Physics, 03.12.2021 03:30 SkyThefly

How far from a very small 140 kg ball would a particle have to be placed so that the ball pulled on the particle just as hard as the earth does

Answers: 1

Show answers

Answers

Answer from: Quest

-first use of geothermal energy occurred more than 10,000 years ago in north america by american paleo-indians

-the first industrial use of geothermal energy began near pisa, italy in late 18th century

-in 1904, italian scientist piero ginori conti invented the first geothermal electric power plant in which steam was used to generate the power

-with the above experiment, the first geothermal plant in usa started in 1922 with a capacity of 250 kilowatts

-however, in 1946 first ground-source geothermal heat pump installed at commonwealth building in portland, oregon

-during the 1960’s, pacific gas and electric began operation of first large scale geothermal power plant in san francisco, producing 11 megawatts

-in 1973, when oil crisis began many countries began looking for renewable energy sources and by 1980’s geothermal heat pumps (ghp) started gaining popularity in order to reduce heating and cooling costs

-as effect of climate change started showing results, governments of various countries joined hands to fight against it, for which kyoto protocol was signed in japan in 1997, laid out emission targets for rich countries and required that they transfer funds and technology to developing countries, 184 countries have ratified it

hope this

Answer from: Quest

dang that balls went far

Answer from: Quest

Ibelieve the answer is d. macula hope my answer has you : )

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

Other questions on the subject: Physics

Physics, 22.06.2019 04:00, jshu1998

Amodel rocket with a mass of 0.212 kg is launched into the air with an initial speed of 84 m/s. how much kinetic energy will the rocket have at a height of 214 m? assume there is no wind resistance. 634 j 303 j

Answers: 2

continue

Physics, 22.06.2019 07:30, luckylady

Select all of the following that are equal to an impulse of 20 unitsa) force= 5, time= 5.5b) force= 25, time= 0.8c) force= 0.1, time= 200d) force= 10, time= 2(its b, c, and d)(also in order to find answer they have to add to get the unit)ex= 25x.8=20 or .1x200=20

Answers: 1

continue

Physics, 22.06.2019 09:50, trobbie817

A250-n box is at rest on a frictionless horizotal floor. then a person shoves the box with a non-zero force of magnitude p. which statement best describes the relationship betweeen p and the motion of the box after it was shoved? -the box will not move when p< 250n but otherwise will accelerate at a constant rate that is directly proportional to p - the box will begin moving at a speed that is directly proportional to p, then will gradually slow to a halt -the box will move with constant velocity at a speed that is directly proportional to p -box will accelerate across the floor at a constant rate that is directly proportional to p

Answers: 1

continue

Physics, 22.06.2019 10:00, lulu8167

What is the temperature in degrees celsius of a substance with a tempature of 49k

Answers: 2

continue

You know the right answer?

How far from a very small 140 kg ball would a particle have to be placed so that the ball pulled on...