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:
268N
Explanation:
The upward force acting on the block are the reaction and the hooked table..
The total normal force acting = normal reaction + 24N
Note that the normal reaction is always equal the weight of the table
Hence the normal force acting in the block is 244.0+24 = 268.0N
Answer:
U = 1 / r²
Explanation:
In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related
F = - dU / dr
this derivative is a gradient, that is, a directional derivative, so we must have
dU = - F. dr
the esxresion for strength is
F = B / r³
let's replace
∫ dU = - ∫ B / r³ dr
in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product
let's evaluate the integrals
U - Uo = -B (- / 2r² + 1 / 2r₀²)
To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)
U = B / 2r²
we substitute the value of B = 2
U = 1 / r²
Answer:
7.401 * 10^(-15) N
Explanation:
30 electrons will have a charge:
30 * -1.6022 * 10^(-19) C
= - 4.806 * 10^(-18) C
The relationship between electric field and electric force is:
E = F/q
This means that force, F, is
|F| = |E|*|q|
|F| = |1540| * |-4.806 * 10^(-18)|
|F| = |-7401.24 * 10^(-18)|
|F| = 7.401 * 10^(-15) N
Work = (force) x (distance) =
(200 N) x (3.5 m) = <em>700 joules</em>
