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GarryVolchara [31]
3 years ago
8

If you want to delay a pulse of light in a laser experiment, you can send the light through a long coil of fiber optic cable. Li

ght travels somewhat slower in the glass core of a fiber than it does in vacuum. We will approximate the speed of light in the fiber as 2.04 x 108 m/s. What length of fiber (in meters) should you use if you want to delay the arrival of light by 557 ns
Physics
1 answer:
Grace [21]3 years ago
8 0

Answer:

d = 113.6 m

Explanation:

For this exercise, the first thing we must notice is that the speed of the laser beam in the fiber is constant, so we can use the uniform motion relationships to find the necessary distance

          v = d / t

let's reduce to SI units

         t = 557 ns = 557 10⁻⁹ s

         d = v t

         d = 2.04 10⁸ 557 10⁻⁹

         d = 1.136 102 m

         d = 113.6 m

This is the distance of the fiber for the laser to arrive with the desired delay

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what device is made of coils of wire with an electrical current passing through it in order to generate a magnetic field
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Part A What will be the equilibrium temperature when a 227 g block of copper at 283 °C is placed in a 155 g aluminum calorimeter
stellarik [79]

Answer:

T = 20.84°C

Explanation:

From the law of conservation of energy:

Heat Lost by Copper Block = Heat Gained by Aluminum Calorimeter + Heat Gained by Water

m_cC_c\Delta T_c = m_wC_w\Delta T_w + m_aC_a\Delta T_a

where,

m_c = mass of copper = 227 g

m_w = mass of water = 844 g

m_a = mass of aluminum = 155 g

C_c = specific heat capacity of calorimeter = 385 J/kg.°C

C_w = specific heat capacity of water = 4200 J/kg.°C

C_a = specific heat capacity of aluminum = 890 J/kg.°C

\Delta T_c = change in temperature of copper = 283°C - T

\Delta T_w = change in temperature of water = T - 14.6°C

\Delta T_a = change in temperature of aluminum = T - 14.6°C

T = equilibrium temperature = ?

Therefore,

(227\ g)(385\ J/kg.^oC)(283^oC-T)=(844\ g)(4200\ J/kg.^oC)(T-14.6^oC)+(155\ g)(890\ J/kg.^oC)(T-14.6^oC)\\\\24732785\ J - (87395\ J/^oC) T = (3544800\ J/^oC) T - 51754080\ J+ (137950\ J/^oC) T-2014070\ J\\\\24732785\ J +51754080\ J+2014070\ J = (3544800\ J/^oC) T+(137950\ J/^oC+(87395\ J/^oC) T\\\\78560935\ J = (3770145\ J/^oC) T\\\\T = \frac{78560935\ J}{3770145\ J/^oC}

<u>T = 20.84°C</u>

8 0
3 years ago
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