Answer:
= 4
Explanation:
To solve this exercise we will use Bohr's atomic model
= - 13.606 / n² [eV]
The transition from level n = 2 to level n = 1 is valid
= - 13.606 [¼ -1/1]
= 10.2045 eV
Bohr's model for atoms with only one electron is
= -13.606 Z² / n²
Where Z is the atomic number of the atom.
In this case the helium atom has an atomic number of Z = 2 from the level n₀ = 2 let's look up to what level it reaches
ΔE = -13.606 [4 /
² - 4/4]
4 /
² = -ΔE / 13.606 + 1
4 /
² = -10.2045 / 13.606 +1 = -0.75 +1
4 /
² = 0.25
= √ 4 / 0.25
= 4
U=RI Ohm's law
then R=U/I
=120/0.08
=2250Ω
hope this helps you
Answer:
1) t=1.743 sec
2)Vo=61.388 m/sec
3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec
4)Vf=17.08 m/s
Explanation:
1)From second equation of motion we get
h=Vit+(1/2)gt^2
here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion
14.9=(0)*t+(1/2)(9.8)t^2
t^2=14.9/4.9
t^2=3.040 sec
t=1.743 sec
2) s=Vo*t
Putting values we get
107=Vo*1.743
Vo=61.388 m/sec
3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec
4)From third equation of motion we know that
Vf^2-Vi^2=2gh
here Vi=0 m/s,h=14.9 m
Vf^2=Vi^2+2gh=0+2(9.8)(14.9)
Vf^2=292.04
Vf=17.08 m/s
Answer:
1. 1. A quantity is completely described by magnitude alone. A quantity Is completely described by a magnitude with a direction.
[a]. scalar, vector
b. vector, scalar
2.2. Speed is a velocity is a quantity and quantity.
a. scalar, vector
[b]. vector, scalar
Answer:
2.23 × 10^6 g of F- must be added to the cylindrical reservoir in order to obtain a drinking water with a concentration of 0.8ppm of F-
Explanation:
Here are the steps of how to arrive at the answer:
The volume of a cylinder = ((pi)D²/4) × H
Where D = diameter of the cylindrical reservoir = 2.02 × 10^2m
H = Height of the reservoir = 87.32m
Therefore volume of cylindrical reservoir = (3.142×202²/4)m² × 87.32m = 2798740.647m³
1ppm = 1g/m³
0.8ppm = 0.8 × 1g/m³
= 0.8g/m³
Therefore to obtain drinking water of concentration 0.8g/m³ in a reservoir of volume 2798740.647m³, F- of mass = 0.8g/m³ × 2798740.647m³ = 2.23 × 10^6 g must be added to the tank.
Thank you for reading.