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Helga [31]
3 years ago
14

Ten narrow slits are equally spaced 3.00 mm apart and illuminated with indigo light of wavelength 444 nm. The width of bright fr

inges can be calculated as the separation between the two adjacent dark fringes on either side. Find the angular widths (in rad) of the second- and fifth-order bright fringes.
Physics
1 answer:
ankoles [38]3 years ago
8 0

Answer:ask yo mama

Explanation:she finished school

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determine the metacentric height of a cylinder of 4 m diameter and height of 4m floating in water with its axis vertical, if per
krok68 [10]

Answer:

-0.383 m

Explanation:

Diameter of cylinder = 4m  therefore  r = 2

height of cylinder ( H ) = 4 m

specific gravity = 0.6 ( assumed )

depth of immersion = 'h'

<u>Determine the metacentric height </u>

weight of cylinder in water = water displaced

= 0.6 * 1000 * πr^2* H = 1000 * πr^2* h

= 0.6 * 4 = h

∴ h = 2.4 m

hence the depth of center of buoyancy from free space = h /2 = 1.2 m

The metacentric height can be calculated using the formula below

Gm = Io / Vsubmerged  - BG

attached below is the remaining solution

6 0
3 years ago
I desperately need help with my Physics Exam I am failing this class will mark Brainliest to whoever helps the most: PART 1 (I d
GREYUIT [131]
  1. Option 2nd is correct

  1. last option is correct
6 0
2 years ago
How are elements and compounds similar
sergejj [24]
Compound and events are pure substance
4 0
3 years ago
A light platform is suspended from the ceiling by a spring. A student with a mass of 90 kg climbs onto the platform. When it sto
Ilya [14]
Refer to the diagram shown.

When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N

By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m

When the spring is stretched by x from the equilibrium position, the restoring force is
F = - k*x.

If damping is ignored, the equation of motion is
F = m * acceleration
or
m \frac{d^{2}x}{dt^{2}} = -kx \\ \frac{d^{2}x}{dt^{2}} + \frac{k}{m} x = 0

Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)

x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0

The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32

The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.

Answer: 
0.32 m (single amplitude), or
0.64 m (double amplitude)

6 0
3 years ago
Assume the acceleration of the object is a(t) = −9.8 meters per second per second. (Neglect air resistance.) With what initial v
lukranit [14]

Answer:

Vi = 94.64 m/s

Explanation:

I order to find out the initial velocity of the object, we can use third equation of motion:

2ah = Vf² - Vi²

where,

a = acceleration = -9.8 m/s²

h = maximum height covered by object = 460 m - 3 m = 457 m

Vf = Final Velocity = 0 m/s (since, object momentarily stops at highest point)

Vi = Initial Velocity = ?

Therefore,

2(-9.8 m/s²)(457 m) = (0 m/s)² - Vi²

Vi = √8957.2 m²/s²

<u>Vi = 94.64 m/s</u>

3 0
3 years ago
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