Answer:
Tarzan will be moving at 7.4 m/s.
Explanation:
From the question given above, the following data were obtained:
Height (h) of cliff = 2.8 m
Initial velocity (u) = 0 m/s
Final velocity (v) =?
NOTE: Acceleration due to gravity (g) = 9.8 m/s²
Finally, we shall determine how fast (i.e final velocity) Tarzan will be moving at the bottom. This can be obtained as follow:
v² = u² + 2gh
v² = 0² + (2 × 9.8 × 2.8)
v² = 0 + 54.88
v² = 54.88
Take the square root of both side
v = √54.88
v = 7.4 m/s
Therefore, Tarzan will be moving at 7.4 m/s at the bottom.
The correct answer would be True!
1) n=1 -> n=2 : delta E = -5+11 = 6. The answer is D
<span>2) n=1 -> n=3 : delta E = -2+11 = 9. The answer is B </span>
<span>3) n=1 -> n=4 : delta E = -1+11 = 10. No solution available </span>
<span>4) n=1 -> infinity delta E = 11. The answer is A </span>
<span>5) not absorbed would be C, as there is no transition with delta E of 8. </span>
F=mg=Gm1m2/r^2
g=Gm2/r^2
g=2Gm2/(2r)^2=2Gm2/4r^2=Gm2/2r^2
So since there is half times the gravity on this unknown planet that has twice earth's mass and twice it's radius, then the person can jump twice as high. 1.5*2= 3m high