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grin007 [14]
2 years ago
9

A penny falls from a windowsill which is 25.0 m above the sidewalk. How much time does a passerby on the sidewalk below have to

move out of the way before the penny hits the ground?
Physics
1 answer:
Yakvenalex [24]2 years ago
3 0
T = √(h)/(0.5)(9.81)
t = √(25)/(4.905)
t = √5.1
t = 2.26 seconds

hope this helps and have a great day :)



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30 (kg)

Explanation:

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What would the speed of each particle be if it had the same wavelength as a photon of yellow light (????=575.0 nm)? Proton (mass
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Answer:

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Neutron: v=0.688 m/s

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De Broglie equation allows you to calculate the “wavelength” of an electron or any other particle or object of mass m that moves with velocity v:

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We know that the wavelength of the particle is 575 nm (575×10⁻⁹m), so we find the velocity v for each particle:

λ=\frac{h}{mv}

v=h÷(mλ)

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m=1.673×10⁻²⁴ g · \frac{1kg}{1000g}=1.673×10⁻²⁷ kg

v=h÷(mλ)

v=6.626×10⁻³⁴\frac{kg.m^2}{s}÷(1.673×10⁻²⁷ kg×575×10⁻⁹m)

v=0.689 m/s

<u>Neutron:</u>

m=1.675×10⁻²⁴ g · \frac{1kg}{1000g}=1.675×10⁻²⁷ kg

v=h÷(mλ)

v=6.626×10⁻³⁴\frac{kg.m^2}{s}÷(1.675×10⁻²⁷ kg×575×10⁻⁹m)

v=0.688 m/s

<u>Electron:</u>

m= 9.109×10⁻²⁸ g · \frac{1kg}{1000g}=9.109×10⁻³¹ kg

v=h÷(mλ)

v=6.626×10⁻³⁴\frac{kg.m^2}{s}÷(9.109×10⁻³¹ kg×575×10⁻⁹m)

v=1265.078 m/s

<u>Alpha particle:</u>

m=6.645×10⁻²⁴ g · \frac{1kg}{1000g}=6.645×10⁻²⁷ kg

v=h÷(mλ)

v=6.626×10⁻³⁴\frac{kg.m^2}{s}÷(6.645×10⁻²⁷ kg×575×10⁻⁹m)

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Answer:

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then

K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)

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⇒  U = 5.2581 J

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