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Dafna11 [192]
2 years ago
14

how many years would it take to reach the star from earth, as measured by observers on the spacecraft

Physics
1 answer:
Assoli18 [71]2 years ago
3 0

In other words, it would take Deep Space 1 more than 81,000 years to travel the 4.24 light-years between Earth and Proxima Centauri at its top speed of 56,000 km/h. In relation to human history, that would be more than 2,700 generations.

Nearly 40 trillion kilometers, or 4.4 light-years, separate us from Alpha Centauri. The NASA-Germany Helios probes, the fastest spacecraft to date to be launched into orbit, flew at a speed of 250,000 kilometers per hour. The probes would need 18,000 years to travel at such pace to arrive at the sun's nearest neighbor. The calculations reveal that it is almost impossible to reach the nearest star in a human lifetime, even with the most futuristic technologies.

Learn more about Light year here-

brainly.com/question/1302132

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The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that
pav-90 [236]

Answer:

The permittivity of rubber is  \epsilon  = 8.703 *10^{-11}

Explanation:

From the question we are told that

     The  magnitude of the point charge is  q_1 =  70 \ nC  =  70 *10^{-9} \  C

      The diameter of the rubber shell is  d = 32 \ cm  =  0.32 \ m

       The Electric field inside the rubber shell is  E =  2500 \ N/ C

The radius of the rubber is  mathematically evaluated as

              r =  \frac{d}{2} =  \frac{0.32}{2}  =  0.16 \ m

Generally the electric field for a point  is in an insulator(rubber) is mathematically represented as

         E =  \frac{Q}{ \epsilon }  *  \frac{1}{4 *  \pi r^2}

Where \epsilon is the permittivity of rubber

    =>     E  *  \epsilon  *  4 * \pi *  r^2 =  Q

   =>      \epsilon  =  \frac{Q}{E *  4 *  \pi *  r^2}

substituting values

            \epsilon  =  \frac{70 *10^{-9}}{2500 *  4 *  3.142 *  (0.16)^2}

            \epsilon  = 8.703 *10^{-11}

7 0
3 years ago
How much heat is needed to warm 0.072kg of gold from 20 celsius and 90 celsius if the specific heat of gold 136 joules
dybincka [34]

Heat supplied to the gold will raise the temperature of the gold from 20 degree Celsius to 90 degree Celsius.

Mass of the gold (m) = 0.072 kg

Temperature change (ΔT) = 90 - 20 = 70 degree Celsius

Specific heat capacity of the gold (c) = 136 J/kg C

Heat supplied = m × c × ΔT

Heat supplied = 0.072 × 136 × 70

Heat supplied = 685.44 Joules

Hence, the heat supplied to the gold to raise the temperature from 20 degree Celsius to 90 degree Celsius = 685.44 Joules

5 0
3 years ago
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