Answer:
a) v = 7.69 10³ m / s, b) T = 92.6 min
Explanation:
a) For this exercise we use the centripetal acceleration ratio, which in itself assumes a circular orbit, is equal to the acceleration of gravity
a = v² / r
v =
the distance to the ISS is
r = R_earth + d
r = 6400 10³ + 400 10³
r = 6800 10³ m
we calculate
v =
Ra (8.69 6800 103)
v =
v = 7.687 10³ m / s
the result with the correct significant figures
v = 7.69 10³ m / s
b) The speed of the ISS is constant, so we can use the uniform motion relationships
v = d / t
if distance is the orbit distance
d = 2π r
time is called period
v = 2π r / T
T = 2π r / v
let's calculate
T = 2π 6800 10³ /7,687 10³
T = 5.558 10³ s
let's reduce the period to minutes
T = 5.558 10³ s (1 min / 60s)
T = 9.26 10¹ min
T = 92.6 min
Answer:
The pulling force that the man should apply to create an upward acceleration of
is 
Explanation:
Hi
As it shows in the drawing at the end, we have that the total mass of the man plus the platform is
, then the force downward
is
.
Due the man needs to do a pulling force upward capable of lifting himself and the platform with an acceleration of
, this force should create an acceleration greater than gravity by
. then
. Therefore the force should be
.
Finally, as we have a pulley arrangement connected to the platform, it gives the man a mechanical advantage, so he has to do only half of that upward force
, therefore
.
Answer:
Hertz is the answer for this question
Answer:
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Answer:
Knowing we only have one load applied in just one direction we have to use the Hooke's law for one dimension
ex = бx/E
бx = Fx/A = Fx/π
Using both equation and solving for the modulus of elasticity E
E = бx/ex = Fx / π
ex
E = 
Apply the Hooke's law for either y or z direction (circle will change in every direction) we can find the change in radius
ey =
(бy - v (бx + бz)) =
бx
=
= 
Finally
ey = Δr / r
Δr = ey * r = 10 * 
Δd = 2Δr = 
Explanation: