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Pilobolus is a genus of fungi commonly found on dung and known for launching its spores a large distance for a sporangiophore on

ly 1 cm tall. To achieve this, Pilobolus accelerates its spores (m = 10−8 kg) to 7.0 m/s in 2.0 μs. For calculations, use g = 10 m/s2.
How much work is done by the Pilobolus head when it ejects the spore?

If the spores are shot out horizontally at the maximum speed, how far away from the fungi do they land? Neglect air resistance.

A. 5.0 m
B. 1.0 m
C. 10.0 m
D. 0 m

1 answer:

Softa [3K]1 month ago

3 0

1) The work performed on the spores is $2.45\cdot 10^{-7}J$

2) The spores land 0.32 m away

Explanation:

1)

According to the work-energy principle, the work done on the spores equals their change in kinetic energy:

$W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$

where

W is work done

m represents the spore's mass

v is the final velocity

u is the initial velocity

Given:

spore mass = $m=10^{-8} kg$

initial velocity u = 0 (starting from rest)

final velocity v = 7.0 m/s

Calculating the work:

$W=\frac{1}{2}(10^{-8})(7.0)^2-0=2.45\cdot 10^{-7}J$

2)

The spores follow projectile motion, moving along a parabolic trajectory made of two motions:

- Constant speed horizontally

- Vertically accelerated due to gravity

Considering vertical motion first, using kinematics:

$s=ut+\frac{1}{2}at^2$

where

s = 0.01 m (shooting height)

u = 0 (no initial vertical velocity, horizontal ejection)

t = time of flight

g = acceleration due to gravity $a=g=10 m/s^2$

Solving for t:

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(0.01)}{10}}=0.045 s$

Now for horizontal displacement, when velocity is constant:

$d=v_x t$

where

horizontal velocity = $v_x = 7.0 m/s$

time t = 0.045 s

Calculating distance d:

$d=(7.0)(0.045)=0.32 m$

Thus, the spores land 0.32 meters away.

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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