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guajiro [1.7K]
2 years ago
11

g A high-energy photon turns into and electron and a positron. (A positron has exactly the same mass as the electron, but opposi

te charge.) What is the longest wavelength that the photon can have for this process to occur
Physics
1 answer:
yawa3891 [41]2 years ago
5 0

Answer:

2 m = E / c^2      where m is mass of electron

E = h v     where v is the frequency ( nu) of the incident photon

E = h c / y      where y is the incident wavelength (lambda)

2 m = h / (c y)

y = h / (2 m c)      wavelength required

y = 6.62 * 10E-34 / (2 * 9.1 * 10E-31 * 3 * 10E8)  m

y = 3.31 / 27.3 E-11 m

y = 1.21 E -12 m   = .0121 Angstrom units

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if an object is moving at a constant speed in a changing direction, What is the acceleration of the object (zero or not zero)​
Andreas93 [3]
  • We know, acceleration is the change of velocity by time.
  • Velocity is the speed of an object which also indicates the direction.
  • Hence, acceleration is both dependant upon the speed as well as the direction.
  • So, if an object is moving at a constant speed in a changing direction, the acceleration will also change. It will not be zero.
  • An example is that of uniform circular motion.

Answer:

if an object is moving at a constant speed in a changing direction, the acceleration of the object will not be zero.

4 0
2 years ago
I don't understand this question at all, can I please get some help?
beks73 [17]
V^2/R=180W
v=root 180R
4 0
3 years ago
The scientific method _____.
Lubov Fominskaja [6]
The answer is D I took the test
3 0
2 years ago
A rock is thrown horizontally from a high building at 33.8 m/s. What is the magnitude of its velocity 4.25 s later?
Alex17521 [72]
<h2>Answer:53.63ms^{-2}</h2>

Explanation:

The equations of motion used in this question is v=u+at

When a object is projected horizontally from a sufficiently height,the x-component of acceleration remains zero because there is no force that drags the object in x direction.

But,due to gravity,the object accelerates downward at a rate of 9.8ms^{-2}.

In X-Direction,

Given that initial velocity=u_{x}=33.8ms^{-1}

Using v=u+at,

v_{x}=33.8+(0)4.25=33.8ms^{-1}

In Y-Direction,

Given that initial velocity=u_{x}=0ms^{-1}

Using v=u+at,

v_{y}=0+(9.8)4.25=41.65ms^{-1}

v=\sqrt{v_{x}^{2}+v_{y}^{2}}

v=\sqrt{1142.44+1734.72}=\sqrt{2877.163}=53.63ms^{-1}

7 0
3 years ago
If you push any floating object down from equilibrium and release it, it bobs up and down. That looks like an oscillation, so le
GarryVolchara [31]

Answer:

  F_{y} = ( ρ_fluid g A) y

Explanation:

This exercise can be solved in two parts, the first finding the equilibrium force and the second finding the oscillating force

for the first part, let's write Newton's equilibrium equation

        B₀ - W = 0

        B₀ = W

        ρ_fluid g V_fluid = W

the volume of the fluid is the area of ​​the cube times the height it is submerged

      V_fluid = A y  

For the second part, the body introduces a quantity and below this equilibrium point, the equation is

        B - W = m a

        ρ_fluid g A (y₀ + y) - W = m a

        ρ_fluid g A y + (ρ_fluid g A y₀ -W) = m a

       ρ_fluid g A y + (B₀-W) = ma

the part in parentheses is zero since it is the force when it is in equilibrium

      ρ_fluid g A y = m a

      this equation the net force is

      F_{y} = ( ρ_fluid g A) y

we can see that this force varies linearly the distance and measured from the equilibrium position

8 0
2 years ago
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