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Andrej [43]
3 years ago
8

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.

Physics
1 answer:
Mashcka [7]3 years ago
7 0

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

F_m = Force on the laser sensor due to motion = 129×2.84 = 366.36 N

F_g = Force due to gravity on the laser = mg = 129×9.81 = 1265.49 N

Tension in the cable

T=\sqrt{F_m^2+F_h^2}\\\Rightarrow T=\sqrt{366.36^2+1265.49^2}\\\Rightarrow T=1317.52\ N

∴ The tension in the cable towing the sensor is 1317.52 Newton

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Answer:

m=1.01\times 10^6\ kg

Explanation:

Given that,

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P=\dfrac{mg}{A}\\\\m=\dfrac{PA}{g}\\\\m=\dfrac{101325\times 10^2}{10}\\\\m=1.01\times 10^6\ kg

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Volcanologists and seismologist are specialzed in what
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The answer is:

They are specialized in geology and geophysics

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5 0
3 years ago
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Answer:

Person B

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A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the magnitude of the dri
nordsb [41]

Complete Question

A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm2.) mm/s

Answer:

The drift velocity is v  = 0.0002808 \ m/s

Explanation:

From the question we are told that

    The current on the copper is  I  = 20 \ A

     The cross-sectional area is  A =  5.261 \ mm^2 =  5.261 *10^{-6} \ m^2

The number of copper atom in the wire is  mathematically evaluated

      n  =  \frac{\rho *  N_a}Z}

Where \rho is the density of copper with a value \rho =  8.93 \ g/m^3

          N_a is the Avogadro's number with a value N_a  = 6.02 *10^{23}\ atom/mol

         Z  is the molar mass of copper with a value  Z =  63.55 \ g/mol

So

     n  =  \frac{8.93 * 6.02 *10^{23}}{63.55}

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Given the 1 atom is equivalent to 1 free electron then the number of free electron is  

         N  = 8.46 * 10^{28}  \  electrons

The current through the wire is mathematically represented as

         I  =  N * e * v * A

substituting values

        20 =  8.46 *10^{28} * (1.60*10^{-19}) * v *  5.261 *10^{-6}

=>     v  = 0.0002808 \ m/s

       

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