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lubasha [3.4K]
3 years ago
11

An electron is accelerated within a particle accelerator using a 100 MV electric potential. The 100 MeV electron moves along an

evacuated tube of length 4 m, fixed to the laboratory frame.
What length of the tube would be measured by an observer moving with the electron?
Physics
1 answer:
Delicious77 [7]3 years ago
8 0

Answer:

The length of the tube is 3.92 m.

Explanation:

Given that,

Electric potential = 100 MV

Length = 4 m

Energy = 100 MeV

We need to calculate the value of \gamma

Using formula of relativistic energy

E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)

Put the value into the formula

1.6\times10^{-15}= 9.1\times`10^{-31}\times9\times10^{16}(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)

(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)=\dfrac{1.6\times10^{-15}}{9.1\times10^{-31}\times9\times10^{16}}

Here, \gamma-1=(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)

\gamma-1=0.01953

\gamma=0.01953+1

\gamma=1.01953

We need to calculate the length

Using formula of length

L'=\dfrac{L}{\gamma}

Put the value into the formula

L'=\dfrac{4}{1.01953}

L'=3.92\ m

Hence, The length of the tube is 3.92 m.

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To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105
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One uniform ladder of mass 30 kg and 10 m long rests against a frictionless vertical wall and makes an angle of 60o with the flo
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Answer:

   μ = 0.37

Explanation:

For this exercise we must use the translational and rotational equilibrium equations.

We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive

let's write the rotational equilibrium

           W₁  x/2 + W₂ x₂ - fr y = 0

where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances

             cos 60 = x / L

where L is the length of the ladder

              x = L cos 60

            sin 60 = y / L

           y = L sin60

the horizontal distance of man is

            cos 60 = x2 / 7.0

            x2 = 7 cos 60

we substitute

         m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0

         fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60

let's calculate

         fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)

         fr = (735 + 2450) / 8.66

         fr = 367.78 N

the friction force has the expression

         fr = μ N

write the translational equilibrium equation

         N - W₁ -W₂ = 0

         N = m₁ g + W₂

         N = 30 9.8 + 700

         N = 994 N

we clear the friction force from the eucacion

        μ = fr / N

        μ = 367.78 / 994

        μ = 0.37

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