Physics, 05.03.2021 14:50 Demaryaflie

Physics, calculating gravity between two objects. please work it out for me

Answers: 2

Show answers

Answers

Answer from: Quest

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j

Answer from: Quest

You need to be more specific. i really don't know what a "motion map" is.you can draw a graph of an object's position, distance, displacement, speed,velocity, or acceleration versus time, or you can draw a graph of the object's x-position versus its y-position and not give any information about time at all.if the graph presents position or displacement against time, then constant velocity is a straight line . . i.e. a line that has constant slope.if the graph shows speed against time, then constant velocityis a horizontal piece of line.if the graph shows acceleration against time, then constant velocity is a piece of line that lies on the x-axis . . it's value is zero.

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

false

hope this : d

Other questions on the subject: Physics

Physics, 22.06.2019 04:00, binodkharal2048

Which configuration would produce an electric current? a) rotate a coil of copper wire between two magnets. b) connect a wire between a copper and zinc strip sitting in a beaker of water. c) connect a wire to the (+) positive end of a battery and the other end to a light bulb. d) connect a wire to the negative end of a battery and the other end to a light bulb.

Answers: 2

continue

Physics, 22.06.2019 06:00, 20alyssityh224

Ineed the answer to question 1 , i’m desperate

Answers: 3

continue

Physics, 22.06.2019 18:00, emory238

What structural formula represents 4 electrons shared between two atoms?

Answers: 1

continue

Physics, 22.06.2019 21:20, dee0828

16. a brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800 n on him. the mass of the losing player plus equipment is 90.0 kg, and he is accelerating at 1.20m/s2 size 12{1 "." "20"" m/s" rsup { size 8{2} } } {} backward. (a) what is the force of friction between the losing player’s feet and the grass? (b) what force does the winning player exert on the ground to move forward if his mass plus equipment is 110 kg? (c) draw a sketch of the situation showing the system of interest used to solve each part. for this situation, draw a free-body diagram and write the net force equation.

Answers: 2

continue

You know the right answer?

Physics, calculating gravity between two objects. please work it out for me
...