Physics, 14.02.2020 20:12 sidthesciencekidz

Ships A and B leave port together. For the next two hours, ship Atravels at 20 mph in a direction 30° west of north while theship B travels 20° east of north at 25 mph.
a. What is the distance between the two ships two hours after theydepart?
b. What is the speed of ship A as seen by ship B?

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Answer from: jocelyngm

The answers to the question are

a. The distance between the two ships two hours after they depart is 39.1 miles

b. The speed of ship A as seen by ship B is 12.14 m/h south and 15.32 m/h east or Ship B sees ship A has having a magnitude velocity of 19.5 m/h ∡38.4 ° South East

Explanation:

Firstly we list out the known variables as follows

Ship A velocity = 20 m/h

Ship A direction = 30 ° West of North

Ship B velocity = 25 m/h

Ship B direction = 20 ° East of North

Distance covered after 2 hours for ship A = 20×2 = 40 miles

Distance covered after 2 hours for ship B = 25×2 = 50 miles

Dx component of ship A = 40×sin (30) = -20 miles and Dy = 40×cos (30) = 34.64 miles

While for ship B, we have

Dx = 50×sin (20) = 17.1 miles and Dy = 50×cos (20) = 46.98 miles

Therefore, the distance between the two ships is

((Ship B Dx - Ship A Dx)^2 +Ship B Dy - Ship A Dy)^2)^0.5

=((17.1-(-20))^2 + (46.98-34.64)^2)^0.5 = 39.1 miles

The speed of ship A as seen by ship B can be found by

The vy of ship A relative B = 20 cos 50 = 12.86 m/h

Therefore relative to ship B ship A is moving at a speed of 12.86 m/h - 25 m/h = -12.14 m/h north and 20 sin 50 or 15.32 m/h east

or 12.14 m/h south and 15.32 m/h east having a magnitude of 19.5 m/h 38.4 ° South East

This is solved graphically by the attached diagram, the required vector is the line between the green and blue dots

Ships A and B leave port together. For the next two hours, ship Atravels at 20 mph in a direction 30

Answer from: Quest

i believe the answer is a hope it and is correct

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

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Ships A and B leave port together. For the next two hours, ship Atravels at 20 mph in a direction 30...

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