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.
Daniddmelo says it right there, don't know why he got reported.
The potential energy (PE) is mass x height x gravity. So it would be 25 kg x 4 m x 9.8 = 980 joules. The child starts out with 980 joules of potential energy. The kinetic energy (KE) is (1/2) x mass x velocity squared. KE = (1/2) x 25 kg x 5 m/s2 = 312.5 joules. So he ends with 312.5 joules of kinetic energy. The Energy lost to friction = PE - KE. 980- 312.5 = 667.5 joules of energy lost to friction.
Please don't just copy and paste, and thank you Dan cause you practically did it I just... elaborated more? I dunno.
The magnetic field direction and direction of induced current in a wire are related by the right hand grip rule. Since the magnetic field was upwards, the thumb points upwards and the fingers curl around it. When viewed from above, it is seen as a current flowing in the counter clockwise direction.
Answer:
W = F * s
Work done equals applied force * distance traveled
Apparent weight = M g (1 - sin θ) since some of applied force will lighten sled
μ = coefficient of kinetic friction
F cos θ = force applied to motion of sled
s = distance traveled
[μ M g (1 - sin θ)] cos θ * s = work done in moving sled
Note that F = μ M g if applied force is in the horizontal direction