The solution would be like this for this specific problem:
Given:
diffraction grating
slits = 900 slits per centimeter
interference pattern that
is observed on a screen from the grating = 2.38m
maxima for two different
wavelengths = 3.40mm
slit separation .. d =
1/900cm = 1.11^-3cm = 1.111^-5 m <span>
Whenas n = 1, maxima (grating equation) sinθ = λ/d
Grant distance of each maxima from centre = y ..
<span>As sinθ ≈ y/D y/D =
λ/d λ = yd / D </span>
∆λ = (λ2 - λ1) = y2.d/D - y1.d/D
∆λ = (d/D) [y2 -y1]
<span>∆λ = 1.111^-5m x [3.40^-3m] / 2.38m .. .. ►∆λ = 1.587^-8 m</span></span>
No, you can't aim where you see the fish, because refraction will change the position of the fish and make it appear in a different position from where it actually is.
Hope this helps!
Actual Mechanical Advantage(AMA) = Weight / Force
Here, Weight = 764 N
Force = 255 N
Substitute the values in to the expression,
AMA = 764 / 255
AMA = 2.99
After rounding-off to the nearest tenth value, it would be 3
Finally, option C would be your answer.
Hope this helps!
Answer:
Resistance =330 Ω
Tolerance = 33 Ω
Explanation:
see attached resistor color code table
The first stripe is orange, which means the leftmost digit is a 3.
The second stripe is orange , which means the next digit is a 3.
The third stripe is brown. Since brown is 1, it means add one zero to the right of the first two digits.
The resistance is:
orange-orange-brown= 330 Ω
The tolerance is:
The fourth color band indicates the resistor's tolerance. Tolerance is the percentage of error in the resistor's resistance.
silver is 10%
A 330 Ω resistor has a silver tolerance band.
<em>Tolerance = value of resistor x value of tolerance band </em>
= 330 Ω x 10% = 33 Ω
330 Ω stated resistance +/- 33 Ω tolerance means that the resistor could range in actual value from as much as 363 Ω to as little as 297 Ω.
Answer:
F = 7.68 10¹¹ N, θ = 45º
Explanation:
In this exercise we ask for the net electric force. Let's start by writing the configuration of the charges, the charges of the same sign must be on the diagonal of the cube so that the net force is directed towards the interior of the cube, see in the attached numbering and sign of the charges
The net force is
F_ {net} = F₂₁ + F₂₃ + F₂₄
bold letters indicate vectors. The easiest method to solve this exercise is by using the components of each force.
let's use trigonometry
cos 45 = F₂₄ₓ / F₂₄
sin 45 = F_{24y) / F₂₄
F₂₄ₓ = F₂₄ cos 45
F_{24y} = F₂₄ sin 45
let's do the sum on each axis
X axis
Fₓ = -F₂₁ + F₂₄ₓ
Fₓ = -F₂₁₁ + F₂₄ cos 45
Y axis
F_y = - F₂₃ + F_{24y}
F_y = -F₂₃ + F₂₄ sin 45
They indicate that the magnitude of all charges is the same, therefore
F₂₁ = F₂₃
Let's use Coulomb's law
F₂₁ = k q₁ q₂ / r₁₂²
the distance between the two charges is
r = a
F₂₁ = k q² / a²
we calculate F₂₄
F₂₄ = k q₂ q₄ / r₂₄²
the distance is
r² = a² + a²
r² = 2 a²
we substitute
F₂₄ = k q² / 2 a²
we substitute in the components of the forces
Fx =
Fx =
( -1 + ½ cos 45)
F_y = k \frac{q^2}{a^2} ( -1 + ½ sin 45)
We calculate
F₀ = 9 10⁹ 4.25² / 0.440²
F₀ = 8.40 10¹¹ N
Fₓ = 8.40 10¹¹ (½ 0.707 - 1)
Fₓ = -5.43 10¹¹ N
remember cos 45 = sin 45
F_y = - 5.43 10¹¹ N
We can give the resultant force in two ways
a) F = Fₓ î + F_y ^j
F = -5.43 10¹¹ (i + j) N
b) In the form of module and angle.
For the module we use the Pythagorean theorem
F =
F = 5.43 10¹¹ √2
F = 7.68 10¹¹ N
in angle is
θ = 45º