The Hubble Space Telescope is a joint ESA/NASA project and was launched in 1990 by the Space Shuttle mission STS-31 into a low-Earth orbit 569 km above the ground. During its lifetime Hubble has become one of the most important science projects ever. Hope this helps! ~ Autumn :)
Explanation:
Given that,
Initial velocity, u = 11.3 m/s
Angle above the horizontal, 
Time of flight :

Horizontal distance traveled is given by :
x = ut
x = 11.3 m/s × 1.32 s
x = 14.916 m
Maximum height is given by :

Hence, time of flight is 1.32 s, horizontal distance is 14.916 m and maximum height is 2.14 m.
Answer:
It will travel 350 meters each second.
Explanation:
The unit rate, 350 m/s, tells us that the jet will travel 350 meters per every second elapsed.
I think the answer is "<span>The ball that went out of the park shows more work because the distance was greater."</span>
Answer:
11.07Hz
Explanation:
Check the attachment for diagram of the standing wave in question.
Formula for calculating the fundamental frequency Fo in strings is V/2L where;
V is the velocity of the wave in string
L is the length of the string which is expressed as a function of its wavelength.
The wavelength of the string given is 1.5λ(one loop is equivalent to 0.5 wavelength)
Therefore L = 1.5λ
If L = 3.0m
1.5λ = 3.0m
λ = 3/1.5
λ = 2m
Also;
V = √T/m where;
T is the tension = 0.98N
m is the mass per unit length = 2.0g = 0.002kg
V = √0.98/0.002
V = √490
V = 22.14m/s
Fo = V/2L (for string)
Fo = 22.14/2(3)
Fo = 22.14/6
Fo = 3.69Hz
Harmonics are multiple integrals of the fundamental frequency. The string in question resonates in 2nd harmonics F2 = 3Fo
Frequency of the wave = 3×3.69
Frequency of the wave = 11.07Hz