Based on the equation KE = 1/2(m)(v^2), Kinetic Energy can be measured based on velocity. If an object has a large velocity, it have a larger kinetic energy than if the velocity is small.
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This question is incomplete because the options are missing; here is the complete question:
A runner starts at point A, runs around a 1-mile track, and finishes the run back at point A. Which of the following statements is true?
A. The runner's displacement is 1 mile.
B. The runner's displacement is zero.
C. The distance the runner covered is zero.
D. The runner's speed was zero.
The answer to this question is B. The runner's displacement is zero
Explanation:
Displacement always implies a change of position; this means an object or individual moves from point A to point B, and therefore the original position is different from the final position. Additionally, in displacement, other related factors such as the total distance the body moved and the direction of movement. In the case presented, it can be concluded there was no displacement or the displacement is zero because even when the runner moved and ran two miles, he returned to the initial position, and without a change in the position, there is no displacement.
Answer:
The average speed for the whole journey is 49.5 miles per hour.
Explanation:
Step 1 :
Here, both the ways, he covers the same distance. Then, the formula to find average speed is
= 2xy / (x+y)
Step 2 :
x ----> Rate at which he travels from New York to Washington
x = 45
y ----> Rate at which he travels from New York to Washington
y = 55
Step 3 :
So, the average speed is
= (2 ⋅ 45 ⋅ 55) / (45 + 55)
= 4950 / 100
= 49.5
Answer:
second is the SI unit of time
Answer:
a) Diffusion coefficient, D = 1.5 in/hr
b) Mean jump frequency, f = 0.0833 Hz
Explanation:
a) The relationship between the diffusion coefficient, time and mean displacement and can be given by the expression:
..........(1)
Where <r> = mean displacement
D = Diffusion coefficient
t = time = 12 hrs
sum of the squares of the distance divided by 100 is 36 in2.
<r>²= 36 in²
Substituting these values into equation (1) above
b) Mean jumping distance, <r> = 0.1 inches
Applying equation (1) again
Where D = 1.5 in/hr
The mean jump frequency, f = 1/t
f = 1/12
f = 0.0833 Hz
