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Chapter 35: Problem 68

At what speed will the length of a meter stick look \(90.0 \mathrm{~cm} ?\)

Short Answer

Expert verified
Answer: Approximately \(1.35 \times 10^8 \mathrm{\frac{m}{s}}\).

Step by step solution

01

Write down the length contraction formula

The length contraction formula is given by: $$ L = L_0\sqrt{1 - \frac{v^2}{c^2}} $$ where \(L\) is the contracted length, \(L_0\) is the original length (in this case, 1 meter or 100 cm), \(v\) is the relative speed of the observer to the meter stick, and \(c\) is the speed of light.
02

Plug in the given values

We want the length to be \(90.0 \mathrm{~cm}\), so substitute it into the equation: $$ 90 = 100\sqrt{1 - \frac{v^2}{c^2}} $$
03

Solve for \(v^2\)

Divide both sides of the equation by 100, and then square both sides to get rid of the square root: $$ \frac{9}{10} = \sqrt{1 - \frac{v^2}{c^2}} $$ $$ \left(\frac{9}{10}\right)^2 = 1 - \frac{v^2}{c^2} $$
04

Solve for \(v\)

Rearrange the equation to have \(v^2\) terms on the left and the constants on the right, then take the square root: $$ v^2 = c^2 \left(1 - \left(\frac{9}{10}\right)^2\right) $$ $$ v = c \sqrt{1 - \left(\frac{9}{10}\right)^2} $$
05

Calculate the speed

Now, substitute the speed of light, \(c = 3 \times 10^8 \mathrm{\frac{m}{s}}\), into the equation and compute the value of \(v\): $$ v = (3 \times 10^8 \mathrm{\frac{m}{s}}) \sqrt{1 - \left(\frac{9}{10}\right)^2} $$ $$ v \approx 1.35 \times 10^8 \mathrm{\frac{m}{s}} $$ Therefore, the length of the meter stick will look \(90.0 \mathrm{~cm}\) when it is moving at a speed of approximately \(1.35 \times 10^8 \mathrm{\frac{m}{s}}\).

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Most popular questions from this chapter

The most important fact we learned about aether is that: a) It was experimentally proven not to exist. b) Its existence was proven experimentally. c) It transmits light in all directions equally. d) It transmits light faster in longitudinal direction. e) It transmits light slower in longitudinal direction.

A square of area \(100 \mathrm{~m}^{2}\) that is at rest in the reference frame is moving with a speed \((\sqrt{3} / 2) c\). Which of the following statements is incorrect? a) \(\beta=\sqrt{3} / 2\) b) \(\gamma=2\) c) To an observer at rest, it looks like another square with an area less than \(100 \mathrm{~m}^{2}\) d) The length along the moving direction is contracted by a factor of \(\frac{1}{2}\)

Consider a positively charged particle moving at constant speed parallel to a current-carrying wire, in the direction of the current. As you know (after studying Chapters 27 and 28), the particle is attracted to the wire by the magnetic force due to the current. Now suppose another observer moves along with the particle, so according to him the particle is at rest. Of course, a particle at rest feels no magnetic force. Does that observer see the particle attracted to the wire or not? How can that be? (Either answer seems to lead to a contradiction: If the particle is attracted, it must be by an electric force because there is no magnetic force, but there is no electric field from a neutral wire; if the particle is not attracted, you see that the particle is, in fact, moving toward the wire.)

In an elementary-particle experiment, a particle of mass \(m\) is fired, with momentum \(m c\), at a target particle of mass \(2 \sqrt{2} m\). The two particles form a single new particle (completely inelastic collision). Find: a) the speed of the projectile before the collision b) the mass of the new particle c) the speed of the new particle after the collision

A rocket ship approaching Earth at \(0.90 c\) fires a missile toward Earth with a speed of \(0.50 c,\) relative to the rocket ship. As viewed from Earth, how fast is the missile approaching Earth?
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