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madam [21]
3 years ago
5

The two uniform, slender rods B1and B2, each of mass 2kg, are pinned together at P, and then B1is suspended from a pin at O. (Th

is arrangement is called a double pendulum.) The counterclockwise couple C0, having moment 150????⋅m, is applied to B2beginning at t=0. Find the angular accelerations of B1and B2upon application of the couple, and the force exerted on B2at P.
Physics
1 answer:
Bezzdna [24]3 years ago
3 0

Answer:

hello the diagram relating to this question is attached below

a) angular accelerations : B1 = 180 rad/sec,  B2 = 1080 rad/sec

b) Force exerted on B2 at P = 39.2 N

Explanation:

Given data:

Co = 150 N-m ,

<u>a) Determine the angular accelerations of B1 and B2 when couple is applied</u>

at point P ; Co = I* ∝B2'

                150  = ( (2*0.5^2) / 3 ) * ∝B2

∴ ∝B2' = 900 rad/sec

hence angular acceleration of B2 = ∝B2' + ∝B1 = 900 + 180 = 1080 rad/sec

at point 0 ; Co = Inet * ∝B1

                  150 = [ (2*0.5^2) / 3  + (2*0.5^2) / 3  + (2*0.5^2) ] * ∝B1

∴ ∝B1 = 180 rad/sec

hence angular acceleration of B1 =  180 rad/sec

<u>b) Determine the force exerted on B2 at P</u>

T2 = mB1g + T1  -------- ( 1 )

where ; T1 = mB2g  ( at point p )

                 = 2 * 9.81 = 19.6 N

back to equation 1

T2 = (2 * 9.8 ) + 19.6 = 39.2 N

<u />

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3 years ago
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Answer:

 λ = 3 10⁻⁷ m,   UV laser

Explanation:

The diffraction phenomenon is described by the expression

         a sin θ = m λ

let's use trigonometry

         tan θ = y / L

as in this phenomenon the angles are small

        tan θ = \frac{sin \ \theta}{cos \ \theta} = sin θ

        sin θ = y / L

we substitute

      a y / L = m  λ

let's apply this equation to the initial data

       a  0.04 / L = 1 600 10⁻⁹

       a / L = 1.5 10⁻⁵

now they tell us that we change the laser and we have y = 0.04 m for m = 2

      a 0.04 / L = 2  λ

       a / L = 50  λ

we solve the two expression is

         1.5 10⁻⁵ = 50  λ

          λ = 1.5 10⁻⁵ / 50

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mixas84 [53]

Answer:

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Explanation:

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         E = h c / λ

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Let's calculate

          E₀ = 6.6310⁻³⁴ 3 10⁸/1000 10⁻⁹

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This is the energy emitted by a photon let's use a proportions rule to find the number emitted in P = 100 w

                #_photon = P / E₀

               #_photon = 100 / 19.89 10⁻²⁰

              #_photon = 5 10²⁰ photons / s

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3 years ago
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Answer:

No, it is not attracted.

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