1) 0.0011 rad/s
2) 7667 m/s
Explanation:
1)
The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:
where
is the angular displacement of the object
t is the time elapsed
is the angular velocity
In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is
rad
And the time taken is
Therefore, the angular velocity of the telescope is
2)
For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation
where
v is the linear velocity
is the angular velocity
r is the radius of the circular orbit
In this problem:
is the angular velocity of the Hubble telescope
The telescope is at an altitude of
h = 600 km
over the Earth's surface, which has a radius of
R = 6370 km
So the actual radius of the Hubble's orbit is
Therefore, the linear velocity of the telescope is:
Element 2 explanation just trust
Answer:
200 km\h
or 0.621 mp\h its the same speed just different mesuarements
Answer:
a) v2=4147.72 m/s
b) stotal=5.53x10^6 m
Explanation:
a) the length from the center of the earth is equal to:
L1=1x10^6+((6.37/2)x10^6)=4.18x10^6 m
the velocity is 5.14 km/s=5.14x10^3 m/s
the farthest distance is equal to:
L2=2x10^6+((6.37/2)x10^6)=5.18x10^6 m
As the angular momentum is conserved, we have to:
I1=I2
m*L1*v1=m*L2*V2, where m is the mass of satelite
clearing v2:
v2=(L1*V1)/L2=(4.18x10^6*5.14x10^3)/5.18x10^6=4147.72 m/s
b) Using the Newton 3rd law:
vf^2=vi^2+2as
where:
a=g=9.8 m/s^2
vf=0
vi=5.14 km/s
s=?
Clearing s:
s=(vf^2-vi^2)/(2g)=((0-(5.14x10^3)^2)/(2*9.8)=1.35x10^6 m
the total distance is equal to:
stotal=s+L1=1.35x10^6+4.18x10^6=5.53x10^6 m
