Hello,
Here is your answer:
The proper answer to this question is option B "<span>chemical properties".
Here is how:
</span>Electrons determine a atoms chemical properties mainly what its made of.
Your answer is B.
If you need anymore help feel free to ask me!
Hope this helps!
Answer:
<h2> 4kg</h2>
Explanation:
Step one:
given
length of rod=2m
mass of object 1 m1=1kg
let the unknown mass be x
center of mass<em> c.m</em>= 1.6m
hence 1kg is 1.6m from the <em>c.m</em>
and x is 0.4m from the <em>c.m</em>
Taking moment about the <em>c.m</em>
<em>clockwise moment equals anticlockwise moments</em>
1*1.6=x*0.4
1.6=0.4x
divide both sides by 0.4 we have
x=1.6/0.4
x=4kg
The mass of the other object is 4kg
Answer:
R1 = 5.13 Ω
Explanation:
From Ohm's law,
V = IR............... Equation 1
Where V = Voltage, I = current, R = resistance.
From the question,
I = 2 A, R = R1
Substitute into equation 1
V = 2R1................ Equation 2
When a resistance of 2.2Ω is added in series with R1,
assuming the voltage source remain constant
R = 2.2+R1, and I = 1.4 A
V = 1.4(2.2+R1)................. Equation 3
Substitute the value of V into equation 3
2R1 = 1.4(2.2+R1)
2R1 = 3.08+1.4R1
2R1-1.4R1 = 3.08
0.6R1 = 3.08
R1 = 3.08/0.6
R1 = 5.13 Ω
A) We want to find the work function of the potassium. Apply this equation:
E = 1243/λ - Φ
E = energy of photoelectron, λ = incoming light wavelength, Φ = potassium work function
Given values:
E = 2.93eV, λ = 240nm
Plug in and solve for Φ:
2.93 = 1243/240 - Φ
Φ = 2.25eV
B) We want to find the threshold wavelength, i.e. find the wavelength such that the energy E of the photoelectrons is 0eV. Plug in E = 0eV and Φ = 2.25eV and solve for the threshold wavelength λ:
E = 1243/λ - Φ
0 = 1243/λ - Φ
0 = 1243/λ - 2.25
λ = 552nm
C) We want to find the frequency associated with the threshold wavelength. Apply this equation:
c = fλ
c = speed of light in a vacuum, f = frequency, λ = wavelength
Given values:
c = 3×10⁸m/s, λ = 5.52×10⁻⁷m
Plug in and solve for f:
3×10⁸ = f(5.52×10⁻⁷)
f = 5.43×10¹⁴Hz
Answer:

Explanation:
The speed of light is given by
and
hence

Speed of light is given by
