1) The work done on the spores is ![2.45\cdot 10^{-7}J](https://tex.z-dn.net/?f=2.45%5Ccdot%2010%5E%7B-7%7DJ)
2) The spores land 0.32 m away
Explanation:
1)
According to the work-energy theorem, the work done on the ejected spores is equal to the change in kinetic energy of the spores:
![W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2](https://tex.z-dn.net/?f=W%3DK_f%20-%20K_i%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2-%5Cfrac%7B1%7D%7B2%7Dmu%5E2)
where
W is the work done
m is the mass of one spore
v is the final velocity
u is the initial velocity
For the spores in this problem, we have:
is hte mass
u = 0 is the initial velocity (they start from rest)
v = 7.0 m/s is the final velocity
Substituting, we find the work done:
![W=\frac{1}{2}(10^{-8})(7.0)^2-0=2.45\cdot 10^{-7}J](https://tex.z-dn.net/?f=W%3D%5Cfrac%7B1%7D%7B2%7D%2810%5E%7B-8%7D%29%287.0%29%5E2-0%3D2.45%5Ccdot%2010%5E%7B-7%7DJ)
2)
The motion of the spores is a projectile motion, so it follows a parabolic path which consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction
We start by considering the vertical motion: this is a free fall motion, so we can use the suvat equation
where
s = 1 cm = 0.01 m is the vertical displacement (the height of the fungi)
u = 0 is the initial vertical velocity of the spore (it is ejected horizontally)
t is the time of flight
is the acceleration of gravity
Solving for t, we find:
Now we can consider the horizontal uniform motion. Since the velocity is constant, the horizontal distance travelled is given by :
where:
is the horizontal velocity, which is constant
t = 0.045 s is the time of flight
Solving for d, we find:
So, the spores land 0.32 m away.
Learn more about work:
brainly.com/question/6763771
brainly.com/question/6443626
Learn more about projectile motion:
brainly.com/question/8751410
#LearnwithBrainly