Physics, 15.10.2020 14:01 olivernolasco23

Two students are discussing how the speed of the car compares to the speed of the truck when both vehicles are in front of the house. Student 1 says, "The distance traveled by the car and the truck is the same, and the time is the same, so they must have the same speed." Student 2 says, "I don’t see how that can be. The car catches up to the truck, so the car has to be going faster." a. Which aspects of Student 1’s reasoning, if any, are correct? Support your answer in terms of relevant features of your graphs in part (a).
b. Which aspects of Student 2’s reasoning, if any, are correct? Support your answer in terms of relevant features of your graphs in part (a).
c. Derive an expression for the acceleration of the car. Express your answer in terms of D and vt
d. Determine the time at which the speed of the car is equal to the speed vt of the truck. Express your answer in terms of tD. Justify your answer.

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Answer from: anme3c9

a) Student 1 is right in the way he calculates the speed cid d of the truck, since going at constant speed

b) Student 2 is right that for the car to reach the truck it must have some relationship,

c) t = √d/a

Explanation:

In this exercise you are asked to analyze the movement in a mention using kinematics.

a) Student 1 is right in the way he calculates the speed cid d of the truck, since going at constant speed he can use the equation

v = d / t

b) Student 2 is right that for the car to reach the truck it must have some relationship, which is given by

v = v₀ + a t

x = v₀ t + ½ a t²

c) let's use the equations

v = d / t

v = v₀ + at

d / t = vo + at

If we can assume that the car starts from rest, so v₀ = 0

d / t = a t

t² = d / a

t = √d / a

if the speed of the car is not zero the result is a little more complicated

d = vo t + a t²

at² + vo t - d = 0

let's solve the quadratic equation

t = [-v₀ ± √ (v₀² + 4a d)] / 2a

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

Answer from: Quest

Physical science b final exam bernoulli’smovepowercompound translucent the sun i was having such a pain for this searching every question so i hope i made someone happy

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