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Troyanec [42]
2 years ago
7

A kinesin that is transporting a secretory vesicle uses approximately 80 ATP molecules/s. Each ATP provides a kinesin molecule w

ith an energy of about 0.8 × 10-19 J. If the velocity of the kinesin is 800 nm/s, can you determine the force the kinesin is exerting, if you assume that all the ATP energy is used (100% efficiency)? If you can, find it and give your answer in newtons. If not, answer with 0.
Physics
1 answer:
Firdavs [7]2 years ago
7 0

Answer:

The force is  F  =   8*10^{-12} \ N

Explanation:

From the question we are told that

     The rate at which ATP molecules are used is R =  80 ATP/ s

       The energy provided by a single ATP is  E_{ATP} =  0.8 *  10^{-19} J

       The velocity of the kinesin is  v  =  800 nm/s =  800*10^{-9} m/s

The power provided by the ATP in one second is  mathematically represented as

       P =  E_{ATP}  *  R

substituting values

       P =  80 * 0.8*10^{-19 }

       P =  6.4 *10^{-18}J/s

Now  this power is mathematically represented as

       P  =  F *  v

Where  F  is  the force the kinesin is exerting

  Thus  

          F  =   \frac{P}{v}

substituting values

            F  =   \frac{6.4*0^{-18}}{800 *10^{-9}}

           F  =   8*10^{-12} \ N

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Answer:

Part a)

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Explanation:

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As we know that

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2 years ago
A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg·m2. If the arms are pulled in so t
Serggg [28]

a) 6.25 rad/s

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The angular momentum is given by:

L=I\omega

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I is the moment of inertia

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Since the angular momentum must be conserved, we can write

L_1 = L_2\\I_1 \omega_1 = I_2 \omega_2

where we have

I_1 = 2.25 kg m^2 is the initial moment of inertia

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I_2 = 2.25 kg m^2 is the final moment of inertia

\omega_2 is the final angular speed

Solving for \omega_2, we find

\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(2.25 kg m^2)(5.00 rad/s)}{1.80 kg m^2}=6.25 rad/s

b) 28.1 J and 35.2 J

The rotational kinetic energy is given by

K=\frac{1}{2}I\omega^2

where

I is the moment of inertia

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Applying the formula, we have:

- Initial kinetic energy:

K=\frac{1}{2}(2.25 kg m^2)(5.00 rad/s)^2=28.1 J

- Final kinetic energy:

K=\frac{1}{2}(1.80 kg m^2)(6.25 rad/s)^2=35.2 J

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Also, W = F.d

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F is the force applied by the engine of car

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So, the force of the car engine increased the car’s kinetic energy. Hence, this is the required solution.

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