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andrezito [222]
3 years ago
11

How to do this, i'm completely lost

Physics
1 answer:
vaieri [72.5K]3 years ago
8 0
There are two torques t1 and t2 on the beam due to the weights, one torque t3 due to the weight of the beam, and one torque t4 due to the string.

You need to figure out t4 to know the tension in the string.

Since the whole thing is not moving t1 + t2 + t3 = t4.

torque t = r * F * sinФ = distance from axis of rotation * force * sin (∡ between r and F)

t1 =3.2 * 44g 
t2 = 7 * 49g 
t3 = 3.5 * 24g 

t4 = t1 + t2 + t3 = 5570,118

The t4 also is given by:

t4 = r * T * sin Ф

r = 7
Ф = 32°
T: tension in the string

T = t4 / (r * sinФ)

T = t4 / (7 * sin(32°)) 

T = 1501,6 N

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it produces

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a lot of waste

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At the point 0 we have that,

L_i=mvl

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At the end with the collition we have

L_f=(I_b+I_s)\omega

Substituting

L_f=(ml^2+\frac{10ml^2}{3})\omega

L_f=\frac{13}{10}ml^2w

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Applying conservative energy equation we have

\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'

\frac{1}{2}(ml^2+\frac{10ml^2}{3})(\frac{1}{2l})^2=mg(l-lcos\theta)+\frac{10}{2}mg(l-lcos\theta)

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for a certain chemical reaction, the reactants contain 52 kj of potential energy and the products contain 32 kj how much energy
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Answer:

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x-4 | x^3+2x^2-16x-32

    -  x^3-4x^2             <-- (x-4)(x^2)

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              6x^2-16x-32

           -  6x^2-24x     <-- (x-4)(6x)

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