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ipn [44]
3 years ago
9

Two forces of 83 pounds and 56 pounds act simultaneously on an object. The resultant forms an angle of 52° with the 56-pound for

ce. Find the angle formed between the two original forces, to the nearest 10th of a degree.

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

Answer:

95.9°

Explanation:

The diagram illustrating the action of the two forces on the object is given in the attached photo.

Using sine rule a/SineA = b/SineB, we can obtain the value of B° as shown in the attached photo as follow:

a/SineA = b/SineB,

83/Sine52 = 56/SineB

Cross multiply to express in linear form

83 x SineB = 56 x Sine52

Divide both side by 83

SineB = (56 x Sine52)/83

SineB = 0.5317

B = Sine^-1(0.5317)

B = 32.1°

Now, we can obtain the angle θ, between the two forces as shown in the attached photo as follow:

52° + B° + θ = 180° ( sum of angles in a triangle)

52° + 32.1° + θ = 180°

Collect like terms

θ = 180° - 52° - 32.1°

θ = 95.9°

Therefore, the angle between the two forces is 95.9°

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Need help finding the average speed.
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Explanation:

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  • 2.07 + 0. 74 + 1.33 = 4.14. 4.14 / 3 = 1.38
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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}

     n  = 8.46 * 10^{28}  \  atoms /m^3

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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