Hi there!
We can begin by calculating the time taken to reach its highest point (when the vertical velocity = 0).
Remember to break the velocity into its vertical and horizontal components.
Thus:
0 = vi - at
0 = 16sin(33°) - 9.8(t)
9.8t = 16sin(33°)
t = .889 sec
Find the max height by plugging this time into the equation:
Δd = vit + 1/2at²
Δd = (16sin(33°))(.889) + 1/2(-9.8)(.889)²
Solve:
Δd = 7.747 - 3.873 = 3.8744 m
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:
Applied force, F = 18 N
Coefficient of static friction, μs = 0.4
Coefficient of kinetic friction, μs = 0.3
θ = 27°
Let N be the normal reaction of the wall acting on the block and m be the mass of block.
Resolve the components of force F.
As the block is in the horizontal equilibrium, so
F Cos 27° = N
N = 18 Cos 27° = 16.04 N
As the block does not slide so it means that the syatic friction force acting on the block balances the downwards forces acting on the block .
The force of static friction is μs x N = 0.4 x 16.04 = 6.42 N .... (1)
The vertically downward force acting on the block is mg - F Sin 27°
= mg - 18 Sin 27° = mg - 8.172 ... (2)
Now by equating the forces from equation (1) and (2), we get
mg - 8.172 = 6.42
mg = 14.592
m x 9.8 = 14.592
m = 1.49 kg
Thus, the mass of block is 1.5 kg.
Think of it like this, gravity has to pull harder on the heavier object to make them fall at the same rate , but doesn't have to pull as hard for the lighter object , thus is why sometimes heavier objects fall faster then lighter ones