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Alika [10]
3 years ago
11

Need physics help! (there’s a picture)

Physics
1 answer:
Nezavi [6.7K]3 years ago
5 0

Answer:

i. Subtract v₀² from both of the equation.

ii. Divide both side of the equation by 2

iii. Divide both side of the equation by (x – x₀)

a = (v² – v₀²) / 2(x – x₀)

Explanation:

v² = v₀² + 2a(x – x₀)

To solve for a in the expression above, do the following:

i. Subtract v₀² from both of the equation.

v² – v₀² = v₀² – v₀² + 2a(x – x₀)

v² – v₀² = 2a(x – x₀)

ii. Divide both side of the equation by 2

(v² – v₀²) /2= 2a(x – x₀)/2

(v² – v₀²) /2 = a(x – x₀)

iii. Divide both side of the equation by (x – x₀)

(v² – v₀²) /2 ÷ (x – x₀) = a(x – x₀)/(x – x₀)

(v² – v₀²) /2 × 1/ (x – x₀) = a

a = (v² – v₀²) / 2(x – x₀)

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A 6.9-kg wheel with geometric radius m has radius of gyration computed about its mass center given by m. A massless bar at angle
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Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is  \alpha =2.538 \  rad/s^2

Explanation:

From the question we are told that

  The mass of the wheel is  m =  6.9  kg

   The radius is  r =  0.69 \  m

    The radius of gyration is  k_G = 0.4\  m

    The angle is  \theta = 47^o

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Generally given that the wheels rolls without slipping on the flat stationary ground surface, it implies that  point A is the  center of rotation.

  Generally the moment of  inertia about A is mathematically represented as

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Here I_G is the moment of inertia about G with respect to the radius of gyration  which is mathematically represented as

    I_G =  M *  k_G

=>I_a = k_G*  M + M* r^2

=>I_a =0.4 * 6.9  + 6.9 * 0.69^2

=>I_a =6.045 \  kg \cdot m^2

Generally the torque experienced by the wheel  is mathematically represented as

       \tau =  F *  cos (47)

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=>     \tau =  15.34 \ kg \cdot m^2 \cdot s^{-2}

Generally this torque is also mathematically represented as

     \tau = I_a * \alpha

=>   15.34  =  6.045 * \alpha

=>   \alpha =2.538 \  rad/s^2

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