Answer:
she is pulling with 40 N force
Explanation:
The ring does not move means that the forces are equal.
Let's call Jo's force x
We have the quation
140 = x + 100
x = 40
As we know that
here we know that
now from above equation we have
so image will form on left side of lens at a distance of 15 cm
This image will be magnified and virtual image
Ray diagram is attached below here
<span>v = 25.0 km/</span><span>h = 25*5/18 m/s = 6.94 m/s
</span><span>centripetal force = mv²/r = 1275*6.94²/40 = 1537.18 N </span>
<span>f2 = f0/4
The gravity from the planet can be modeled as a point source at the center of the planet with all of the planet's mass concentrated at that point. So the initial condition for f0 has the satellite at a distance of 2r, where r equals the planet's radius.
The expression for the force of gravity is
F = G*m1*m2/r^2
where
F = Force
G = Gravitational constant
m1,m2 = masses involved
r = distance between center of masses.
Now for f2, the satellite has an altitude of 3r and when you add in the planet's radius, the distance from the center of the planet is now 4r. When you compare that to the original distance of 2r, that will show you that the satellite is now twice as far from the center of the planet as it was when it started. So let's compare the gravitational attraction, before and after.
f0 = G*m1*m2/r^2
f2 = G*m1*m2/(2r)^2
f2/f0 = (G*m1*m2/(2r)^2) / (G*m1*m2/r^2)
The Gm m1, and m2 terms cancel, so
f2/f0 = (1/(2r)^2) / (1/r^2)
f2/f0 = (1/4r^2) / (1/r^2)
And the r^2 terms cancel, so
f2/f0 = (1/4) / (1/1)
f2/f0 = (1/4) / 1
f2/f0 = 1/4
f2 = f0*1/4
f2 = f0/4
So the gravitational force on the satellite after tripling it's altitude is one fourth the original force.</span>
Answer:
Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.
Explanation:
Since Juan is closer to the center and Kuri is away from the center so we can say that Juan will move smaller distance in one complete revolution
As we know that the distance moved in one revolution is given as
also the time period of revolution for both will remain same as they move with the time period of carousel
Now we can say that the speed is given as
so Juan will have less tangential speed. so correct answer will be
Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.
