Answer:
the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³
Explanation:
Given the data in the question;
we make use of the following expression;
hall Voltage VH = IB / ned
where I = 2.25 A
B = 0.685 T
d = 0.107 mm = 0.107 × 10⁻³ m
e = 1.602×10⁻¹⁹ C
VH = 2.59 mV = 2.59 × 10⁻³ volt
n is the electron density
so from the form; VH = IB / ned
VHned = IB
n = IB / VHed
so we substitute
n = (2.25 × 0.685) / ( 2.59 × 10⁻³ × 1.602×10⁻¹⁹ × 0.107 × 10⁻³ )
n = 1.54125 / 4.4396226 × 10⁻²⁶
n = 3.4716 × 10²⁵ m⁻³
Therefore, the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³
Answer:
W = 1.06 MJ
Explanation:
- We will use differential calculus to solve this problem.
- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.
- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.
- Now develop and expression of Force required:
F = p*V*g
F = 1000*(2*0.5*x*8*dx)*g
F = 78480*x*dx
- Now, the work done is given by:
W = F.s
- Where, s is the distance from top of hose to the differential volume:
s = (5 - x)
- We have the work as follows:
dW = 78400*x*(5-x)dx
- Now integrate the following express from 0 to 3 till the tank is empty:
W = 78400*(2.5*x^2 - (1/3)*x^3)
W = 78400*(2.5*3^2 - (1/3)*3^3)
W = 78400*13.5 = 1058400 J
Answer:
Explanation:
Gamma-rays have the smallest wavelengths and the most energy of any other wave in the electromagnetic spectrum. Gamma-rays can kill living cells, a fact which medicine uses to its advantage, using gamma-rays to kill cancerous cells. Also, industry (sterilization and disinfection) and the nuclear industry.
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