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WARRIOR [948]
2 years ago
9

A large solar panel on a spacecraft in Earth orbit produces 1.0 kW of power when the panel is turned toward the sun. What power

would the solar cell produce if the spacecraft were in orbit around Saturn, 9.5 times as far from the sun?
Physics
1 answer:
Mandarinka [93]2 years ago
3 0

Answer:

e*P_s = 11 W

Explanation:

Given:

- e*P = 1.0 KW

- r_s = 9.5*r_e

- e is the efficiency of the panels

Find:

What power would the solar cell produce if the spacecraft were in orbit around Saturn

Solution:

- We use the relation between the intensity I and distance of light:

                                  I_1 / I_2 = ( r_2 / r_1 ) ^2

- The intensity of sun light at Saturn's orbit can be expressed as:

                                  I_s = I_e * ( r_e / r_s ) ^2

                                  I_s = ( 1.0 KW / e*a) * ( 1 / 9.5 )^2

                                  I_s = 11 W / e*a

- We know that P = I*a, hence we have:

                                  P_s = I_s*a

                                  P_s = 11 W / e

Hence,                       e*P_s = 11 W

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The answer to this question is "Greater than". As per the OSHA or Organizational Health and Health Organization, a noise level of 95 DB or decibels is greater than the lowest level at which hearing protection is required in 85 decibels and the person's exposure should be limited only to six hours or less than of it. 
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The weight of spaceman Speff at the surface of planet X, solely due to its gravitational pull, is 389 N. If he moves to a distan
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Answer:

mass of the planet X = 5.6 × 10²³ kg.

Explanation:

According to Newtons law of universal gravitation,

F = GM₁M₂/r²

Where F = gravitational force, M₁ = mass of the speff, M₂ = mass of the planet X, G = gravitational constant r = distance between the speff and the planet X

making M₂ The subject of the equation above,

M₂ = Fr²/GM₁ .......................... equation 2

Where F = 24.31 N, r = 1.08×18⁴km ⇒( convert to m ) =1.08 × 10⁴  × 1000 m

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A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wal
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(a) 18.9 m/s

The motion of the stone consists of two independent motions:

- A horizontal motion at constant speed

- A vertical motion with constant acceleration (g=9.8 m/s^2) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

v=18.9 m/s

(b) 22.2 m/s

We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

v_y^2 - u_y^2 = 2ah (1)

where

u_y = 16.4 m/s is the initial vertical velocity

v_y is the vertical velocity at height h

a=g=-9.8 m/s^2 is the acceleration due to gravity (negative because it is downward)

At the top of the parabolic path, v_y = 0, so we can use the equation to find the maximum height

h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m

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h = \frac{13.7}{2}=6.9 m

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v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s

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v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s

(c) 17.4% faster

We said that the speed at the top of the trajectory (part a) is

v_1 = 18.9 m/s

while the speed at half of the maximum height (part b) is

v_2 = 22.2 m/s

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\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s

And so, in percentage,

\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%

So, the stone in part (b) is moving 17.4% faster than in part (a).

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<u>Answer:</u>

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<u>Explanation: </u>

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