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Lostsunrise [7]
3 years ago
12

Slow moving vehicles must display ___________ emblem at the rear to warn of their low speed.

Physics
2 answers:
Naddik [55]3 years ago
7 0

Answer:

Reflective emblem

Explanation:

A slow moving vehicle must display a reflective emblem at the rear to warn of their low speed. The sign of emblem is triangular in shape indicating that any sad incident have happened. Slow moving vehicles are generally prone to accident because of low visibility so they must display reflective emblem at rear.

KIM [24]3 years ago
6 0
<span>Slow moving vehicles must display reflective emblem at the rear to warn of their low speed.

The sign consists of a triangle and the primary purpose is to prevent any sad incident to happen, as slow moving vehicle (SMV) can cause serious accidents due to the less visibility at various times, therefore the sign must be visible enough.  </span>
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a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro
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Answer:

\theta=53.13^o

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

V_x=V_{ox}=V_ocos\theta

V_y=V_{oy}-gt=V_osin\theta-gt

x=V_{ox}t

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\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}

\displaystyle y_{max}=\frac{V_{oy}^2}{2g}

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

x_{max}=3y_{max}

\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}

Using the formulas for V_{ox}, V_{oy}:

\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}

Simplifying

4cos\theta sin\theta=3sin^2\theta

Dividing by sin\theta

4cos\theta=3sin\theta

Rearranging

tan\theta=\frac{4}{3}

\theta=arctan\frac{4}{3}

\theta=53.13^o

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