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bezimeni [28]
3 years ago
6

Two power lines run parallel for a distance of 220 m and are separated by a distance of 40.0 cm. If the current in each of the t

wo lines is 130 A and if they run in opposite directions, determine the magnitude and direction of the force each wire exerts on the other. magnitude
Physics
1 answer:
daser333 [38]3 years ago
5 0

Answer:

The magnitude of force is 1.86 N and the direction of force is towards the other wire.

Explanation:

Given:

Current flowing through each power line, I = 130 A

Distance between the two power lines, d = 40 cm = 0.4 m

Length of power lines, L = 220 m

The force exerted by the power lines on each other is given by the relation:

F = \frac{\mu_{0}LII }{2\pi d}

Substitute the suitable values in the above equation.

F = \frac{4\pi\times10^{-7}\times220\times130\times130 }{2\pi\times0.4}

F = 1.86 N

Since the direction of current flowing through the power lines are opposite to each other, so the force is attractive in nature. Hence, the direction of force experienced by the power lines on each other is towards the each other.

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A force causes a mass of 4 kg to have an acceleration of 8 m/s2. Suppose something causes the mass to be one-quarter of its orig
solniwko [45]

Explanation:

From Newton's second law:

F = ma

Given that m = 4 kg and a = 8 m/s²:

F = (4 kg) (8 m/s²)

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32 N = (1 kg) a

a = 32 m/s²

So the acceleration increases by a factor of 4.

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3 years ago
Fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.20 mm cos[(
bezimeni [28]

Answer

given,

y(x,t)= 2.20 mm cos[( 7.02 rad/m )x+( 743 rad/s )t]

length of the rope = 1.33 m

mass of the rope = 3.31 g

comparing the given equation from the general wave equation

y(x,t)= A cos[k x+ω t]

A is amplitude

now on comparing

a) Amplitude  = 2.20 mm

b) frequency =

     f = \dfrac{\omega}{2\pi}

     f = \dfrac{743}{2\pi}

          f = 118.25 Hz

c) wavelength

        k= \dfrac{2\pi}{\lambda}

        \lambda= \dfrac{2\pi}{k}

        \lambda= \dfrac{2\pi}{7.02}

        \lambda= 0.895\ m

d) speed

         v = \dfrac{\omega}{k}

         v = \dfrac{743}{7.02}

                v = 105.84 m/s

e) direction of the motion will be in negative x-direction

f) tension

  T = \dfrac{v^2\ m}{L}

  T = \dfrac{(105.84)^2\times 3.31 \times 10^{-3}}{1.33}

      T = 27.87 N

g) Power transmitted by the wave

  P = \dfrac{1}{2}m\ v \omega^2\ A^2

  P = \dfrac{1}{2}\times 0.00331\times 105.84\times 743^2\ 0.0022^2

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3 years ago
an object is thrown off a cliff with intial pward velocity of 15m/s it strikes the ground with downward velocity of 25 m/s what
Genrish500 [490]

Answer:

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Explanation:

Given:

v₀ = 15 m/s

v = -25 m/s

a = -10 m/s²

Find: Δy

v² = v₀² + 2aΔy

(-25 m/s)² = (15 m/s)² + 2 (-10 m/s²) Δy

Δy = 20 m

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3 years ago
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