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rewona [7]
2 years ago
13

A proton moves along the x-axis in the laboratory with velocity uy = 0.6c. An observer moves with a velocity of v=0.8c along the

direction of motion of the proton. a. What is the total energy of the proton in the lab frame? b. What kinetic energy does the observer measure for the proton? c. What momentum (in units of MeV/c) does the observer measure for the protons?
Physics
1 answer:
Dima020 [189]2 years ago
3 0

Explanation:

Given that,

Velocity of the proton in lab frame u_{x} = 0.6c

Velocity of the observer v= 0.8c

We need to calculate the velocity of the proton with respect to the observer

Using formula of velocity

u'=\dfrac{v-u_{x}}{1-\dfrac{u_{x}v}{c^2}}

u'=\dfrac{0.8c-0.6c}{1-\dfrac{0.6c\times0.8c}{c^2}}

u'=c(\dfrac{0.8-0.6}{1-(0.8\times0.6)})

u'=0.385c

(a). We need to calculate the total energy of the proton in the lab frame

Using formula of kinetic energy

K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u_{x}^2}{c^2}}}-1)

Where, Proton mass energy = m₀c²

Put the value into the formula

K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.6c)^2}{c^2}}}-1)

K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.6)^2}}-1)

K.E=234.57\ MeV

(b). We need to calculate the kinetic energy of the proton in the observer

Using formula of kinetic energy

K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u'^2}{c^2}}}-1)

K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.385)^2}{c^2}}}-1)

K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.385)^2}}-1)

K.E=78.366\ MeV

(c). We need to calculate the momentum of the proton with respect to observer

Using formula of momentum

P_{obs}=\dfrac{m_{0}u'^2}{\sqrt{1-\dfrac{u'^2}{c^2}}}

We know that,

Proton mass energy = m₀c²

m_{0}=\dfrac{938.28}{c^2}

P_{obs}=\dfrac{\dfrac{938.28}{c^2}\times(0.385c)^2}{\sqrt{1-\dfrac{(0.385c)^2}{c^2}}}

P_{obs}=\dfrac{938.28\times(0.385)^2}{\sqrt{1-0.385^2}}

P_{obs}=150.69\ MeV/c

Hence, This is required solution.

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Answer:

0.75%

Explanation:

Measured value of melting point of potassium thiocyanate = 174.5 °C

Actual value of melting point of potassium thiocyanate = 173.2 °C

<em>Error in the reading = |Experimental value - Theoretical value|</em>

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47 m

Explanation:

Data obtained from the question include the following:

Length of dry leg 1 (L1) = 40 m

Length of dry leg 2 (L2) = 25 m

Length of swimming course (L) =..?

The length of the swimming course can be obtained by using pythagoras theory as shown below:

L² = L1² + L2²

L² = 40² + 25²

L² = 1600 + 625

L² = 2225

Take the square root of both side.

L = √2225

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The membrane that surrounds a certain type of living cell has a surface area of 5.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A
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Answer:

2.1\times 10^{-12} c

Explanation:

We are given that

Surface area of membrane=5.3\times 10^{-9} m^2

Thickness of membrane=1.1\times 10^{-8} m

Assume that membrane behave like a parallel plate capacitor.

Dielectric constant=5.9

Potential difference between surfaces=85.9 mV

We have to find the charge resides on the outer surface of membrane.

Capacitance between parallel plate capacitor is given by

C=\frac{k\epsilon_0 A}{d}

Substitute the values then we get

Capacitance between parallel plate capacitor=\frac{5.9\times 8.85\times 10^{-12}\times 5.3\times 10^{-9}}{1.1\times 10^{-8}}

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Hence, the charge resides on the outer surface=2.1\times 10^{-12} c

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