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alex41 [277]
3 years ago
12

A hungry lion is looking for food. It went 5 miles (mi) before it spotted a zebra to stalk. It took this lion 0.2 hours to get t

o the zebra. What is the lion’s speed?
Question 1 options:

1 mi


1 mi/h


25 mi


25 mi/h
Physics
1 answer:
Helen [10]3 years ago
5 0

To solve this exercise, the following speed equation must be applied:

 V = d / t (mi/hr)

 V: speed (mi/hr)

 d: distance (mi)

 t: time (s)

We know the distance (d= 5 miles) and the time (t= 0.2 hours), so we proceed to replace these values in the formula to calculate the lion’s speed:

 V = d / t

 V = 5 mi/0.2 hr

 V = 25 mi/hr

 The lion’s speed is 25 mi/hr, so the last option is the right answer

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Answer: 90\°

Explanation:

The Compton Shift \Delta \lambda in wavelength when the photons are scattered is given by the following equation:

\Delta \lambda=\lambda_{c}(1-cos\theta)     (1)

Where:

\lambda_{c}=2.43(10)^{-12} m is a constant whose value is given by \frac{h}{m_{e}c}, being h the Planck constant, m_{e} the mass of the electron and c the speed of light in vacuum.

\theta) the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through 180\°:

\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))     (2)

\Delta \lambda_{max}=\lambda_{c}(1-(-1))    

\Delta \lambda_{max}=2\lambda_{c}     (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)      (4)

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)      (5)

If we want \frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}, 1-cos\theta   must be equal to 1:

1-cos\theta=1   (6)

Finding \theta:

1-1=cos\theta

0=cos\theta  

\theta=cos^{-1} (0)  

Finally:

\theta=90\°    This is the scattering angle that will produce \frac{1}{4}\Delta \lambda_{max}      

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