Answer:
Explanation:
refractive index of ember = sin of angle of incidence / sin of angle of refraction
= sin 35 / sin24
= .5735 / .4067
= 1.41
This is refractive index of ember with respect to water
refractive index of ember with respect to water
= wμe = μe / μw
μe = wμe x μw
= 1.33 x 1.41
= 1.87
refractive index of ember with respect to air = 1.87 .
Answer:
Explanation:
First, It's important to remember F = ma, and in this problem m = 13.3 kg
This can be reduced to a simple system of equations problem. Now if they are both going the same way then we add them, while if they are going the opposite way we subtract them. So let's call them F1 and F2, with F1 arger than F2. Now, When we add them together F1+F2 = (.723 m/s^2)*13.3kg and then when we subtract them, and have the larger one pushing toward the east, let's call F1 the larger one, F1-F2 = (.493 m/s^2)*13.3kg.
Can you solve this system of equations seeing them like this, or do you need more help?
Answer:
I = 1.875 A
Explanation:
For this exercise we use Ampere's law
∫ B . ds = μ₀ I
We use a circular path around the wire whereby B and ds are parallel, whereby the dot product is reduced to the algebraic product
ds = 2π dr
B (2πr) = μ₀ I
I = B 2π R /μ₀
r= 7.5 cm = 0.075 m
calculate
I = (50 μ₀ /π) 2π 0.075 /μ₀
I = 1.875 A
Answer:
3.1×10⁻¹¹ N
Explanation:
Use Coulomb's law:
F = k q₁ q₂ / r²
F = (9×10⁹) (6.0×10⁻¹⁰) (2.3×10⁻¹⁵) / (0.02 m)²
F = 3.1×10⁻¹¹
