Answer:
total number of electron in 1 litter is 3.34 ×
electron
Explanation:
given data
mass per mole = 18 g/mol
no of electron = 10
to find out
how many electron in 1 liter of water
solution
we know molecules per gram mole is 6.02 ×
molecules
no of moles is 1
so
total number of electron in water is = no of electron ×molecules per gram mole × no of moles
total number of electron in water is = 10 × 6.02 ×
× 1
total number of electron in water is = 6.02×
electron
and
we know
mass = density × volume ..........1
here we know density of water is 1000 kg/m
and volume = 1 litter = 1 ×
m³
mass of 1 litter = 1000 × 1 × 
mass = 1000 g
so
total number of electron in 1 litter = mass of 1 litter × 
total number of electron in 1 litter = 1000 × 
total number of electron in 1 litter is 3.34 ×
electron
Answer:(10.69, 11.436)
Explanation:
Given
initial height of ball is 2 m
height of basket is 3.05 m
Launching angle

y=1.05
equation of trajectory of ball is given by

for x=12.27

u=10.69
for x=11.73

u=11.436 m/s
Thus range of speed is (10.69, 11.436)
The answer is Hypothesis because she can predict that she needs a bigger bag
Answer: the constant angular velocity of the arms is 86.1883 rad/sec
Explanation:
First we calculate the linear velocity of the single sprinkler;
Area of the nozzle = π/4 × d²
given that d = 8mm = 8 × 10⁻³
Area of the nozzle = π/4 × (8 × 10⁻³)²
A = 5.024 × 10⁻⁵ m²
Now total discharge is dived into 4 jets so discharge for single jet will be;
Q_single = Q / n = 0.006 / 4 = 1.5 × 10⁻³ m³/sec
So using continuity equation ;
Q_single = A × V_single
V_single = Q_single/A
we substitute
V_single = (1.5 × 10⁻³) / (5.024 × 10⁻⁵)
V_single = 29.8566 m/s
Now resolving the forces as shown in the second image,
Vt = Vcos30°
Vt = 29.8566 × cos30°
Vt = 25.8565 m/s
Finally we calculate the angular velocity;
Vt = rω
ω_single = Vt / r
from the given diagram, radius is 300mm = 0.3m
so we substitute
ω_single = 25.8565 / 0.3
ω_single = 86.1883 rad/sec
Therefore the constant angular velocity of the arms is 86.1883 rad/sec