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Taya2010 [7]
3 years ago
10

A subatomic particle created in an experiment exists in a certain state for a time of before decaying into other particles. Appl

y both the Heisenberg uncertainty principle and the equivalence of energy and mass to determine the minimum uncertainty involved in measuring the mass of this short-lived particle.
Physics
1 answer:
dem82 [27]3 years ago
4 0

Answer:

Δm Δt> h ’/ 2c²

Explanation:

Heisenberg uncertainty principle, stable uncertainty of energy and time, with the expressions

     ΔE Δt> h ’/ 2

     h’= h / 2π

to relate this to the masses let's use Einstein's relationship

      E = m c²

let's replace

     Δ (mc²) Δt> h '/ 2

the speed of light is a constant that we can condense exact, so

      Δm Δt> h ’/ 2c²

     

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Elena-2011 [213]
Magma develops within the mantle or the crust when the temperature- pressure conditions favor the molten state. It rises toward the Earths surface when it is less dense than the surrounding rock or when a structural zone allows movement.
Hope this answer helps!:)
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3 years ago
Jake calculates that the frequency of a wave is 230 hertz, and its wave is moving at 460 m/s. What is the wavelength of the wave
Harrizon [31]
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7 0
3 years ago
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William tell must split the apple atop his son's head from a distance of 24 m. when he aims directly at the apple, the arrow is
mezya [45]
Let x = the angle of elevation for shooting the arrow.

Assume
g = 9.8 m/s²
No wind resistance
The vertical launch velocity is 25.1 sin(x) m/s
The horizontal velocity is 25.1 cos(x) m/s

The time of flight is
24/[25.1 cos(x)] s = 0.9562 sec(x) s

Therefore
0.5*[0.9562 sec(x)]*(9.8)  = 25.1 sin(x)
4.6854 = 25.1* sin(x)cos(x)
sin(2x) = 0.3733
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8 0
4 years ago
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What hanging mass will stretch a 3.0-m-long, 0.32 mm - diameter steel wire by 1.3 mm ? The Young's modulus of steel is 20×10^10
raketka [301]

Answer:

0.71 kg

Explanation:

L = length of the steel wire = 3.0 m

d = diameter of steel wire = 0.32 mm = 0.32 x 10⁻³ m

Area of cross-section of the steel wire is given as

A = (0.25) πd²

A = (0.25) (3.14) (0.32 x 10⁻³)²

A = 8.04 x 10⁻⁸ m²

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Y = Young's modulus of steel = 20 x 10¹⁰ Nm⁻²

m = mass hanging

F = weight of the mass hanging

Young's modulus of steel is given as

Y = \frac{FL}{A\Delta L}

20\times 10^{10} = \frac{F(3)}{(8.04\times 10^{-8})(1.3\times 10^{-3})}

F = 6.968 N

Weight of the hanging mass is given as

F = mg

6.968 = m (9.8)

m = 0.71 kg

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4 years ago
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erma4kov [3.2K]

a. magical i hope it help

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3 years ago
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