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Luden [163]
3 years ago
7

A space station sounds an alert signal at time intervals of 1.00 h . Spaceships A and B pass the station, both moving at 0.400c0

relative to the station but in opposite directions.
Part A
How long is the time interval between signals according to an observer on A?
Part B
How long is the time interval between signals according to an observer on B?
Part C
At what speed must A move relative to the station in order to measure a time interval of 2.00 hbetween signals?
Physics
1 answer:
Ann [662]3 years ago
6 0

Answer:

(A). The the time interval between signals according to an observer on A is 1.09 h.

(B). The time interval between signals according to an observer on B is 1.09 h.

(C). The speed is 0.866c.

Explanation:

Given that,

Time interval = 1.00 h

Speed = 0.400 c

(A). We need to calculate the the time interval between signals according to an observer on A

Using formula of time

\Delta t=\dfrac{\Delta t_{0}}{\sqrt{1-(\dfrac{v}{c})^2}}

Put the value into the formula

\Delta t=\dfrac{1.00}{\sqrt{1-(\dfrac{0.400c}{c})^2}}

\Delta t=\dfrac{1.00}{\sqrt{1-(0.400)^2}}

\Delta t=1.09\ h

(B). We need to calculate the time interval between signals according to an observer on B

Using formula of time

\Delta t=\dfrac{\Delta t_{0}}{\sqrt{1-(\dfrac{v}{c})^2}}

Put the value into the formula

\Delta t=\dfrac{1.00}{\sqrt{1-(\dfrac{0.400c}{c})^2}}

\Delta t=\dfrac{1.00}{\sqrt{1-(0.400)^2}}

\Delta t=1.09\ h

(C). Here, time interval of 2.00 h between signals.

We need to calculate the speed

Using formula of speed

\Delta t=\dfrac{\Delta t_{0}}{\sqrt{1-(\dfrac{v}{c})^2}}

Put the value into the formula

2.00=\dfrac{1.00}{\sqrt{1-(\dfrac{v}{c})^2}}

\sqrt{1-(\dfrac{v}{c})^2}=\dfrac{1.00}{2.00}

1-(\dfrac{v}{c})^2=(\dfrac{1.00}{2.00})^2

(\dfrac{v}{c})^2=\dfrac{3}{4}

v=\dfrac{\sqrt{3}}{2}c

v=0.866c

Hence, (A). The the time interval between signals according to an observer on A is 1.09 h.

(B). The time interval between signals according to an observer on B is 1.09 h.

(C). The speed is 0.866c.

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You want to produce three 2.00-mm-diametercylindrical wires, each with a resistance of 1.00 Ω at room temperature. One wire is g
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Answer:

(a) L =  128.75 m

(b) L =  182.56 m

(c) L =  114.28 m

(d) Mass of Gold = 7.68 kg = 7680 gram

(e) Cost of Gold Wire = $ 307040

Explanation:

The resistance of the wire is given as:

R = ρL/A

where,

R = Resistance

ρ = resistivity

L = Length

A = cross-sectional area

(a)

For Gold Wire:

ρ = 2.44 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (2.44 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.44 x 10⁻⁸ Ω.m)

<u>L =  128.75 m</u>

<u></u>

(b)

For Copper Wire:

ρ = 1.72 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (1.72 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(1.72 x 10⁻⁸ Ω.m)

<u>L =  182.56 m</u>

<u></u>

(c)

For Aluminum Wire:

ρ = 2.75 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (2.75 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.75 x 10⁻⁸ Ω.m)

<u>L =  114.28 m</u>

<u></u>

(d)

Density = Mass/Volume

Mass = (Density)(Volume)

Volume of Gold = AL = (3.14 x 10⁻⁶ m²)(128.75 m) = 4.04 x 10⁻⁴ m³

Therefore,

Mass of Gold = (1.9 x 10⁴ kg/m³)(4.04 x 10⁻⁴ m³)

<u>Mass of Gold = 7.68 kg = 7680 gram</u>

<u></u>

(e)

Cost of Gold Wire = (Unit Price of Gold)(Mass of Gold)

Cost of Gold Wire = ($ 40/gram)(7680 grams)

<u>Cost of Gold Wire = $ 307040</u>

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