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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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The angular width of localizers varies between ______ and ______ degrees in order to provide a signal width of approximately 700
natali 33 [55]

Answer: 10 and 35 degrees

Explanation: Localizers width below 10 degree and 35 degree signal arc is unreliable and considered unusable for navigation and as a result, aircrafts may loose alignment

5 0
3 years ago
use the pendulum equation to calcuate the period of a 1.50 pendulum. Remeber that the vaule of "g" is 9.81 m/s² please help ​
kati45 [8]

Answer:

L = 3.51

Explanation:

Pendulum equation is T = 2pi\sqrt{L/9.81}

T = 1.5 and we are solving for L

1.5=2\pi \sqrt{L/9.81}

square both sides to get 2.25 = 2\pi ( L/9.81)

multiply both sides by 9.81 then divide by 2 and 3.14 as a substitue for pi. The answer should be about 3.51 in length

L = 3.51

If this helps, mark me brainliest pls

3 0
3 years ago
The a992 steel rod bc has a diameter of 50 mm and is used as a strut to support the beam. determine the maximum intensity w of t
EastWind [94]

Answer:

w = 11.211 KN/m

Explanation:

Given:

diameter, d = 50 mm

F.S = 2

L = 3

Due to symmetry, we have:

Ay = By = \frac{w * 6}{2} = 3w

P_c_r = 3w * F.S = 3w * 2.0 = 6w

I = \frac{\pi}{64} (0.05)^4 = 3.067*10^-^7

To find the maximum intensity, w, let's take the Pcr formula, we have:

P_c_r = \frac{\pi^2 E I}{(KL)^2}

Let's take k = 1

E = 200*10^9

Substituting figures, we have:

6w = \frac{\pi^2 * 200*10^9 * 3.067*10^-^7}{(1 * 3)^2}

Solving for w, we have:

w = \frac{67266.84}{6}

w = 11211.14 N/m = 11.211 KN/m

Since Area, A= pi * (0.05)²

\sigma _c_r = \frac{w}{A}

\sigma _c_r = \frac{11.211}{\pi (0.05)^2} = 1.4 MPA < \sigma y. This means it is safe

The maximum intensity w = 11.211KN/m

3 0
3 years ago
Which quantities below of a solid object on this planet are NOT the same as on Earth?
AysviL [449]

Answer:

Weight, acceleration when it falls vertically, are not same as that of earth.

Explanation:

Weight of the object is given by the product of mass of the object and the acceleration due to gravity of the planet.

So, the weight of object is not same as that on earth.

The mass is defined as the amount of matter contained in the object.

So, the mass of the object is same as that of earth.  

The volume of the object is defined as the space occupied by the object.

So, the volume of the object is same as that of earth.  

The density is defined as the ratio of mass of the object to its volume.

So, the density of the object is same as that of earth.  

The acceleration due to gravity on a planet depends on the mass of planet and radius of planet.

So, the acceleration is not same as that of earth.

The color of the object is its characteristic.

It is same as that of earth.

8 0
2 years ago
What is the oldest animal on earth?
Lilit [14]

Answer:

e3f3ewfeewfewgwgewggegegeggegeggege

Explanation:

6 0
3 years ago
Read 2 more answers
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