solve number 4 and 3 How far would you have to stretch a spring with K = 1,5kN/m

Question

solve number 4 and 3

How far would you have to stretch a spring with K = 1,5kN/m for it to store 200 J of energy Space explorer land on a planet that has the same mass as Earth, but find their weight twice as much as they would on Earth, What is the planet's radius Earth's distance from the Sun varies from 1.47 X 10^11 mat perihelion to 1.52 * 10^11 mat aphelion because Its orbit is not quite circular. Find the change in potential energy as Earth goes from perihelion to aphelion. A 28-kg child sits at one end of a 4.0m-long seesaw. Where should his 65-Kg father sit so that the center of the mass will be also at the center of the seesaw 6. An alpha particle (^4 He) strikes a stationary Uranium nucleus (^235 U) head-on and elastically. What fraction of the alpha's kinetic energy is transferred to the Uranium

Explanation / Answer

Part 3:

Weight W = mass* acceleration due to gravity

Weight onearth WE = m* g

Weight on planet WP = m* g'

But WP / WE = 2 => g' = 2* g ------------------- (1)

Acceleration due togravity g = G* M / R2

where M and R are mass and radius of particular planet.

on earth g = G* M / RE2

on the planet g' = G *M / RP2

Dividing g'/ g = (G * M / RP2) / (G * M / RE2)

RE2 / RP2 = 2

Radius of planet RP = RE / 2

i.e. radius of planet RP = 6.4* 106 / 2

radius of planet RP = 4.52 *106 m

Part 4:

The Earth's distance from the sun at perihelion is r1 = 1.47*1011m

The Earth's distance from the sun at aphelion is r2 = 1.52*1011m

The change in potential energy as Earth goes from perihelion to aphelion is

U = GM*M1*(1/r1 -1/r2)

Here, G = 6.67 * 10-11Nm2/kg2, M = 5.9742 * 1024kg and M1 = 1.98*1030 kg

Hence,

The change in potential energy as Earth goes from perihelion to aphelion is U = 157.7*1030 J

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