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Anni [7]
3 years ago
12

Car A is traveling north on a straight highway and car B is traveling west on a different straight highway. Each car is approach

ing the intersection of these highways. At a certain moment, car A is 0.4 km from the intersection and traveling at 75 km/h while car B is 0.3 km from the intersection and traveling at 70 km/h. How fast is the distance between the cars changing at that moment? km/h
Physics
1 answer:
Inessa05 [86]3 years ago
7 0

Answer:

102 km/h

Explanation:

x = distance of car A at any time from the intersection

x₀ = distance of car A at some time = 0.4 km

v_{A} = Speed of car A = 75 km/h

y = distance of car B at any time from the intersection

y₀ = distance of car B at some time = 0.3 km

v_{B} = Speed of car B = 70 km/h

d = distance between the two cars at any time

d₀ = distance between the two cars at some time

v = rate of change of distance between the cars

Using Pythagorean theorem

d²₀ = x₀² + y₀²

d²₀ = 0.4² + 0.3²

d₀ = 0.5 m

Distance between the two cars at any time is given using Pythagorean theorem as

d² = x² + y²

Taking derivative both side relative to "t"

2d \left ( \frac{dd}{dt} \right ) = 2x ( \frac{dx}{dt} \right ) + 2y ( \frac{dy}{dt} \right )

d_{o} v = x_{o} ( \frac{dx}{dt} \right ) + y_{o} ( \frac{dy}{dt} \right )

(0.5) v = (0.4) v_{A} + (0.3) v_{B}

(0.5) v = (0.4) (75) + (0.3) (70)

v = 102 km/h

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olga2289 [7]

Answer:

electricfield at (0,0,0) is Et = 2 k q / a²

Explanation:

For the first part see the diagram , the field lines start from the positive charge and reach the negative charge, notice that no line should cross, some lines go to infinity

For the second part we use that the electric field is a vector quantity and therefore we add the field of each charge, using the equation

     E = k q / r²

Point (0,0,0)

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    Et = 2 k q / a²

Point (0,0, R)

We use the same equations, but with another distance, for the charge -q the distance is R = R+a and for the charge + q the distance is R = R-a

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     E2 = kq / (R-a)²

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     Et= kq {[(R-a)² + (R + a)²] / [(R + a)² (R-a)²]}

     Et= kq {2 (R² + a²) / [(R + a)² (R-a)²]}

If we use the condition that  R> a we can despise in the patents "a"

     (R² + a²) = R² (1+ a² / R²) ≈ R²

     (R + a)² = R² (1 + a / R)² ≈ R²

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Substituting in the total electric field

     Et = kq {2 R²) / [R²R²]}

     Et =kq 2 / R²

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Answer:

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