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juin [17]
3 years ago
10

In an "atom smasher," two particles collide head on at relativistic speeds. If the velocity of the first particle is 0.741c to t

he left, and the velocity of the second particle is 0.543c to the right (both as measured in the lab rest frame), how fast are the particles moving with respect to each other?
Physics
1 answer:
USPshnik [31]3 years ago
5 0

Answer:

W_x = 0.9156\ c

Explanation:

given,

velocity of particle 1 = 0.741 c to left

velocity of second particle = 0.543 c to right

relative velocity between the particle = ?

for the relative velocity calculation we have formula

W_x = \dfrac{|u_x - v_x|}{1-\dfrac{u_xv_x}{c^2}}

u_x = 0.543 c

v_x = - 0.741 c

W_x = \dfrac{0.543 c - (-0.741 c)}{1-\dfrac{(0.543 c)(-0.741 c)}{c^2}}

W_x = \dfrac{0.543 c +0.741 c)}{1+\dfrac{(0.543)(0.741)c^2}{c^2}}

W_x = \dfrac{1.284c}{1+0.402363}

W_x = 0.9156\ c

Relative velocity of the particle is W_x = 0.9156\ c

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A wire is stretched between two posts. Another wire is stretched between two posts that are twice as far apart. The tension in t
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Since the second wire is twice as far apart as the first, it will be L2 = 2L1

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V1 = √T/M1... 1

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Since V1 = 249m/s, M1 = m/L1 M2 = m/L2 = m/2L1

The equations will now become

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249²/V2² = TL1/m×m/2TL1

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Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center
kiruha [24]

Answer:

The angular  velocity is w_f =  1.531 \ rad/ s

Explanation:

From the question we are told that

     The mass of each astronauts is  m =  50 \ kg

      The initial  distance between the two  astronauts  d_i  =  7 \  m

Generally the radius is mathematically represented as r_i  =  \frac{d_i}{2} = \frac{7}{2}  =  3.5 \  m

      The initial  angular velocity is  w_1 = 0.5 \  rad /s

       The  distance between the two astronauts after the rope is pulled is d_f =  4 \  m

Generally the radius is mathematically represented as r_f  =  \frac{d_f}{2} = \frac{4}{2}  =  2\  m

Generally from the law of angular momentum conservation we have that

           I_{k_1} w_{k_1}+ I_{p_1} w_{p_1} = I_{k_2} w_{k_2}+ I_{p_2} w_{p_2}

Here I_{k_1 } is the initial moment of inertia of the first astronauts which is equal to I_{p_1} the initial moment of inertia of the second astronauts  So

      I_{k_1} = I_{p_1 } =  m *  r_i^2

Also   w_{k_1 } is the initial angular velocity of the first astronauts which is equal to w_{p_1} the initial angular velocity of the second astronauts  So

      w_{k_1} =w_{p_1 } = w_1

Here I_{k_2 } is the final moment of inertia of the first astronauts which is equal to I_{p_2} the final moment of inertia of the second astronauts  So

      I_{k_2} = I_{p_2} =  m *  r_f^2

Also   w_{k_2 } is the final angular velocity of the first astronauts which is equal to w_{p_2} the  final angular velocity of the second astronauts  So

      w_{k_2} =w_{p_2 } = w_2

So

      mr_i^2 w_1 + mr_i^2 w_1 = mr_f^2 w_2 + mr_f^2 w_2

=>   2 mr_i^2 w_1 = 2 mr_f^2 w_2

=>   w_f =  \frac{2 * m * r_i^2 w_1}{2 * m *  r_f^2 }

=>    w_f =  \frac{3.5^2 *  0.5}{  2^2 }

=>   w_f =  1.531 \ rad/ s

       

3 0
2 years ago
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