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Papessa [141]
4 years ago
12

suppose that a particle at rest with mass m decays into a photon and another particle with mass m/2 , which recoilds in a direct

ion opposite to the photon. Using the conservation of energy and momentum find the momentum of the recoiling particle
Physics
1 answer:
Nady [450]4 years ago
5 0

Answer:

p_r=\frac{h}{\lambda}

Explanation:

To find the momentum of the recoiling particle you can use the momentum formula for a photon:

p=\frac{h}{\lambda}

before the decay the momentum is zero. Hence, after the decay the momentum of the photon plus the momentum of the recoiling particle must be zero:

p_a=p_b\\\\\frac{h}{\lambda}-\frac{m}{2}v=0\\\\p_r=\frac{m}{2}v=\frac{h}{\lambda}

where pr is the momentum of the recoiling particle.

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Answer:

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4 years ago
Residential building codes typically require the use of 12-gauge copper wire (diameter 0.2053 cm) for wiring receptacles. Such c
AysviL [449]

Answer:

a) E = 4.26 W

b) E' = 6.724 W

c) copper wire is the safer option to use.

Explanation:

Given:

Diameter of the 12 gauge copper wire, d = 0.2053 cm

Thus, Radius of the 12 gauge copper wire, r = 0.2053 cm

/ 2 = 0.10265 cm  = 0.10265 × 10⁻² m

Now.

the area (A) comes out as

A = π × (0.10265 × 10⁻²)²

A = 3.3103 × 10⁻⁶ m²

Length of the copper wire, L = 2.10 m

a) The resisitivity (ρ) of copper = 1.68 × 10⁻⁸ ohm m

Now,

the resistance of the copper , R = ρL/A

or

R = (1.68 × 10⁻⁸ × 2.1) / ( 3.3103 × 10⁻⁶)

or

R = 0.01065 ohm

The Energy (E) is given as,

E = I²R

where, I is the current

I = 20.0 A

on substituting the values, we get

E = 20.0² × 0.01065

E = 4.26 W

(b) For the aluminium

Resisitivity, ρ' = 2.65 × 10⁻⁸  ohm m

Now, the resistance of the aluminium wire, R' = (ρ' × L) / A

Since the cross-section of the aluminium wire is same as the copper wire

thus,

R = (2.65 × 10⁻⁸ × 2.1) / ( 3.3103 × 10⁻⁶)

or

R = 0.0168 ohm

Therefore,

The Rate of energy produced by the aluminium wire, E' = I²R'

or

E' = 20.0² ×  0.0168

or

E' = 6.724 W

(c) From the above results, we can conclude that the power consumed or the rate of energy produced by the aluminium wire is more.

Hence, copper wire is the safer option to use.

8 0
4 years ago
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3 0
3 years ago
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Aliun [14]
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W=Fd \cos \theta
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6 0
3 years ago
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