Answer:
the intensity of the sun on the other planet is a hundredth of that of the intensity of the sun on earth.
That is,
Intensity of sun on the other planet, Iₒ = (intensity of the sun on earth, Iₑ)/100
Explanation:
Let the intensity of light be represented by I
Let the distance of the star be d
I ∝ (1/d²)
I = k/d²
For the earth,
Iₑ = k/dₑ²
k = Iₑdₑ²
For the other planet, let intensity be Iₒ and distance be dₒ
Iₒ = k/dₒ²
But dₒ = 10dₑ
Iₒ = k/(10dₑ)²
Iₒ = k/100dₑ²
But k = Iₑdₑ²
Iₒ = Iₑdₑ²/100dₑ² = Iₑ/100
Iₒ = Iₑ/100
Meaning the intensity of the sun on the other planet is a hundredth of that of the intensity on earth.
<span>9000 Pascals
Looking on the internet, it appears that a human can only suck about 3 feet of water. So let's convert that measurement into a few more convenient units.
cmH2O = 36 * 2.54 = 91.44 cmH2O
cmHg = 91.44 * 0.73555912101486 = 67.26 mmHg
Pascal = 91.44 * 98.0665 = 8967 Pascals
PSI = 91.44 * 0.0142233 = 1.3 psi
Since we're dealing with science and the metric system is the most common system used in science, I'd recommend an answer of 9000 Pascals.</span>
Answer:
Closest to the dog.
Explanation:
Sounds are louder the closer you are to them.
Answer:
The average induced emf in the coil is 0.0286 V
Explanation:
Given;
diameter of the wire, d = 11.2 cm = 0.112 m
initial magnetic field, B₁ = 0.53 T
final magnetic field, B₂ = 0.24 T
time of change in magnetic field, t = 0.1 s
The induced emf in the coil is calculated as;
E = A(dB)/dt
where;
A is area of the coil = πr²
r is the radius of the wire coil = 0.112m / 2 = 0.056 m
A = π(0.056)²
A = 0.00985 m²
E = -0.00985(B₂-B₁)/t
E = 0.00985(B₁-B₂)/t
E = 0.00985(0.53 - 0.24)/0.1
E = 0.00985 (0.29)/ 0.1
E = 0.0286 V
Therefore, the average induced emf in the coil is 0.0286 V
Answer:
8.5 Ω
Explanation:
La resistencia de un material es directamente proporcional a su longitud e inversamente proporcional al área de la sección transversal.
La fórmula de la resistencia (R) viene dada por:
R = ρL/A
Donde ρ es la resistividad del material, L es la longitud del material y A es el área de la sección transversal del material.
Dado que:
L = 1 km = 1000 m, A = 2 mm² = 2 * 10⁻⁶ m², ρ (cobre) = 1.7 * 10⁻⁸ Ωm
Sustituyendo da:
R = 1,7 * 10⁻⁸ * 1000/2 * 10⁻⁶
R = 8.5 Ω
