What is the reversal experienced by an image formed in a plane mirror called?
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The average force of the water droplets is the force given by the impact
per second of the droplets on the limestone floor.
- The average force exerted on the limestone floor is approximately <u>1.6013 × 10⁻² N</u>
Reasons:
The given parameters are;
Volume of a droplet = 10 ml = 1 × 10⁻⁵ m³
Height from which the water falls, <em>h </em>= 5 meters
Rate at which the water falls = 10 per minute
Required:
The average force exerted on the floor by the water droplets.
Solution:
According to Newton's Second Law of motion, we have;
Force = Rate of change of momentum
Momentum = Mass × Velocity
Mass of a droplet of water = Volume × Density
Density of water = 997 kg/m³
Mass of a droplet = 1 × 10⁻⁵ m³ × 997 kg/m³ = 0.00997 kg
The velocity just before the droplet reaches the ground, v = √(2·g·h)
Where;
g = Acceleration due to gravity ≈ 9.81 m/s²
Which gives;
v = √(2 × 9.81 m/s² × 5 m) ≈ 9.905 m/s
The rate of change in momentum per minute = 1
Therefore;
ΔMomentum = Mass × ΔVelocity
Considering the 10 drops per minute, we have;
ΔMomentum = 10 × 0.0097 kg × 9.905 m/s = 0.960785 kg·m/s
ΔTime = 1 minute = 60 seconds
Therefore;
- The average force exerted on the limestone floor by the droplets of water is
≈ <u>1.6013 × 10⁻² N</u>
Learn more about Newton's Second Law of motion and force exerted water here:
Answer:
The answer is 12.27 g (NH4)3PO4
Explanation:
Step 1: balance the chemical equation:
H3PO4 + 3NH3 → (NH4)3PO4
step 2: the molar masses of each of the reagents and products are calculated using the periodic table:
H3PO4:
3 atoms of H: 3x1=3 g/mol
1 atom of P: 1x30.97=30.97 g/mol
4 atoms of O: 4x16=64 g/mol
Molar mass=3+30.97+64=97.97 g/mol
3NH3:
3 atoms of N: 3x14=42 g/mol
9 atoms of H: 9x1=9 g/mol
Molar mass=42+9=51 g/mol
(NH4)3PO4:
3 atoms of N: 3x14=42 g/mol
12 atoms of H: 12x1=12 g/mol
1 atom of P: 1x30.97=30.97 g/mol
4 atoms of O: 4x16=64 g/mol
Molar mass=42+12+30.97+64=148.97 g/mol
we make a rule of three to calculate the amount of ammonium phosphate:
51 g NH3------------------148.97 g (NH4)3PO4
4.2 g NH3-----------------x g (NH4)3PO4
Clearing the x, we have: