Answer:
31.25 m
25m/sec
Explanation:
Given :-
Time = 5sec
V = 0 (in going up)
U = 0 (in comming down)
Find :-
H and U by which it is thrown up
Since the total time is 5 sec ,therefore half time will be taken to go up and another half will be taken to go down .
We know that ,
V = U + gt
0 = U - 10*2.5
U = 25 m/sec
Also,
V² = U² +2gs
0 = 625 - 20s
s = 625/20 = 31.25 m
Use the law of universal gravitation, which says the force of gravitation between two bodies of mass <em>m</em>₁ and <em>m</em>₂ a distance <em>r</em> apart is
<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²
where <em>G</em> = 6.67 x 10⁻¹¹ N m²/kg².
The Earth has a radius of about 6371 km = 6.371 x 10⁶ m (large enough for a pineapple on the surface of the earth to have an effective distance from the center of the Earth to be equal to this radius), and a mass of about 5.97 x 10²⁴ kg, so the force of gravitation between the pineapple and the Earth is
<em>F</em> = (6.67 x 10⁻¹¹ N m²/kg²) (1 kg) (5.97 x 10²⁴ kg) / (6.371 x 10⁶ m)²
<em>F</em> ≈ 9.81 N
Notice that this is roughly equal to the weight of the pineapple on Earth, (1 kg)<em>g</em>, where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity, so that [force of gravity] = [weight] on any given planet.
This means that on this new planet with twice the radius of Earth, the pineapple would have a weight of
<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / (2<em>r</em>)² = 1/4 <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²
i.e. 1/4 of the weight on Earth, which would be about 2.45 N.
Answer:
weightlessness, condition experienced while in free-fall, in which the effect of gravity is canceled by the inertial (e.g., centrifugal) force resulting from orbital flight. ... Excluding spaceflight, true weightlessness can be experienced only briefly, as in an airplane following a ballistic (i.e., parabolic) path.
Answer:
I think it's a because it goes thru it and reflects
-- Bob covered a distance of (32m + 45m) = 77 meters.
-- His displacement is the straight-line distance and direction
from his starting point to his ending point.
The straight-line distance is
D = √(32² + 45²)
D = √(1,024 + 2,025)
D = √3,049 = 55.22 meters
The direction is the angle whose tangent is (32/45) south of east.
tan⁻¹(32/45) = tan⁻¹(0.7111...) = 35.42° south of east.
