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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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Answer:

a) -35.6°C

b) 237.4 K

Explanation:

To convert temperature from degree celsius to degree fahrenheit, use the formula below:

T_c=\frac{5}{9}(T_f-32)

a) Therefore to convert -32°F to celsius, substitute it into the celsius

\begin{gathered} T_c=\frac{5}{9}(-32-32) \\  \\ T_c=\frac{5}{9}(-64) \\  \\ T_c=-35.6^0C \end{gathered}

b) To covert to the Kelvin scale, use the formula below

\begin{gathered} T_k=T_c+273 \\  \\ T_k=-35.6+273 \\  \\ T_k=237.4K \\  \end{gathered}

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1 year ago
A 15kg dog jumps out of a 40kg canoe. If the dogs velocity is 1.2m/s, what is the velocity of the canoe?
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Answer:

v = -0.45 m/s

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6 0
2 years ago
1 lb equals how many grams
shusha [124]

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A 41.0 g marble moving at 2.30 m/s strikes a 25.0 g marble at rest. What is the speed of each marble immediately after the colli
Gre4nikov [31]

Answer:

speed of each marble after collision will be 1.728 m/sec

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6 0
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