If a box with a mass of 8.0 kg is sitting on a frictionless surface and experiences an acceleration of 4.0 m/s2 to the right, wh
1 answer:
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Answer:
Explanation:
Yes , the horizontal distance travelled by the ball will depend upon the height of release .
When a ball is thrown at some angle from a height , it has two components , the vertical component and horizontal component . The ball goes in horizontal direction due to its horizontal component . Its vertical component has no role to play . But the horizontal range covered by the body thrown
depends upon the duration of time in which it remains in air . The longer it remains in air , the greater distance it can cover horizontally .
Horizontal distance covered = t x horizontal velocity
If V be the velocity of throw and Vx be its horizontal component
Horizontal distance covered = t x Vx
Now t depends upon the height . If height rises , time of fall will increase so horizontal distance covered will increase .
If h be the height from which the body is thrown , Vy be the vertical upward component of initial velocity
from the relation
s = ut + 1/2 at²
h = - Vy t + 1/2 at²
As h increases , t will increase and therefore horizontal distance covered will increase. If the ball has only horizontal velocity initially , Vy = 0
h = 1/2 gt²
Horizontal distance covered = t x Vx
=
From this expression also
Horizontal distance covered is proportional to .
Answer:
(a) g = 8.82158145.
(b) 7699.990192m/s.
(c)5484.3301s = 1.5234 hours.(extremely fast).
Explanation:
(a) Strength of gravitational field 'g' by definition is
, here G is Gravitational Constant, and r is distance from center of earth, all the values will remain same except r which will be radius of earth + altitude at which ISS is in orbit.
r = 6721,000 meters, putting this value in above equation gives g = 8.82158145.
(b) We have to essentially calculate centripetal acceleration that equals new 'g'.
here g is known, r is known and v is unknown.
plugging in r and g in above and solving for unknown gives V = 7699.990192m/s.
(c) S = vT, here T is time period or time required to complete one full revolution.
S = earth's circumfrence , V is calculated in (B) T is unknown.
solving for unknown gives T = 5484.3301s = 1.5234hours.