-- The overall <em>distance</em> he travels is (100m + 30m + 70m) = <em>200 meters</em>.
-- His <em>displacement </em>when he arrives at his front door is
D = (100m East) + (30m West) + (70m East)
D = (100m + 70m)East + (30m)West
D = (170m East) + (30m West)
<em>D = 140 meters East </em>
It's interesting to notice that his displacement is 60 meters shorter than the distance he walked.
That's because there's a stretch of 30 meters somewhere in the middle that he actually covered <em>three times</em>.
Two of those times added to the distance his shoes covered (2x30m=60m), but they cancelled out of the displacement.
His front door is 140 meters East of school. He walked 60m farther than that, going back and forth over the 30m piece.
Three quarters remain un-decayed
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:upper right hand side i think
Explanation:
The advertisement is deciribing the car’s acceleration, as it gains speed when moving from 0-60.
