Explanation:
The given data is as follows.
= 0, = 0, = 0
= 17.0 m, = 2.10 sec
As the force P is constant and the mass "m" of the tool is constant then it means that the acceleration "a" will also be constant.
Now,
=
17.0 = 2.205a
a =
Also, we know that
F =
m =
So, m =
= 1.66 kg
Since, the tool is subject to its weight W and is in free fall. Hence,
10.0 m =
g = 2.411
Hence, weight of tool in Newtonia is as follows.
W = mg
=
= 4.00 N
Hence, weight of the tool on Newtonia is 4.00 N.
And, weight of the tool on the Earth is as follows.
W =
= 23.62 N
Hence, weight of the tool on Earth is 23.62 N.
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
Answer:
Explanation:
work done=force * displacement
=500 N * 40 m
=2000 joule