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Tpy6a [65]
3 years ago
15

Two spacecraft are both 10 million kilometers from a star. The total power output of the star is 4 x 1025 W. Spacecraft 1 has a

solar panel with a radius of 18 m. Spacecraft 2 has a solar panel with a radius of 6 m. What is the ratio of the power collected by the solar panel of Spacecraft 1 to the power collected by the solar panel of Spacecraft 2. a. 0.11 b. 0.06 c. 9 d. 18 e. 1 f. 0.33 g. 3
Physics
2 answers:
Ksenya-84 [330]3 years ago
5 0

The concept of power is given by the relationship between intensity and area, that is to say that power is defined as

P = A*I

Our values are given under the condition of,

r_1 = 18m

r_2 = m

The power is proportional to the Area, and in turn, we know that the Area of a circle is the product between \pi  times the radius squared, therefore the power is proportional to the radius squared.

\text{Power} \propto r^2

For both panels we would have to

\frac{\text{Power by panel 1}}{\text{Power by panel 2}} = \frac{r_1^2}{r_2^2}

\frac{P_1}{P_2} = (\frac{18}{6})^2

\frac{P_1}{P_2} = 9

Therefore the correct option is option C.9

balandron [24]3 years ago
5 0

Answer:

the correct answer is option (c) 9

Explanation:

solution;

Given data;

radius of spacecraft 1 = 18m

radius of spacecraft 1 = 6m

The total power output of the star = 4 x 1025 W

power is given by the formula;

P = I * A

where I is light intensity and A is the area

but power can be said to be proportional to area.

Since area = πr² therefore, power is proportional to r²

The equation becomes,

power of panel 1 /power of panel 2 = r₁²/r₂²

                                                          = 18²/6²

                                                          = 324/36

                                Ratio                     = 9

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Answer with Explanation:

We are given that

Diameter of fighter plane=2.3 m

Radius=r=\frac{d}{2}=\frac{2.3}{2}=1.15 m

a.We have to find the angular velocity in radians per second if it spins=1200 rev/min

Frequency=\frac{1200}{60}=20 Hz

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Angular velocity=\omega=2\pi f

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7 0
4 years ago
Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is
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Complete Question

A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.

If the acceleration of the projective is : a = c/s m/s​2

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?

Answer:

The value of the constant is  c = 28853.78 \ m^2 /s^2

Explanation:

From the question we are told that

         The acceleration is  a =  \frac{c}{s}\   m/s^2

         The  initial position of the projectile is s= 1.5m

         The final position of the projectile is s_f =  3 \ m

          The velocity is  v = 200 \ m/s

     Generally  time  =  \frac{ds}{dv}

   and  acceleration is a =  \frac{v}{time }

so

            a = v  \frac{dv}{ds}

 =>        vdv  =  a ds

             vdv  = \frac{c}{s}  ds

integrating both sides

           \int\limits^a_b  vdv  = \int\limits^c_d \frac{c}{s}  ds

Now for the limit

          a =  200 m/s

             b = 0 m/s  

         c = s= 3 m

          d =s_f= 1.5 m

So we have  

           \int\limits^{200}_{0}  vdv  = \int\limits^{3}_{1.5} \frac{c}{s}  ds

              [\frac{v^2}{2} ] \left | 200} \atop {0}} \right.  = c [ln s]\left | 3} \atop {1.5}} \right.

            \frac{200^2}{2}  =  c ln[\frac{3}{1.5} ]

=>           c = \frac{20000}{0.69315}

              c = 28853.78 \ m^2 /s^2

     

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