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Leya [2.2K]
3 years ago
11

A uranium nucleus is traveling at 0.94 c in the positive direction relative to the laboratory when it suddenly splits into two p

ieces. Piece A is propelled in the forward direction with a speed of 0.43 c relative to the original nucleus. Piece B is sent backward at 0.35 c relative to the original nucleus. Part A Find the velocity of piece A as measured by an observer in the laboratory. Do the same for piece B.
Physics
1 answer:
Anettt [7]3 years ago
6 0

Answer:

A   u = 0.36c      B u = 0.961c

Explanation:

In special relativity the transformation of velocities is carried out using the Lorentz equations, if the movement in the x direction remains

     u ’= (u-v) / (1- uv / c²)

Where u’ is the speed with respect to the mobile system, in this case the initial nucleus of uranium, u the speed with respect to the fixed system (the observer in the laboratory) and v the speed of the mobile system with respect to the laboratory

The data give is u ’= 0.43c and the initial core velocity v = 0.94c

Let's clear the speed with respect to the observer (u)

      u’ (1- u v / c²) = u -v

      u + u ’uv / c² = v - u’

      u (1 + u ’v / c²) = v - u’

      u = (v-u ’) / (1+ u’ v / c²)

Let's calculate

      u = (0.94 c - 0.43c) / (1+ 0.43c 0.94 c / c²)

      u = 0.51c / (1 + 0.4042)

      u = 0.36c

We repeat the calculation for the other piece

In this case u ’= - 0.35c

We calculate

       u = (0.94c + 0.35c) / (1 - 0.35c 0.94c / c²)

       u = 1.29c / (1- 0.329)

       u = 0.961c

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Answer:

θ  = 6.3 *10³ revolutions

Explanation:

Angular acceleration of the drill

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ωf= ω₀ + α*t  Formula (1)

Where:  

α : Angular acceleration (rad/s²)  

ω₀ : Initial angular speed ( rad/s)  

ωf : Final angular speed ( rad

t :  time interval (s)

Data

ω₀ = 0

ωf = 350000 rpm = 350000 rev/min

1 rev = 2π rad

1 min= 60 s

ωf = 350000 rev/min =350000*(2π rad/60 s)

ωf = 36651.9 rad/s

t = 2.2 s

We replace data in the formula (2) :

ωf= ω₀ + α*t

36651.9 = 0 + α* (2.2)

α = 36651.9 / (2.2)

α = 17000 rad/s²

Revolutions made by the drill

We apply the equations of circular motion uniformly accelerated

ωf²= ω₀ ²+ 2α*θ Formula (2)

Where:  

θ : Angle that the body has rotated in a given time interval (rad)

We replace data in the formula (2):  

(ωf)²= ω₀²+ 2α*θ

(36651.9)²= (0)²+ 2( 17000 )*θ

θ = (36651.9)²/ (34000 )

θ  = 39510.64 rad = 39510.64 rad* (1 rev/2πrad)

θ  = 6288.31 revolutions

θ  = 6.3 *10³ revolutions

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Explanation:

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1. A kangaroo hops 84 m to the east in 7 seconds.
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Explanation:

Given parameters:

Distance hopped  = 84m

Displacement  = 84m due east

Time  = 7s

Unknown:

Speed of kangaroo  = ?

Velocity of kangaroo  = ?

Solution:

To solve this problem,

    Speed  = \frac{distance}{time }   = \frac{84}{7}  = 12m/s

  Velocity  = \frac{displacement}{time}   = \frac{84}{7}   = 12m/s due east

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