7.According to the periodic table, how many total particles are there in the nucleus of a Titanium, Ti, atom?
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Answer:
The journey started from point A, with a speed of 1 m/s for 10 seconds after which the it became stationary at 10 meters from the start point, for 10 seconds. The journey continued at a higher speed of 2 m/s for 5 seconds and halted (became stationary) again at 20 meters which was the maximum distance reached from the start point
The return journey to the start point started at the 20 meter mark and lasted for 5 seconds, at a speed of 4 m/s
The total distance travelled during the journey from the start point A to the final point B is 40 m
Explanation:
From the start point A to point B, we have;
The speed from A to B = 10 m/(10 s) = 1 m/s
The distance traveled from A to B = 10 m
The time it takes to move from A to B = 10 seconds
From the point B to point C, we have;
The distance traveled from B to C = 0 m, (stationary)
The time it remains at point B distance from the start point = 10 seconds
The speed between point B to C = 0 m/(10 s) = 0 m/s
From the point C to point D, we have;
The distance traveled from C to D = 10 m
The time it takes to move from C to D = 5 seconds
The speed between point C and D = 10 m/(5 s) = 2 m/s
From the point D to point E, we have;
The distance traveled from D to E = 0 m, (stationary)
The time it remains at point D distance from the start point = 10 seconds
The speed between point D to E = 0 m/(10 s) = 0 m/s
From the point E to point F, we have;
The distance traveled from E to F = 20 m (return journey starts at point E)
The time it takes to move from E to F = 5 seconds
The speed between point E to F = 20 m/(5 s) = 4 m/s (Return journey)
Therefore, the journey started from point A, with a speed of 1 m/s for 10 seconds after which the it became stationary at 10 meters from the start point, for 10 seconds. The journey continued at a higher speed of 2 m/s for 5 seconds and halted (became stationary) again at 20 meters which was the maximum distance reached from the start point
The return journey to the start point started at the 20 meter mark and lasted for 5 seconds, at a speed of 4 m/s
The total distance moved, 'd', to and from the start point with reference to the graph is given as follows;
d = (From A to B) 10 m + (From B to C) 0 m + (From C to D) 10 m + (From D to E) 0 m + (From E to F) 20 m = 40 m
The total distance travelled in the journey is 40 m
The total displacement, = 10 m + 10 m - 20 m = 0 m