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Papessa [141]
2 years ago
9

In a closed system, the loss of momentum of one object_____ the gain in momentum of another object.

Physics
1 answer:
densk [106]2 years ago
6 0

In a closed system, the loss of momentum of one object  is same as________ the gain in momentum of another object

according to law of conservation of momentum, total momentum before and after collision in a closed system in absence of any net external force, remains conserved . that is

total momentum before collision = total momentum after collision

P₁ + P₂ = P'₁ + P'₂

where P₁ and P₂ are momentum before collision for object 1 and object 2 respectively.

P'₁ - P₁  = - (P'₂ -  P₂)

so clearly gain in momentum of one object is same as the loss of momentum of other object

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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:

Proton: v=0.689 m/s

Neutron: v=0.688 m/s

Electron: v=1265.078 m/s

Alpha particle: v=0.173 m/s

Explanation:

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:

λ=\frac{h}{mv}

h is the Planck constant: 6.626×10⁻³⁴\frac{kg.m^2}{s}

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}

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<u>Proton:</u>

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)

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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

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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)

v=0.173 m/s

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