Imagine you are in a swimming pool 30m deep. Assuming you know that water is denser than air, you would know that the 30m of water above you will carry more weight, and press down on your body. Say you were in a swimming pool 60m deep, you would be sandwiched between 30m of water pressing down on you, and the upthrust created by the 30m of water below you.
In a building 30m up, the pressure will be regulated, as you are in a building. The floor will be strong enough to support the weight of the body, and the body will not recoil into itself.
Answer:19.32 m/s
Explanation:
Given
initial speed of car(u)=4.92 m/s
acceleration(a)=
Speed of car after 4.5 s
using equation of motion
v=u+at
v=19.32 m/s
Displacement of the car after 4.5 s
s=54.54 m
Explanation:
It is given that, the position of a particle as as function of time t is given by :
Let v is the velocity of the particle. Velocity of an object is given by :
![v=\dfrac{d[(8t+9)i+(2t^2-8)j+6tk]}{dt}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7Bd%5B%288t%2B9%29i%2B%282t%5E2-8%29j%2B6tk%5D%7D%7Bdt%7D)
So, the above equation is the velocity vector.
Let a is the acceleration of the particle. Acceleration of an object is given by :
![a=\dfrac{d[8i+4tj+6k]}{dt}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bd%5B8i%2B4tj%2B6k%5D%7D%7Bdt%7D)
At t = 0, 
Hence, this is the required solution.
The hypothesis because its very hard to make and it confounds me
Answer:
2.32 s
Explanation:
Using the equation of motion,
s = ut+g't²/2............................ Equation 1
Where s = distance, u = initial velocity, g' = acceleration due to gravity of the moon, t = time.
Note: Since Onur drops the basket ball from a height, u = 0 m/s
Then,
s = g't²/2
make t the subject of the equation,
t = √(2s/g')...................... Equation 2
Given: s = 10 m, g' = 3.7 m/s²
Substitute this value into equation 2
t = √(2×10/3.7)
t = √(20/3.7)
t = √(5.405)
t = 2.32 s.
