1110 atm
Let's start by calculating how many cm deep is 36,000 feet.
36000 ft * 12 in/ft * 2.54 cm/in = 1097280 cm
Now calculate how much a column of water 1 cm square and that tall would mass.
1097280 cm * 1.04 g/cm^3 = 1141171.2 g/cm^2
We now have a number using g/cm^2 as it's unit and we desire a unit of Pascals ( kg/(m*s^2) ).
It's pretty obvious how to convert from g to kg. But going from cm^2 to m is problematical. Additionally, the s^2 value is also a problem since nothing in the value has seconds as an unit. This indicates that a value has been omitted. We need something with a s^2 term and an additional length term. And what pops into mind is gravitational acceleration which is m/s^2. So let's multiply that in after getting that cm^2 term into m^2 and the g term into kg.
1141171.2 g/cm^2 / 1000 g/kg * 100 cm/m * 100 cm/m = 11411712 kg/m^2
11411712 kg/m^2 * 9.8 m/s^2 = 111834777.6 kg/(m*s^2) = 111834777.6 Pascals
Now to convert to atm
111834777.6 Pa / 1.01x10^5 Pa/atm = 1107.2750 atm
Now we gotta add in the 1 atm that the atmosphere actually provides (but if you look closely, you'll realize that it won't affect the final result).
1107.274 atm + 1 atm = 1108.274 atm
And finally, round to 3 significant figures since that's the accuracy of our data, giving 1110 atm.
Answer:
6 m/s²
Explanation:
From the question given above, the following data were obtained:
Velocity (v) = 30 m/s
Time (t) = 5 s
Acceleration (a) =..?
Acceleration is defined mathematically as:
Acceleration (a) = Velocity (v) /time (t)
a = v /t
With the above formula, we can obtain the acceleration of the object as follow:
Velocity (v) = 30 m/s
Time (t) = 5 s
Acceleration (a) =..?
a= v/t
a= 30/5
a = 6 m/s²
Therefore, the acceleration of the object is 6 m/s² due East.
Answer: Electron flow / Electric current
Formula:
F = ma
F: force (N) m: mass (kg) a: acceleration (m/s^2)
Solution:
F = ma
F = 1200 × 1.89
= 2268N
