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AleksAgata [21]
3 years ago
7

Read the scenario. A car starts 10 m north of a reference point. It moves at a constant velocity over the next 5 s, reaching a p

osition of 10 m south of the reference point. What is the car’s average velocity? 2 m/s south 4 m/s north 0 m/s 4 m/s south
Physics
1 answer:
Gennadij [26K]3 years ago
4 0

Answer:

4 m/s south

Explanation:

This is a pretty easy one.

Assume that the car has a reference point ragged x

We also know that the car starts moving at a point 10 m North of X. It moves steadily at a uniform velocity and ended at a position 10 m south of X.

If x is the reference point, we can infer that the car moves a total distance of, 10 m + 10 m.

Thus, the total distance moved by the car is 20 m.

It is also stated that it achieves that distance in 5 seconds.

Velocity is defined as the ratio of distance with respect to the tome taken, i.e V = d/t

Then, the velocity is

V = 20 / 5 = 4 m/s

Considering the fact that the car left and moved towards the south. We can say that it moved 4 m/s South.

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Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms f
vampirchik [111]

Answer:

\large \boxed{\text{30 rev/s}}

Explanation:

This question is based on the Law of Conservation of Angular Momentum.

Angular momentum (L) equals the moment of inertia (I) times the angular speed (ω).

L = Iω

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I₁ω₁ = I₂ω₂

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Calculation:

\begin{array}{rcl}I_{1}\omega_{1} &= &I_{2}\omega_{2}\\\text{3.5 kg$\cdot$m$^{2}$}\times \text{6.0 rev/s} &= &\text{0.70 kg$\cdot$m$^{2}$}\times\omega_{2}\\\text{21 rev/s} &= &0.70\omega_{2}\\\omega_{2} & = & \dfrac{\text{21 rev/s}}{0.70}\\\\&=&\textbf{30 rev/s}\\\end{array}\\\text{The skater's final rotational speed is $\large \boxed{\textbf{30 rev/s}}$}

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8. A 2 kg flower pot weighing 20 N falls from a window ledge.
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The force of the air resistance is 4 N.

The given parameters;

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  • weight of the flower pot, W = 20 N

Let the air resistance = F

Apply Newton's second law of motion to determine the force of the air resistance acting upward to oppose the motion of the pot falling downwards.

\Sigma F = ma\\\\W - F = ma\\\\a = \frac{W - F}{m} \\\\8 = \frac{20 - F}{2} \\\\20 - F = 16\\\\F = 20 - 16\\\\F = 4 \ N

Thus, the force of the air resistance is 4 N.

Learn more here: brainly.com/question/19887955

8 0
3 years ago
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