Froghopper insects have a typical mass of around 12.5 mg and can jump to a height of 42.3 cm. the takeoff velocity is achieved as the insect flexes its legs over a distance of approximately 2.00 mm. assume that the jump is vertical and that the froghopper undergoes constant acceleration while its feet are in contact with the ground. ignore air resistance. what is the acceleration of the insect during the time of the jump (before it leaves the ground)?

2065.005 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

Now H-h = 42.3 - 0.2 = 42.1 cm = 0.421 m

The final velocity will be the initial velocity

Acceleration of the frog is 2065.005 m/s²

well, their are quite a few differences !

explanation:

first difference is a main sequence star is made of carbon, while a neutron star is made of neutrons. another difference is a main sequence star still has hydrogen to burn, while a neutron star is a remnant of a supernova.

a main sequence star is what’s left over from a low-mass star death, while a neutron star is what’s left from the death of a high-mass star. main sequence star and a neutron star are considered the same thing, except a main sequence star spins.