Answer:
the mass of the air in the classroom = 2322 kg
Explanation:
given:
A classroom is about 3 meters high, 20 meters wide and 30 meters long.
If the density of air is 1.29 kg/m3
find:
what is the mass of the air in the classroom?
density = mass / volume
where mass (m) = 1.29 kg/m³
volume = 3m x 20m x 30m = 1800 m³
plugin values into the formula
1.29 kg/m³ = <u> mass </u>
1800 m³
mass = 1.29 kg/m³ ( 1800 m³ )
mass = 2322 kg
therefore,
the mass of the air in the classroom = 2322 kg
a) The wind is generated because there are different values of pressure in the amtophera. That is, it is generated due to a pressure difference between two atmospheric points. Generally the movement is performed when the air travels from the highest pressure point, to the lowest pressure point. This is also a direct cause of different types of wind speeds.
b) If the cloud moves from one direction to another, it will indicate that from the starting point the pressure is higher, and the point towards which it is directed, the pressure is lower. If we place this on a Cartesian plane with reference to the cardinal points, we can know the approximate place or area where the pressures are different.
Answer:
<h3>14.97m/s</h3>
Explanation:
Given
Initial velocity of the car u = 8m/s
Distance travelled by the rider S = 40m
Acceleration a = 2m/s²
Required
rider's velocity after the acceleration v
Using the equation of motion
v² = u²+2as
v² = 8²+2(2)(40)
v² = 64+160
v² = 224
v = √224
v = 14.97m/s
Hence the rider's velocity after the acceleration is 14.97m/s
I don’t think I’m right but I want to say 500 m/s
The x- and y-coordinates are 9142.57 m and -304.425 m
<u>Explanation:</u>
As the motion of the shell is in a plane (two dimensional space) and the acceleration is that due to gravity which is vertically downward, we resolve initial velocity of the shell 
in horizontal and vertical directions. If the initial velocity of the shell is making angle with the horizontal, the horizontal component of initial velocity will be
As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of 
and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by
For motion with constant acceleration, we know
Along the horizontal, x-axis, we might write this as
Measuring distances relative to the firing point means
we know that,
or,
By applying the values, we get,
The acceleration of gravity is vertically downward and is 
, hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion
we know, 
and 
, so,
y = 11701.8 - 4.9(2450.25)= 11701.8 - 12006.225 = - 304.425 m
