Explanation:
Given that,
Radius of circular path, r = 5 m
Centripetal acceleration, 
(a) Let v is the astronaut’s speed. The formula for the centripetal acceleration is given by :



v = 18.5 m/s
(b) Let T denotes the time period. It is given by :


T = 1.69 s
Let N is the number of revolutions. So,

So, the number of revolutions per minute is 35.5
(c) T = 1.69 seconds
Hence, this is the required solution.
You will use the height of the bridge from the ground.
Solution:
Formula to be used is y=Viy(t)+g(t^2)/2
Where:
Vi=initial velocity which is 0 m/s
y=10 m
Gravitational acceleration or g =9.8m/s^2
T= time you need
Substitute all the given to the formula
10m=(0m/s)(t)+(9.8m/s^2)(t^2)/2
10mx2=9.8m/s^2(t^2)
Now isolate the variable you want to find which is T or time
10mx2/9.8m/s^2=t^2
20m/9.8m/s^2=t^2
Square root of 2.04= square root of t^2
T=1.43 secs
The answer is 1.43 seconds
Most directly on equador and least directly at poluses
Answer:
9. A 1500kg car traveling +6m/s with a 2000kg truck at rest. The vehicles collide, but do not stick together. The car has a velocity -3m/s after the collision. What is the velocity of the truck? a. What type of collision occurred above?
Answer:
28,400 N
Explanation:
Let's start by calculating the pressure that acts on the upper surface of the hatch. It is given by the sum of the atmospheric pressure and the pressure due to the columb of water, which is given by Stevin's law:

On the lower part of the hatch, there is a pressure equal to

So, the net pressure acting on the hatch is

which acts from above.
The area of the hatch is given by:

So, the force needed to open the hatch from the inside is equal to the pressure multiplied by the area of the hatch:
