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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

An astronaut on a spaceship traveling at a speed of \(0.50 c\) is holding a meter stick parallel to the direction of motion. a) What is the length of the meter stick as measured by another astronaut on the spaceship? b) If an observer on Earth could observe the meter stick, what would be the length of the meter stick as measured by that observer?

Show that momentum and energy transform from one inertial frame to another as \(p_{x}^{\prime}=\gamma\left(p_{x}-v E / c^{2}\right) ; p_{y}^{\prime}=p_{y}\) \(p_{z}^{\prime}=p_{p} ; E^{\prime}=\gamma\left(E-v p_{x}\right) .\) Hint: Look at the derivation for the space-time Lorentz transformation.

Suppose NASA discovers a planet just like Earth orbiting a star just like the Sun. This planet is 35 light-years away from our Solar System. NASA quickly plans to send astronauts to this planet, but with the condition that the astronauts would not age more than 25 years during this journey. a) At what speed must the spaceship travel, in Earth's reference frame, so that the astronauts age 25 years during this journey? b) According to the astronauts, what will be the distance of their trip?

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 spaceship is traveling at two-thirds of the speed of light directly toward a stationary asteroid. If the spaceship turns on it headlights, what will be the speed of the light traveling from the spaceship to the asteroid as observed by a) someone on the spaceship? b) someone on the asteroid?
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