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chubhunter [2.5K]
2 years ago
5

Your friend, Tim, is playing with his sled. He ties a rope to his sled and attaches it to Lar's snowmobile. Tim has a mass of 71

kg, the tension in the rope is 220 N (horizontally), and a coefficient of kinetic friction is 0.1. Unfortunately, Tim can only hang on for 8 s.
(a) What is Tim's maximum speed?
Physics
1 answer:
kykrilka [37]2 years ago
7 0

The maximum speed of Tim is 16.95 m/s.

The given parameters:

  • Mass of the rope, m = 71 kg
  • Tension on the rope, T = 220 N
  • Coefficient of kinetic friction, = 0.1
  • Time of motion, t = 8 s

<h3>What is Newton's second law of motion?</h3>
  • Newton's second law of motion states that, the force applied to an object is directly proportional to the product of mass and acceleration of the object.

The net force on Tim is calculated by applying Newton's second law of motion as follows;

T - \mu _k F_n = ma\\\\T - \mu _k F_n = m\frac{v}{t} \\\\T - \mu_k mg = m\frac{v}{t} \\\\t(\frac{T - \mu_k mg}{m} )= v\\\\8 (\frac{220 \ -\  0.1 \times 71 \times 9.8}{71} ) =v \\\\ 16.95 \ m/s = v

Thus, the maximum speed of Tim is 16.95 m/s.

Learn more about net horizontal force here: brainly.com/question/21684583

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Two identical charges,2.0m apart,exert forces of magnitude 4.0 N on each other.What is the value of either charge?
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Answer:

\large \boxed{42\, \mu \text{C}}$

Explanation:

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\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}

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