Answer:
Explanation:
distance of shuttle from centre of the earth = radius of the orbit
= 6300 + 300 = 6600 km
= 6600 x 10³
Formula of time period of the satellite
T = 2π R /v₀ , v₀ is orbital velocity
v₀ = √gR , ( if height is small with respect to radius )
T = 2π R /√gR
= 2π√ R /√g
= 2 x 3.14 x √ 6600 x 10³ / √9.8
= 2 x 3.14 x 256.9 x 10 / 3.13
= 5154.41 s
= 5154.41 / 60 minutes
= 85.91 m
85.9 minutes.
2 ) No of sunrise per day = no of rotation per day
= 24 x 60 / 85.9
= 16.76
or 17 sunrises.
Answer:
the tension is 18513N
Explanation:
Given that
mass = 1683kg
acceleration = 1.2m/s^2
acceleration due to gravity = 9.8m/s^2
T-mg = ma
T = ma + mg
T = m(a +g)
T = 1683 kg(1.20 m/s2 + 9.8)
T = 1683 (11)
T = 18513N
therefore, the tension is 18513N
The force of gravity on earth is towards the center of it
In the downward direction
-- We know that the y-component of acceleration is the derivative of the
y-component of velocity.
-- We know that the y-component of velocity is the derivative of the
y-component of position.
-- We're given the y-component of position as a function of time.
So, finding the velocity and acceleration is simply a matter of differentiating
the position function ... twice.
Now, the position function may look big and ugly in the picture. But with the
exception of 't' , everything else in the formula is constants, so we don't even
need any fancy processes of differentiation. The toughest part of this is going
to be trying to write it out, given the text-formatting capabilities of the wonderful
envelope-pushing website we're working on here.
From the picture . . . . . y (t) = (1/2) (a₀ - g) t² - (a₀ / 30t₀⁴ ) t⁶
First derivative . . . y' (t) = (a₀ - g) t - 6 (a₀ / 30t₀⁴ ) t⁵ = (a₀ - g) t - (a₀ / 5t₀⁴ ) t⁵
There's your velocity . . . /\ .
Second derivative . . . y'' (t) = (a₀ - g) - 5 (a₀ / 5t₀⁴ ) t⁴ = (a₀ - g) - (a₀ /t₀⁴ ) t⁴
and there's your acceleration . . . /\ .
That's the one you're supposed to graph.
a₀ is the acceleration due to the model rocket engine thrust
combined with the mass of the model rocket
'g' is the acceleration of gravity ... 9.8 m/s² or 32.2 ft/sec²
t₀ is how long the model rocket engine burns
Pick, or look up, some reasonable figures for a₀ and t₀
and you're in business.
The big name in model rocketry is Estes. Their website will give you
all the real numbers for thrust and burn-time of their engines, if you
want to follow it that far.
