Physics, 12.02.2020 02:16 gigi1103

The ends of a massless rope are attached to two stationary objects (e. g., two trees or two cars) so that the rope makes a straight line. For this situation, which of the following statements are true? Check all that apply. 1. The tension in the rope is everywhere the same. 2. The magnitudes of the forces exerted on the two objects by the rope are the same.3. The forces exerted on the two objects by the rope must be in opposite directions.4. The forces exerted on the two objects by the rope must be in the direction of the rope.

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Answer from: jholland03

1. The tension in the rope is everywhere the same.

2. The magnitudes of the forces exerted on the two objects by the rope are the same.

3. The forces exerted on the two objects by the rope must be in opposite directions.

Explanation:

"Massless ropes" do not have a "net force" which means that it is able to transmit the force from one end of the rope to the other end, perfectly. It is known for its property of having a total force of zero. In order to attain this property, the magnitude of the forces exerted on the two stationary objects by the rope are the same and in opposite direction. So this explains number 2 & 3 answers.

Since the objects that are held by the rope are stationary, then this means that the tension in the rope is also stationary. This means that the tension in the rope everywhere is the same (provided that the rope is still or in a straight line, as stated in the situation above, and is being held by two points). So, this explains number 1.

Answer from: Quest

answer: electrons

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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The ends of a massless rope are attached to two stationary objects (e. g., two trees or two cars) so...