Physics, 08.02.2021 19:00 graciemartinez9

Two trucks are traveling east on I-40 when the Walmart truck (61 mph, 57 ft long) decides to pass the FedEx truck (53 mph, 60 ft long). A 'pass' starts when the front of the Walmart truck is 45 ft behind the rear of the FedEx truck and ends when the rear of the Walmart truck is 45 ft in front of the front of the FedEx truck. How long does it take for the Walmart truck to pass the FedEx truck?
How far does the Walmart truck travel while passing the FedEx truck?

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Answer from: nevelle

The answer is below

Explanation:

Let us assume that the slower truck which is the FedEx truck travels d mile. The pass started when the Walmart truck was 45 ft behind the FedEx truck and ended when the Walmart truck is 45 ft in front of the FedEx truck.

The additional distance covered by the truck Walmart = 45 ft + 60 ft (length of FedEx truck) + 45 ft = 150 ft

1 mile = 5280 ft

150 ft = 150 ft * 1 mile/5280 ft = 0.0284 mile

Therefore the Walmart truck covered an additional 0.0284 mile. Hence the distance covered by the walmart truck = d + 0.0284

Let us say it took t hours for the pass to be completed. Hence:

For FedEx; 53 mph = d / t. t =d / 53

For Walmart; 61 mph = (d + 0.0284) / t. t = (d + 0.0284) / 61

Since it took the trucks the same time. Hence:

d / 53 = (d + 0.0284) / 61

61d = 53d + 1.506

8d = 1.506

d = 1.506/8

d = 0.1882 mile = 993.75 ft

The FedEx truck traveled 993.75 mile while the Walmart truck traveled 1143.75 ft (993.75 + 150 ft)

The time (t) = d / 53 = 0.1882 / 53

t=0.00355 h = 12.8 seconds

Answer from: Quest

45 is the correct answer

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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Two trucks are traveling east on I-40 when the Walmart truck (61 mph, 57 ft long) decides to pass th...

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