The Earth gets hotter as one travels towards the core, known as the geothermal gradient. The geothermal gradient is the amount that the Earth's temperature increases with depth. ... On average, the temperature increases by about 25°C for every kilometer of depth.
Pitch is directly related to the frequency of the sound. In this item, we are given that the frequency of the sound is higher compared to those which are audible to the human being's ears. The pitch therefore of the dog's whistle is high.
On the other hand, the frequency and the wavelength of a certain wave are inversely proportional. This means that the high frequency wave will have a short wavelength.
Hence, the answer to this item would have to be "high pitch with a short wavelength"
The answer to this item is the second option.
Answer:
Length = 2.32 m
Explanation:
Let the length required be 'L'.
Given:
Resistance of the resistor (R) = 3.7 Ω
Radius of the rod (r) = 1.9 mm = 0.0019 m [1 mm = 0.001 m]
Resistivity of the material of rod (ρ) = ![1.8\times 10^{-5}\ \Omega\cdot m](https://tex.z-dn.net/?f=1.8%5Ctimes%2010%5E%7B-5%7D%5C%20%5COmega%5Ccdot%20m)
First, let us find the area of the circular rod.
Area is given as:
![A=\pi r^2=3.14\times (0.0019)^2=1.13\times 10^{-5}\ m^2](https://tex.z-dn.net/?f=A%3D%5Cpi%20r%5E2%3D3.14%5Ctimes%20%280.0019%29%5E2%3D1.13%5Ctimes%2010%5E%7B-5%7D%5C%20m%5E2)
Now, the resistance of the material is given by the formula:
![R=\rho( \frac{L}{A})](https://tex.z-dn.net/?f=R%3D%5Crho%28%20%5Cfrac%7BL%7D%7BA%7D%29)
Express this in terms of 'L'. This gives,
![\rho\times L=R\times A\\\\L=\frac{R\times A}{\rho}](https://tex.z-dn.net/?f=%5Crho%5Ctimes%20L%3DR%5Ctimes%20A%5C%5C%5C%5CL%3D%5Cfrac%7BR%5Ctimes%20A%7D%7B%5Crho%7D)
Now, plug in the given values and solve for length 'L'. This gives,
![L=\frac{3.7\ \Omega\times 1.13\times 10^{-5}\ m^2}{1.8\times 10^{-5}\ \Omega\cdot m}\\\\L=\frac{4.181}{1.8}=2.32\ m](https://tex.z-dn.net/?f=L%3D%5Cfrac%7B3.7%5C%20%5COmega%5Ctimes%201.13%5Ctimes%2010%5E%7B-5%7D%5C%20m%5E2%7D%7B1.8%5Ctimes%2010%5E%7B-5%7D%5C%20%5COmega%5Ccdot%20m%7D%5C%5C%5C%5CL%3D%5Cfrac%7B4.181%7D%7B1.8%7D%3D2.32%5C%20m)
Therefore, the length of the material required to make a resistor of 3.7 Ω is 2.32 m.
Answer:
I = 21.13 mA ≈ 21 mA
Explanation:
If
I₁ = 5 mA
L₁ = L₂ = L
V₁ = V₂ = V
ρ₁ = 1.68*10⁻⁸ Ohm-m
ρ₂ = 1.59*10⁻⁸ Ohm-m
D₁ = D
D₂ = 2D
S₁ = 0.25*π*D²
S₂ = 0.25*π*(2*D)² = π*D²
If we apply the equation
R = ρ*L / S
where (using Ohm's Law):
R = V / I
we have
V / I = ρ*L / S
If V and L are the same
V / L = ρ*I / S
then
(V / L)₁ = (V / L)₂ ⇒ ρ₁*I₁ / S₁ = ρ₂*I₂ / S₂
If
S₁ = 0.25*π*D² and
S₂ = 0.25*π*(2*D)² = π*D²
we have
ρ₁*I₁ / (0.25*π*D²) = ρ₂*I₂ / (π*D²)
⇒ I₂ = 4*ρ₁*I₁ / ρ₂
⇒ I₂ = 4*1.68*10⁻⁸ Ohm-m*5 mA / 1.59*10⁻⁸ Ohm-m
⇒ I₂ = 21.13 mA
Answer:
White dwarfs are likely to be much more common. The number of stars decreases with increasing mass, and only the most massive stars are likely to complete their lives as black holes. There are many more stars of the masses appropriate for evolution to a white dwarf.