the outermost layer of Earth’s lithosphere that
is found under the oceans and
molded at scattering
centres ono
ceanic ridges, which occur at deviating plate boundaries
Oceanic crust is about 6 km (4 miles) thick.
hope it helps
Answer:
a) 
b) 
c) 
Explanation:
From the exercise we know the initial velocity of the projectile and its initial height

To find what time does it take to reach maximum height we need to find how high will it go
b) We can calculate its initial height using the following formula
Knowing that its velocity is zero at its maximum height



So, the projectile goes 1024 ft high
a) From the equation of height we calculate how long does it take to reach maximum point



Solving the quadratic equation



So, the projectile reach maximum point at t=2s
c) We can calculate the final velocity by using the following formula:


Since the projectile is going down the velocity at the instant it reaches the ground is:

Answer:
38.64 feet
Explanation:
x=x0 + vx0t + 1/2axt2
x= 0 + 0 + 1/2 X 32.17 ft/sec2 X 1.55 sec2
x = 38.64 feet
Answer:
https://young.scot/get-informed/national/gender-identity-terms
Explanation:
The gap between the two flasks is partially evacuated of air creating a near vacuum which significantly reduces heat transfer by conduction or convection