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ycow [4]
2 years ago
14

Velocity (m/s)

Physics
1 answer:
alekssr [168]2 years ago
3 0

Explanation:

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(a) <u>The</u><u> </u><u>segment</u><u> </u>A shows acceleration as velocity increases with the increase in time.

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(b) <u>The</u><u> </u><u>segment</u><u> </u>C shows the object is slowing down as the time increases in segment C, the velocity decreases and afterwards it comes to rest.

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⠀

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(c) The velocity is segment B is <u>4</u><u>0</u><u>m</u><u>/</u><u>s</u><u>.</u> And in the diagram there is no change in velocity.

⠀

⠀

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(d) The acceleration of segment B is <u>zero</u><u>.</u> As there in no change in curve and it is moving with uniform velocity.

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\:

<h2>Thank you!</h2>
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Katyanochek1 [597]

Answer:

The force is F = 1041.7N

Explanation:

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Substituting  1.9kg for M(Mass of the wheel) and \frac{66cm}{2} * \frac{1m}{100cm} = 0.33m for R_A(Radius of wheel)

              I = 1.9 * 0.33^2

                = 0.207kgm^2

The torque on the wheel due to net force is mathematically represented as

                      \tau = FR_B  - F_rR_A

Substituting  135 N for F_r (Force acting on sprocket),\frac{8.7cm}{2} * \frac{1m}{100cm} = 0.0435m for R_B (radius of the chain) and F is the force acting on the sprocket due to the chain which is unknown for now

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This same torque due to the net force is the also the torque that is required to rotate the wheel to have an angular acceleration of \alpha  = 3.70 rad/s^2 and this torque can also be represented mathematically as

                   \tau = \alpha I

Now equating the two equation for torque

                                F (0.0435) - 135 (0.33) = \alpha I    

Making F the subject

                     F = \frac{\alpha I + (135*0.33) }{0.0435}

Substituting values

                  F = \frac{(3.70 * 0.207)  + (135*0.33)}{0.0435}

                       = 1041.7N

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ElenaW [278]

Answer:

The force per unit length (N/m) on the top wire is 16.842 N/m

Explanation:

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current in the second wire, I₂ = 8.0 kA

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Answer:

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Explanation:

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