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9966 [12]
3 years ago
14

Ship A is 32 miles north of ship B and is sailing due south at 16 mph. Ship B is sailing due east at 12 mph. At what rate is the

distance between them changing at the end of 1 hour?
Physics
1 answer:
Zielflug [23.3K]3 years ago
7 0

Answer:

\dfrac{dz}{dt} =-5.6\ mile/h

Explanation:

distance between ship A and B = 32 mile

Ship A velocity in south, dx/dt = -16 mph

Ship B is sailing toward east with speed, dy/st = 12 mph

time = 1 hour

rate of change of distance between them = ?

x be the distance travel after t time

X = 32 + x

Let distance between them be z

now, using Pythagoras theorem to calculate distance between ships after 1 hours

z² = x² + y²

z² = (32 + x)² + 12²

z² = (32 - 16)² + 12²

z = √400

z = 20 miles

now, calculation of rate of change of distnace

z² = (32 + x)² + y²

differentiating both side w.r.t. time

2 z \dfrac{dz}{dt} = 2(32+x)\dfrac{dx}{dt} + 2 y\dfrac{dy}{dt}

z \dfrac{dz}{dt} =(32-16)\dfrac{dx}{dt} +y\dfrac{dy}{dt}

20\times \dfrac{dz}{dt} =16\times (-16) +12\times 12

\dfrac{dz}{dt} =\dfrac{-112}{20}

\dfrac{dz}{dt} =-5.6\ mile/h

hence, the rate is the distance between them changing at the end of 1 hour is equal to \dfrac{dz}{dt} =-5.6\ mile/h

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kirza4 [7]

Answer:

a) Magnetic field strength, B = 2.397 * 10⁻⁷ T

b) Total energy density, U = 4.58 * 10⁻⁸ J/m³

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U = \frac{1}{2} \epsilon_0E^2 +  \frac{1}{2} \frac{B^2}{\mu_0} \\U = \frac{1}{2}* 8.85 * 10^{-12}*71.9^2 +  \frac{1}{2} \frac{(2.397*10^{-7})^2}{4\pi*10^{-7}}\\U = 2.29 * 10^{-8} + 2.29 * 10^{-8}\\U = 4.58 * 10^{-8} J/m^3

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