The space station is 4.41 x 10^5 kg and orbits the earth 6.78 x 10^6 m from the center of earth. The mass of earth is 5.97 x 10^
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Answer:
Tension in the cable is T = 16653.32 N
Explanation:
Give data:
Cross section Area A = 1.3 m^2
Drag coefficient CD = 1.2
Velocity V = 4.3 m/s
Angle made by cable with horizontal =30 degree
Density
Drag force FD is given as
Drag force = 14422.2 N acting opposite to the motion
As cable made angle of 30 degree with horizontal thus horizontal component is take into action to calculate drag force
TCos30 = F_D
T = 16653.32 N
<h3>Answer:</h3>
(a) <u>i₁ = 0.03818 A = 38.18 mA</u>
(b) downward
(c) <u>i₂ = 0.01091 A = 10.91 mA</u>
(d) rightward
(e) <u>i₃ = 0.02727 A = 27.27 mA</u>
(f) leftward
(g) <u>Eₐ = 3.818 Volts</u>
<h3>Question:</h3>The complete question is stated below and the figure is provided in the attachment:
In Fig. 27-47, E 1 = 6.00 V, E 2 = 12.0 V, R1 = 100 Ω,
R2 = 200 Ω, and R3 = 300 Ω. One point of the circuit is grounded
(V = 0). What are the (a) size and (b) direction (up or down) of the
current through resistance 1, the (c) size and (d) direction
(left or right) of the current through resistance 2, and the
(e) size and (f) direction of the current through resistance 3?
(g) What is the electric potential at point A?
<h3></h3><h3>Explanation:</h3>Applying Kirchoff's voltage law in the loops of botg E₁ and E₂, in the clockwise and anti clockwise direction:
E₁ - i₂R₂ - i₁R₁ = 0
E₂ - i₃R₃ - i₁R₁ = 0
If, we apply Kirchhoff's current law at junction A, we get:
i₁ = i₂ + i₃
Using these relations in loop equations, and re-arranging:
E₁ - i₂R₂ - (i₂ + i₃) R₁ = 0 ___________ eqn (1)
E₂ - i₃R₃ - (i₂ + i₃) R₁ = 0 ___________ eqn (2)
Eqn (1) implies:
6 - 200 i₂ - 100 i₂ - 100 i₃ = 0
i₂ = (6 - 100i₃)/300
Eqn (2) implies:
12 - 300 i₃ - 100 i₂ - 100 i₃ = 0
12 - 400 i₃ = 100 i₂
using value of i₃ from eqn (1)
12 - 400 i₃ = (1/3)(6 - 100 i₃)
36 - 1200 i₃ = 6 - 100 i₃
1100 i₃ = 30
<u>i₃ = 0.02727 A</u>
using this value in eqn of i₂:
i₂ = [6 - 100(0.02727)]/300
i₂ = (6 - 2.727)/300
<u>i₂ = 0.01091 A</u>
Since:
i₁ = i₂ + i₃
i₁ = 0.01091 A + 0.02727 A
<u>i₁ = 0.03818 A</u>
<u></u>
(a)
<u>i₁ = 0.03818 A = 38.18 mA</u>
(b)
Since, the value of current is positive, thus it will have the direction that was assumed.
Therefore, its direction will <u>downward</u>
(c)
<u>i₂ = 0.01091 A = 10.91 mA</u>
(d)
Since, the value of current is positive, thus it will have the direction that was assumed.
Therefore, its direction will <u>rightward</u>
(e)
<u>i₃ = 0.02727 A = 27.27 mA</u>
(f)
Since, the value of current is positive, thus it will have the direction that was assumed.
Therefore, its direction will <u>leftward</u>
<u>(g)</u>
With respect to the grounded portion, the potential drop at the resistance 1 will be equal to the potential at A Eₐ.
Therefore,
Eₐ = i₁R₁
Eₐ = (0.03818 A)(100 Ω)
<u>Eₐ = 3.818 Volts</u>