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mina [271]
3 years ago
14

Why should photons emitted by a hotter gas have, on average, shorter wavelengths than photons emitted by a cooler gas?

Physics
1 answer:
harkovskaia [24]3 years ago
7 0

According to wein's law,

wavelength has inverse relation with temperature

so that's why hotter gas has shorter wavelength than cooler gas.

hope this answer is correct ....

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Due to historical difficulty in delivering supplies by plane, one of your colleagues has suggested you develop a catapult for sl
ikadub [295]

Answer:

Please see below as the answer is self-explanatory.

Explanation:

  • We can take the initial velocity vector, which magnitude is a given (67 m/s) and project it along two directions perpendicular each other, which we choose horizontal (coincident with x-axis, positive to the right), and vertical (coincident with y-axis, positive upward).
  • Both movements are independent each other, due to they are perpendicular.
  • In the horizontal direction, assuming no other forces acting, once launched, the supply must keep the speed constant.
  • Applying the definition of cosine of an angle, we can find the horizontal component of the initial velocity vector, as follows:

       v_{avgx} = v_{o}*cos 50 = 67 m/s * cos 50 = 43.1 m/s (1)

  • Applying the definition of average velocity, since we know the horizontal distance to the target, we can find the time needed to travel this distance, as follows:

       t = \frac{\Delta x}{v_{avgx} } = \frac{400m}{43.1m/s} = 9.3 s  (2)

  • In the vertical direction, once launched, the only influence on the supply is due to gravity, that accelerates it with a downward acceleration that we call g, which magnitude is 9.8 m/s2.
  • Since g is constant (close to the Earth's surface), we can use the following kinematic equation in order to find the vertical displacement at the same time t that we found above, as follows:

       \Delta y = v_{oy}  * t - \frac{1}{2} *g*t^{2} (3)

  • In this case, v₀y, is just the vertical component of the initial velocity, that we can find applying the definition of the sine of an angle, as follows:

       v_{oy} = v_{o}*sin 50 = 67 m/s * sin 50 = 51.3 m/s (4)

  • Replacing in (3) the values of t, g, and v₀y, we can find the vertical displacement at the time t, as follows:

       \Delta y = (53.1m/s * 9.3s) - \frac{1}{2} *9.8m/s2*(9.3s)^{2} = 53.5 m (5)

  • Since when the payload have traveled itself 400 m, it will be at a height of 53.5 m (higher than the target) we can conclude that the payload will be delivered safely to the drop site.
4 0
3 years ago
A tennis ball is thrown vertically upward with an initial velocity of +6.2 m/s. What will the ball’s velocity be when it returns
anastassius [24]

Answer:

v_{f}=-6.2 m/s

Explanation:

The ball will rise decreasing its speed until it reaches the highest point where its speed will be zero. From this point the tennis ball will begin to fall again, in the free fall the tennis ball will gain speed but now in the opposite direction. When it returns to the same point where it was launched, its speed will be the same as the one that was launched but with the opposite sign.

v_{f}=-6.2 m/s

We can check this using the equation:

v_{f}^2=v_{i}^2+2gh

where v_{i}=+6.2 m/s

ang h is the height, but because the ball returns to the same point where it started, h =0

then

v_{f}^2=v_{i}^2

v_{f}=v_{i}

the initial and final velocity will be the same in number, but we know that the ball is going in the opposite direction, so the final velocity must have the opposite sign from the initial velocity

so if v_{i}=+6.2 m/s,

v_{f}=-6.2 m/s

4 0
3 years ago
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