Answer:
1.08 s
Explanation:
From the question given above, the following data were obtained:
Height (h) reached = 1.45 m
Time of flight (T) =?
Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:
Height (h) = 1.45 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
1.45 = ½ × 9.8 × t²
1.45 = 4.9 × t²
Divide both side by 4.9
t² = 1.45/4.9
Take the square root of both side
t = √(1.45/4.9)
t = 0.54 s
Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).
Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:
Time (t) taken to reach the height = 0.54 s
Time of flight (T) =?
T = 2t
T = 2 × 0.54
T = 1.08 s
Therefore, it will take the kangaroo 1.08 s to return to the earth.
Hello, love! The answer is True, or T, on Edge2020.
Hope this helped!
~ V.
Answer:
Measurements are an important part of comparing things, as they provide the basis on comparing objects to other objects. Measurements allow us to recognize three hours and see how it's shorter than five hours, without having to observe the hours passing by themselves.
Answer:
For Xenon fluoride, the average bond energy is 132kj/mol
For tetraflouride,the average bond energy is 150.5kj/mol.
For hexaflouride, the average bond energy is 146.5 kj/mol
Explanation:
For xenon fluoride
105/2 = 52.5
For F-F
159/2 = 79.5
Average bond energy of Xe-F = 79.5 + 52.5 = 132kj/mole
For tetraflouride
284/4 = 71
For F-F
159/2 = 79.5
Average bond energy = 79.5 + 71 = 150.5kj/mol
For hexaflouride
402/6 = 67
F-F = 159/2 = 79.5
Average bond energy = 67 + 79.5 = 146.5kj/ mol