Answer:
Hubble measured the velocity of the movement of galaxies by using Hubble's law states that galaxies located farthest from the center of the universe than those closest to the center.
Explanation:
Hubble's Law says that an object's velocity away from an observer is directly proportional to its distance from the observer. In other words, the farther away something is the faster it is moving away from us. The spectrum of a galaxy allows you to measure its redshift.
<u>Answer:</u>
The velocity of the truck is 34 m/s.
<u>Explanation:</u>
The momentum (p) of an object is the product of its mass (m) and its velocity (v) which can be written as:
<em>p = mv</em>
Here in this problem, we are given a truck which has a momentum of 40120 kg and a mass of 1180 kg and we are to find its velocity.
So substituting the given values in the above formula to find the velocity of the truck.
Truck's velocity = 
= 34 m/s.
Average speed = (total distance) / (total time)
Total distance = 59 km + 74 km = 133 km
Total time = 3 hrs + 5 hrs = 8 hrs
Average speed = (133 km) / (8 hrs) = 16.625 km/hr
Answer:
a) The maximum height the ball will achieve above the launch point is 0.2 m.
b) The minimum velocity with which the ball must be launched is 4.43 m/s or 0.174 in/ms.
Explanation:
a)
For the height reached, we use 3rd equation of motion:
2gh = Vf² - Vo²
Here,
Vo = 3.75 m/s
Vf = 0m/s, since ball stops at the highest point
g = -9.8 m/s² (negative sign for upward motion)
h = maximum height reached by ball
therefore, eqn becomes:
2(-9.8m/s²)(h) = (0 m/s)² - (3.75 m/s²)²
<u>h = 0.2 m</u>
b)
To find out the initial speed to reach the hoop at height of 3.5 m, we again use 3rd eqn. of motion with h= 3.5 m - 2.5m = 1 m (taking launch point as reference), and Vo as unknown:
2(-9.8m/s²)(1 m) = (0 m/s)² - (Vo)²
(Vo)² = 19.6 m²/s²
Vo = √19.6 m²/s²
<u>Vo = 4.43 m/s</u>
Vo = (4.43 m/s)(1 s/1000 ms)(39.37 in/1 m)
<u>Vo = 0.174 in/ms</u>
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