Answer:
1.13 mA
Explanation:
Length of wire L = 20.5 cm = 0.205m
Radius of wire r = 2.60/2 = 1.3cm = 0.0130m
Voltage V = 1 × 10³ V
Resistivity of pure silicon p = 2300 Ohms • m
Cross sectional area of the wire
A = pi × r² = pi × (0.013)² = 5.307 × 10 ^-4 m²
Resistance of the material
R = p• L/A
= 2300 • 0.205/5.307 × 10^-4 = 0.888 × 10⁶ Ohms
Using Ohms Law
R = V/ I
I = V/R
I = 10³/0.888 × 10⁶
= 0.001126 A
= 1.13 mA
Answer:
Explanation:
The Hi line of the Balmer series is emitted in the transition from n = 3 to n = 2 i.e. 
and 
The wavelength of Hi line of the Balmer series is given by :
So, the wavelength for this line is 550 nm. Hence, this is the required solution.
The molecule that trap's the sun's energy is chlorophyll.
Answer:
x = 2,864 m
, Ra = 32.1 m
Explanation:
Let's solve this problem in parts, let's start by finding the intensity of the sound in each observer
observer A β = 64 db
β = 10 log Iₐ / I₀
where I₀ = 1 10⁻¹² W / m²
Iₐ = I₀ 10 (β/ 10)
let's calculate
Iₐ = 1 10⁻¹² (64/10)
Iₐ = 2.51 10⁻⁶ W / m²
Observer B β = 85 db
I_b = 1 10-12 10 (85/10)
I_b = 3.16 10⁻⁴ W / m²
now we use that the emitted power that is constant is the intensity over the area of the sphere where the sound is distributed
P = I A
therefore for the two observers
P = Ia Aa = Ib Ab
the area of a sphere is
A = 4π R²
we substitute
Ia 4pi Ra2 = Ib 4pi Rb2
Ia Ra2 = Ib Rb2
Let us call the distance from the observer be to the haughty R = ax, so the distance from the observer A to the haughty is R = 35 ax; we substitute
Ia (35 -x) 2 = Ib x2
we develop and solve
35-x = Ra (Ib / Ia) x
35 = [Ra (Ib / Ia) +1] x
x (11.22 +1) = 35
x = 35 / 12.22
x = 2,864 m
This is the distance of observer B
The distance from observer A
Ra = 35 - x
Ra = 35 - 2,864
Ra = 32.1 m
Answer:
Explanation:
We can use basic trig in a right triangle to solve this problem. Let the x component of the force be 
. We have the following equation:
