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Pani-rosa [81]
3 years ago
6

The intensity of the Sun's light in the vicinity of the Earth is about 1000W/m^2. Imagine a spacecraft with a mirrored square sa

il of dimension 5.5 km. Estimate how much thrust (in newtons) this craft will experience due to collisions with the Sun's photons. [Hint: assume the photons bounce off the sail with no change in the magnitude of their momentum.}
Physics
1 answer:
NeX [460]3 years ago
8 0

Explanation:

Formula to represent thrust is as follows.

             F = \frac{dP}{dt}

                = \frac{2p}{dt}

or,           p = \frac{E}{c}

         \frac{p}{dt} = \frac{W}{c}

                  F = \frac{2IA}{c}

                     = \frac{2 (1000 W/m^{2})(5.5 \times 10^{3} m)^{2}}{3 \times 10^{8} m/s}

                     = 201.67 N

Thus, we can conclude that the thrust is 201.67 N.

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A tug-of-war game is played by five c/hildren: three on one team and two on the other. How much force will the two child team ha
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Moth X is more likely to be eaten by insect killing birds first. This is because Moth Y blends in with the tree trunk more and is hidden from birds. Its wings camouflage with the tree trunk, hiding it from sight. 
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A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh
Elanso [62]

Answer:

The value is  v  =  2.3359 *10^{4} \ m/s

Explanation:

From the question we are told that

  The  initial speed is u =  2.05 *10^{4} \  m/s

 Generally the total energy possessed by the space probe when on earth is mathematically represented as

             T__{E}} =  KE__{i}} +  KE__{e}}

Here  KE_i is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

          KE_i =   \frac{1}{2}  *  m  *  u^2

=>       KE_i =   \frac{1}{2}  *  m  *  (2.05 *10^{4})^2

=>       KE_i =  2.101 *10^{8} \ \ m \ \ J

And  KE_e is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

       KE_e =  \frac{1}{2}  *  m *  v_e^2

Here v_e is the escape velocity from earth which has a value v_e =  11.2 *10^{3} \  m/s

=>    KE_e =  \frac{1}{2}  *  m *  (11.3 *10^{3})^2

=>    KE_e =  6.272 *10^{7} \  \  m  \ \   J

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

        KE_p =  \frac{1}{2}  *  m *  v^2

Generally from the law energy conservation we have that

        T__{E}} =  KE_p

So

       2.101 *10^{8}  m  +  6.272 *10^{7}  m  =   \frac{1}{2}  *  m *  v^2

=>     5.4564 *10^{8} =   v^2

=>     v =  \sqrt{5.4564 *10^{8}}

=>     v  =  2.3359 *10^{4} \ m/s

4 0
2 years ago
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