Answer:
0.336 rad/s
Explanation:
= Angular speed of the turntable = -0.2 rad/s
R = Radius of turntable = 2.9 m
I = Moment of inertia of turntable = 
M = Mass of turn table = 53 kg
= Magnitude of the runner's velocity relative to the earth = 3.6 m/s
As the momentum in the system is conserved we have
The angular velocity of the system if the runner comes to rest relative to the turntable which is the required answer is 0.336 rad/s
<span>373.2 km
The formula for velocity at any point within an orbit is
v = sqrt(mu(2/r - 1/a))
where
v = velocity
mu = standard gravitational parameter (GM)
r = radius satellite currently at
a = semi-major axis
Since the orbit is assumed to be circular, the equation is simplified to
v = sqrt(mu/r)
The value of mu for earth is
3.986004419 Ă— 10^14 m^3/s^2
Now we need to figure out how many seconds one orbit of the space station takes. So
86400 / 15.65 = 5520.767 seconds
And the distance the space station travels is 2 pi r, and since velocity is distance divided by time, we get the following as the station's velocity
2 pi r / 5520.767
Finally, combining all that gets us the following equality
v = 2 pi r / 5520.767
v = sqrt(mu/r)
mu = 3.986004419 Ă— 10^14 m^3/s^2
2 pi r / 5520.767 s = sqrt(3.986004419 * 10^14 m^3/s^2 / r)
Square both sides
1.29527 * 10^-6 r^2 s^2 = 3.986004419 * 10^14 m^3/s^2 / r
Multiply both sides by r
1.29527 * 10^-6 r^3 s^2 = 3.986004419 * 10^14 m^3/s^2
Divide both sides by 1.29527 * 10^-6 s^2
r^3 = 3.0773498781296 * 10^20 m^3
Take the cube root of both sides
r = 6751375.945 m
Since we actually want how far from the surface of the earth the space station is, we now subtract the radius of the earth from the radius of the orbit. For this problem, I'll be using the equatorial radius. So
6751375.945 m - 6378137.0 m = 373238.945 m
Converting to kilometers and rounding to 4 significant figures gives
373.2 km</span>
I don't know eithier sorry im only in the 5th and i have the same work as you
Answer:
1.42 s
Explanation:
The equation for free fall of an object starting from rest is generally written as
where
s is the vertical distance covered
a is the acceleration due to gravity
t is the time
On this celestial body, the equation is
this means that
so the acceleration of gravity on the body is
The velocity of an object in free fall starting from rest is given by
In this case,
g = 20.08 m/s^2
So the time taken to reach a velocity of
v = 28.6 m/s
is
Answer:
The strength coefficient and strain-hardening exponent in the flow curve equation are 92038 and 0.3608.
Explanation:
Given that,
True stress = 37000
True strain = 0.08
Later,
True stress = 55000
True strain = 0.24
Using formula of true stress
We need to make a two equation of true stress
Using formula of true stress
....(I)
....(I)
Dividing equation (II) and (I)
Taking ln in both side
We need to calculate the value of strength coefficient
Put the value of n in equation (I)
Hence, The strength coefficient and strain-hardening exponent in the flow curve equation are 92038 and 0.3608.
