Physics, 21.08.2019 10:30, lw057080

One of the foci for the moon's orbit would be the
(1) sun (3) mars
(2) earth (4) venus

One of the foci for the moon's orbit would be the (1) sun (3) mars (2) earth (4)

Answers: 2

Show answers

Answers

answered: mattydoug4818

The question is oversimplified, and pretty sloppy.

Relative to the Earth . . .
The Moon is in an elliptical orbit around us, with a period of
27.32... days, and with the Earth at one focus of the ellipse.

Relative to the Sun . . .
The Moon is in an elliptical orbit around the Sun, with a period
of 365.24... days, and with the Sun at one focus of the ellipse,
and the Moon itself makes little dimples or squiggles in its orbit
on account of the gravitational influence of the nearby Earth.

I'm sorry if that seems complicated. You know that motion is
always relative to something, and the solar system is not simple.

answered: itzmelanie1

2) Earth

Explanation:

As per Kepler's first law we know that all planets and all satellites revolves in elliptical path with one of its focii as Sun or planet around which satellite and planets are revolving.

Now here we need to know about the focii of moon which is revolving around Earth in its elliptical path.

So as per same law we can say that since moon is revolving around the Earth so its focii must be Earth

So here correct answer would be

2) Earth

answered: Guest

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

answered: Guest

125 m

explanation:

plz (×_×)^when a stone is released the distance traveled by it in the last second is eq

Other questions on the subject: Physics

Physics, 22.06.2019 00:20, pooperjooper

Aparticle of mass m is projected with an initial velocity v0 in a direction making an angle α with the horizontal level ground as shown in the figure. the motion of the particle occurs under a uniform gravitational field g pointing downward. (a) write down the lagrangian of the system by using the cartesian coordinates (x, y). (b) is there any cyclic coordinate(s). if so, interpret it (them) physically. (c) find the euler-lagrange equations. find at least one constant of motion. (d) solve the differential equation in part (c) and obtain x and y coordinates of the projectile as a function of time. (e) construct the hamiltonian of the system, h, and write down the hamilton’s equations (canonical equations) of motion.

Answers: 2

continue

Physics, 22.06.2019 10:10, Fionaauggies

How will the system respond to a temperature increase?

Answers: 1

continue

Physics, 22.06.2019 12:30, jfrjtwrg

Compare the horizontal displacement of the two balls as they travel to the ground. what do you observe?

Answers: 1

continue

Physics, 22.06.2019 17:00, AnNguyen

How much energy is supplied to each coulomb of charge that flows through a 12-v battery?