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Alex Ar [27]
3 years ago
10

Twin A makes a one way trip at 0.6c to a star 12 light years away while twin B stays on Earth. Each twin sends the other a signa

l once a year by his own time. (a) How many signals does A send during the trip
Physics
1 answer:
hram777 [196]3 years ago
7 0

Answer:

8 signals received by twin A during the trip.

Explanation:

Given that,

Distance = 12 light year

Speed = 0.6 c

Time = 1 year

We need to calculate the time by A

Using formula of time

T=t\sqrt{\dfrac{1+\dfrac{v}{c}}{1-\dfrac{v}{c}}}

Put the value into the formula

T=1\sqrt{\dfrac{1+0.6}{1-0.6}}

T=2\ years

Similarly,

The expression for distance cover by A

D=d\sqrt{1-\dfrac{v^2}{c^2}}

D=12\sqrt{1-(0.6)^2}

D=9.6\ ly

We need to calculate the time

Using formula of time

t=\dfrac{D}{v}

t=\dfrac{9.6}{0.6}

t=16\ years

We need to calculate the signals received by twin A

Using formula for number of signals

n=\dfrac{t}{T}

Put the value into the formula

n=\dfrac{16}{2}

n=8\ signals

Hence, 8 signals received by twin A during the trip.

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A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of
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Answer:

4.5\times 10^{-5} T

Explanation:

We are given that

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Magnetic field=B_1=3.5\times 10^{-5} T( vertically downward)

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B_{wire}=B_2=\frac{\mu_0I}{2\pi R}

We have R=29 cm=\frac{29}{100}=0.29 m

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The resultant magnetic field is given by

B=\sqrt{B^2_1+B^2_2}

Substitute the values then we get

B=\sqrt{(3.5\times 10^{-5})^2+(2.76\times 10^{-5})^2}

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The resultant magnitude of magnetic field is same above and below the wire as it is at same distance.

The resultant magnitude of the magnetic field 29 cm below the wire=4.5\times 10^{-5} T

Hence, the resultant magnitude of the magnetic field 29 cm above  the wire=4.5\times 10^{-5} T

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1) A train accelerates from 36 km/hr to 54 km/hr in 10<br> s. Find acceleration?
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Answer: Given:

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Final velocity =54km/h=54x5/18=15m/s

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S=ut +1/2 at^2

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=100+25

=125m

So distance travelled 125m

Hope it helps you

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