If a planet has 3 times the radius of the earth, but has the same density as the earth, what is the gravitational acceleration at the surface of the planet? (g = 9.8 m/s2 )

The Gravitational Acceleration of the Big planet is G = 3g

where G is the gravitational acceleration of the Big Planet and g is the gravitational acceleration of the Earth

So, it becomes G = 3 x 9.8

G = 29.4  

Explanation:

Let's start by calculating the Density of the Earth of mass m anad Radius r

We know that mass is density times the volume

ρ =

m = ρV

we know the volume is V = (please ignore the symbol of Pi)

m = ρ

Calculating the Density of Big Planet

→For Big Planet we know that radius is 3-times the radius of Earth, that is 3r

M = ρ    

So in conclusion, the mass of Earth and the mass of Big planet are related as M = 27m    

Now let's come to the Gravitational Force, we know that gravitational force is directly proportional to the mass of the body and inversely proportional to the radius.

→Suppose for Earth:  Mass = m, Radius = r  

For Big Planet:   Mass = M, Radius = R

:   = g : G

=  

G =

→ Putting the value of R = 3r and M = 27m

G =

By cutting the terms, we get

G = 3g

value of g= 9.8  

G = 3 x  

G = 29.4  

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer

the journeys of marquette, joliet and la salle a) new france expand.

all three of these explorers explored the midwest and the area around the great lakes, which they subsequently claimed for france.

this was a big portion of territory and expanded new france which before then only consisted of quebec province in northeastern canada