Answer:
H = 109.14 cm
Explanation:
given,
Assume ,
Total energy be equal to 1 unit
Balance of energy after first collision = 0.78 x 1 unit
= 0.78 unit
Balance after second collision = 0.78 ^2 unit
= 0.6084 unit
Balance after third collision = 0.78 ^3 unit
= 0.475 unit
height achieved by the third collision will be equal to energy remained
H be the height achieved after 3 collision
0.475 ( m g h) = m g H
H = 0.475 x h
H = 0.475 x 2.3 m
H = 1.0914 m
H = 109.14 cm
Answer:
Sugars...
Explanation:
Several meteorites have been found to carry molecules of sugars that are essential for life. These sugars include Ribose, Arabinose and Xylose. These are found in meteorites that are rich in carbon. These significant discoveries can pave way in finding the origin of life on Earth.
The gravitational force exerted on the satellite is called the centrifugal force, the force keeping it orbiting to the planet. Its formula is F= mass times the square of the velocity all over the radius.Thus,
F = 2400 * 6670^2 * (1/8.92x10^6)
F = 11,970 N
I hope I was able to help you. Have a good day.
<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>
<span>A = area of styrofoam
M = mass of stryofoam = A*h*rho_s
m = mass of swimmer
Total mass = m + M = m + A*h*rho_s
Downward force = g*(total mass) = g*[m + A*h*rho_s]
The slab is completely submerged.
Buoyant force = g*(mass of water displaced) = g*[A*h*rho_w]
Equate these
g*[m + A*h*rho_s] = g*[A*h*rho_w]
m + A*h*rho_s = A*h*rho_w
A*h*[rho_w - rho_s] = m
A = m/[h*(rho_w - rho_s)]</span>
