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marishachu [46]
3 years ago
10

A mail carrier leaves the post office and drives 2.00 km to the north. He then drives in a direction 60.0° south of east for 7.00

km. After dropping off a package, he drives 9.50 km 35.0° north of east to Starbucks. What is the direction relative to the positive x axis?
Physics
1 answer:
den301095 [7]3 years ago
7 0

Answer:

\theta=7^o

Explanation:

<u>Displacement</u>

It is a vector that points to the final point where an object traveled from its starting point. If the object traveled to several points, then the individual displacements must be added as vectors.

The mail carrier leaves the post office and drives 2 km due north. The first displacement vector is

\vec r_1=\ km

Then the carrier drives 7 km in 60° south of east. The displacement has two components in the x and y axis given by

\vec r_2=\ km=\ km

Finally, he drives 9.5 km 35° north of east.

\vec r_3=\ km=\ km

The total displacement is

\vec r_t=\ km+\ km+\ km

\vec r_t=\ km

The direction can be calculated with

\displaystyle tan\theta=\frac{1.39}{11.28}=0.1232

\boxed{\theta=7^o}

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Answer:

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Translational kinetic energy = 200 J

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Explanation:

Rotational kinetic energy is given by

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For a solid sphere,

I = \frac{2}{5}mr^2

where <em>m</em> is its mass and <em>r</em> is its radius.

From the question,

<em>ω</em> = 49 rad/s

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\text{RKE} = \frac{1}{2}\times \frac{2}{5} mr^2\omega^2 = \frac{1}{5} mr^2\omega^2

\text{RKE} = \frac{1}{5} (0.15\ \text{kg})(0.037\ \text{m})^2(49\ \text{rad/s})^2 = 0.099\text{ J}

Translational kinetic energy is given by

\text{TKE} = \frac{1}{2} mv^2

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\text{TKE} = \frac{1}{2} (0.15\ \text{kg})(52\ \text{m/s})^2 = 200\text{ J}

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Learn more about Density here brainly.com/question/1354972

 

#SPJ9

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