A helicopter flies over the arctic ice pack at a constant altitude, towing an airborne 129-kg laser sensor that measures the thi
ckness of the ice (see the drawing). The helicopter and the sensor both move only in the horizontal direction and have a horizontal acceleration of magnitude 2.84 m/s2 . Ignoring air resistance, find the tension in the cable towing the sensor.
1 answer:
Answer:
1317.52 Newton
Explanation:
Mass of sensor = m = 129 kg
Acceleration of the sensor = a = 2.84 m/s²
g = Acceleration due to gravity = 9.81 m/s²
= Force on the laser sensor due to motion = 129×2.84 = 366.36 N
= Force due to gravity on the laser = mg = 129×9.81 = 1265.49 N
Tension in the cable
∴ The tension in the cable towing the sensor is 1317.52 Newton
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Answer:
Explanation:
The difference of electric potential between two points is given by the formula , where <em>d</em> is the distance between them and<em> E</em> the electric field in that region, assuming it's constant.
The electric field formula is , where <em>F </em>is the force experimented by a charge <em>q </em>placed in it.
Putting this together we have , so we need to obtain the electric force the charged ball is experimenting.
On the second drop, the ball takes more time to reach the ground, this means that the electric force is opposite to its weight <em>W</em>, giving a net force . On the first drop only <em>W</em> acts, while on the second drop is <em>N</em> that acts.
Using the equation for accelerated motion (departing from rest) , so we can get the accelerations for each drop (1 and 2) and relate them to the forces by writting:
These relate with the forces by Newton's 2nd Law:
Putting all together:
Which means:
And finally we substitute:
Which for our values means:
<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>