Let l = Q/L = linear charge density. The semi-circle has a length L which is half the circumference of the circle. So w can relate the radius of the circle to L by
<span>C = 2L = 2*pi*R ---> R = L/pi </span>
<span>Now define the center of the semi-circle as the origin of coordinates and define a as the angle between R and the x-axis. </span>
<span>we can define a small charge dq as </span>
<span>dq = l*ds = l*R*da </span>
<span>So the electric field can be written as: </span>
<span>dE =kdq*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) </span>
<span>dE = k*I*R*da*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) </span>
<span>E = k*I*(sin(a)/R I_hat - cos(a)/R^2 j_hat) </span>
<span>E = pi*k*Q/L(sin(a)/L I_hat - cos(a)/L j_hat)</span>
Answer: 0.0138 m^2 = 138 cm^2
Explanation:
The thermal expansion is the term use for the physical phenomena of dilation of the objects when they are exposed to changes in temperature.
The objects dilate when they are heated and contract when they are cooled.
The dilation is proportional to the change in temperatur.
For linear dilation, the proportionality constant is called linear dilation coefficient of the materials, it is named α and is measured in °C ^-1.
ΔL = α * Lo * ΔT, which means that the dilation (or contraction) is proportional to the product of the original length (Lo) and the change of temperature (ΔT).
There is also superficial dilation, for which the dilation is:
ΔA = β * Ao * ΔT, which means that the superficial dilation (or contraction) is proportional to the product of the original area (Ao) and the change of temperature (ΔT).
It is very interesting and important to solve problems that β = 2α, because regularly you will find the values of α for different materials and so, you just to multiply it times 2 to use β.
For this problem:
- Original area, Ao = area of the flat roof at - 10°C = 2.0m * 3.0m = 6.0 m^2.
- α for aluminum = 24 * 10^ -6 °C^-1.
- ΔT = 38°C - (-10°C) = 48°C
So, ΔA = 6.0m^2 * (2 * 24*10^-6 °C&-1) * 48°C = 0.0138 m^2
And that is the area that should stick out in summer to fit the structure during cold winter nights.
You can pass that number to cm^2 to grasp better the idea of this size:
0.0138 m^2 * (100 cm)^2 / m^2 = 138 cm^2
Answer:
age of the site is 15411.75 years old
Explanation:
Given data
plant or animal dies = 14C
time period = 5730 year
carbon = 15.5%
to find out
age (in years) of the ancient site
solution
we know that Final value = Initial value × 
here n is half life passed
so for 15.5%
15.5% = 100% of
0.155 = 1 ×
now take log both side
log 0.155 = log
n = log 0.155 / log 0.5
n = 2.68966
we know here 5730 years in half life
so for 2.68966 half-lives = 2.68966 × 5730 = 15411.7518
age of the site is 15411.75 years old
7. solar flare: f.
8. core: h.
9. chromosphere: b.
10. sunspot: d.
11. corona: c.
12. nuclear fusion: j.
13. photosphere: a.
14. solar wind: g.
15. prominence: e.
16. radiation zone: k.
17. convection zone: i.
