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Serhud [2]
3 years ago
13

Which light is most sensitive to the eyes?

Physics
1 answer:
san4es73 [151]3 years ago
6 0

Answer:

Our eyes are most sensitive to the wavelengths corresponding to the yellow and green colors of the spectrum. Flashy signs and some fire engines are painted in a yellowish-green color to attract our attention.

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A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface.
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Answer:

Kinetic energy of the projectile at the vertex of the trajectory: 900\; {\rm J}.

Work done when firing this projectile: 2500\; {\rm J}.

Explanation:

Since the drag on this projectile is negligible, the horizontal velocity v_{x} of this projectile would stay the same (at 30\; {\rm m\cdot s^{-1}}) throughout the flight.

The vertical velocity v_{y} of this projectile would be 0\; {\rm m\cdot s^{-1}} at the vertex (highest point) of its trajectory. (Otherwise, if v_{y} > 0, this projectile would continue moving up and reach an even higher point. If v_{y} < 0, the projectile would be moving downwards, meaning that its previous location was higher than the current one.)

Overall, the velocity of this projectile would be v = 30\; {\rm m\cdot s^{-1}}\! when it is at the top of the trajectory. The kinetic energy \text{KE} of this projectile (mass m = 2.0\; {\rm kg}) at the vertex of its trajectory would be:

\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (30\; {\rm m\cdot s^{-1}})^{2} \\ &= 900\; {\rm J} \end{aligned}.

Apply the Pythagorean Theorem to find the initial speed of this projectile:

\begin{aligned}v &= \sqrt{(v_{x})^{2} + (v_{y})^{2}} \\ &= \left(\sqrt{900 + 1600}\right)\; {\rm m\cdot s^{-1}} \\ &= 50\; {\rm m\cdot s^{-1}}\end{aligned}.

Hence, the initial kinetic energy \text{KE} of this projectile would be:

\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (50\; {\rm m\cdot s^{-1}})^{2} \\ &=2500\; {\rm J} \end{aligned}.

All that energy was from the work done in launching this projectile. Hence, the (useful) work done in launching this projectile would be 2500\; {\rm J}.

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