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:
Answer:
2(maximum), -2(minimum), -2(maximum).
Explanation:
V(t)= 2πcos πt--------------------------------------------------------------------------------(1).
Therefore, there is a need to integrate v(t) to get S(t).
S(t)= 2×sinπt + C ------------------------------------------------------------------------------(2).
Applying the condition given, we have s(0)= 0.
S(0)= 2sin ×π(0) + C.
Which means that; 0+C= 0. That is; C=0.
S(t)= 2 sin πt.
The mass moves to its highest positions at time,t=half(1/2=.5) and time,t=2.5.
Take note that; sin(π/2) = sin(5π/2) = 1 .
Also, the mass moves to its lowest position at time,t=(3/2); also, sin(3π/2) = -1.
Therefore, we have that 2 maximum; -2 minimum and -2 maximum.
The answer is 17 m because you have to add the 15 m and the 8 m together to get the answer so it will be like this 17x17 = 15x15 + 8x8 got it?
