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Elina [12.6K]
3 years ago
9

A crow is flying horizontally with a constant speed of 2.45 m/s when it releases a clam from its beak. The clam lands on the roc

ky beach 2.30 s later. Consider the moment just before the clam lands. (Neglect air resistance.). . a)What is its horizontal component of velocity? (in m/s). (b) What is its vertical component of velocity? (in m/s)
Physics
1 answer:
Bezzdna [24]3 years ago
3 0
<span>Neglecting air resistance, the horizontal velocity will not have changed (i.e. it remains 2.45 m/s). The vertical velocity starts at 0 m/s and accelerates for 2.30 s (due to gravity). 9.8 = (x-0)/2.3 x = 22.54 Therefore your vertical velocity is 22.54 m/s.
</span>a) It would remain the same, because there is no force, and thus acceleration, in that direction. So, it would just be 2.45 m/s. b) Use<span>v=<span>v0</span>+at</span> Because <span>v0</span><span>=0 in the y direction, it is just v=at It said that t=2.3 s, and a=9.8. Plug that in... v=2.3*9.8=22.54 m/s</span>
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The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .
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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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A ball with a mass of 5 kg is accelerating at 5 m/s/s. What is the force acting on the ball?
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In the writing of ionic chemical formulas the value of each ion's charge is crossed over in the crossover rule.

Rules for naming Ionic compounds

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    The cation (element with a negative charge) is written first in the name then the anion(element with a positive charge) is written second in the name.
  • Second rule
    When the formula unit contains two or more of the same polyatomic ion, that ion is written in parentheses with the subscript written outside the parentheses.
    Example: Sodium carbonate is written as Na₂CO₃ not Na₂(CO)₃
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    If the cation is a metal ion with a fixed charge then the name of the cation will remain the same as the (neutral) element from which it is derived (Example: Na+ will be sodium).
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The oxidation state of each ion is also important, thus in the crossover rule, the value of each ion's charge is crossed over.

Learn more about chemical formulas here:

<u>brainly.com/question/11995171</u>

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