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eduard
3 years ago
13

A heavy meterstick has a mass of 1 kg. When the meterstick is thrown like a spear past you, you measure its momentum to be 2mv.

What do you measure its length to be?
Physics
1 answer:
CaHeK987 [17]3 years ago
4 0

Answer:

Its length is measured to be 0.5 m

Explanation:

From theory of relativity (mass variation), we know that:

m = mo/√(1-v²/c²)

Where, m = relative mass

and, mo = rest mass

The momentum of stick while moving, will be:

P = mv

but, it is given in the form of rest mass as:

P = 2(mo)v

thus, by comparison;

2(mo)v = mv

using value of m from theory of relativity;

2(mo)v = (mo)v/√(1-v²/c²)

√(1-v²/c²) = 1/2     ______ eqn(1)

Now, for relativistic length (L), we have the formula from same theory of relativity;

L = (Lo)√(1-v²/c²)

The rest length (Lo) of meter stick is 1 m, and the remaining term on right side √(1-v²/c²), known as Lorentz Factor, can be given by eqn (1), as equal to 1/2.

Thus,

L = (1 m)(1/2)

<u>L = 0.5 m</u>

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nadya68 [22]

Answer:

F = 147,78*10⁻⁹ [N]

Explanation:

By symmetry the Fy components of the forces acting on charge in point x = 0,7 m canceled each other, and the total force will be twice Fx ( Fx is x axis component of one of the forces .

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tan β  =  0,5/0,7

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d = √0,25 + 0,49

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Now Foce between two charges  is:

F  = K* q₁*q₂/ d²      (1)

Where  K  = 9*10⁹ N*m²/C²

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q₂  = 3,0*10⁻⁹C

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Plugging these values in (1)

F  =  9*10⁹* 2,5* 10⁻⁹*3,0*10⁻⁹ / 0,74       [N*m²/C²]*C*C/m²

F = 91,21 * 10⁻⁹   [N]

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Fx  =  91,21 * 10⁻⁹ *0,81

Fx =73,89*10⁻⁹  [N]

Then total force acting on charge located at x = 0,7 m is:

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F = 2*73,89*10⁻⁹  [N]

F = 147,78*10⁻⁹ [N]

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Answer:

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