses can now be given as follows:
<u>ΔT = 0.02412 s</u>
Distance= speed (multiplied by) time
Answer:
Positive
Explanation:
An ideal fluid is on which is incompressible and non-viscous.
In an ideal fluid element when a fluid element moves from a region of high pressure to region of low pressure then the work done on the fluid element is positive because the force acts in the direction from high pressure towards low pressure and displaces the fluid element, so the work-done is positive.
Answer:
It only depends on the vertical component
Explanation:
Hello!
The horizontal component will tell you how much you travel in that direction.
You could have a large horizontal velocity, but if the vertical velocity is zero, you will never be out of the ground. Similarly, you could have a zero horizontal velocity, but if you have a non-zero vertical velocity you will be some time off the ground. This time can be calculated by two means, one is using the equation of motion (position as a function of time) and the other using the velocity as a fucntion of time.
For the former you must find the time when the position is zero.
Lets consider the origin of teh coordinate system at your feet
y(t) = vt - (1/2)gt^2
We are looking for a time t' for which y(t')=0
0 = vt' - (1/2)gt'^2
vt' = (1/2)gt'^2
The trivial solution is when t'=0 which is the initial position, however we are looking for t'≠0, therefore we can divide teh last equation by t'
v = (1/2)gt'
Solving for t'
t' = (2v/g)
