Physics, 06.12.2019 22:31 peperivera2652738

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

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Answer from: MicarriHarris6122

A) The amount of work done by the head when it ejects the spore ; $2.45 * 10^{-7} J$

B) The spores will land 0.32 m away from the fungi

A)To determine the amount of work done we apply the work-energy theorem

W = 1/2mv^2 - 1/2 mu^2

where ; m = 10⁻⁸⁸ kg

v = 7.0 m/s

u = 0 m/s

∴ W = $1/2 * (10^{-8})(7)^2 - 1/2(10^{-8})(0)^2 = 2.45 * 10^{-7} J$

B) The motion of the spores is follows a parabolic path which consists of two independent motions

i) considering the vertical motion ;

s = ut + 1/2 gt²

where ; s = 0.01 m

g = 10 m/s²

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

ii) considering the horizontal motion to determine how far away from the fungi will the spores land

d = Vₓ * t

Vₓ = 7 m/s

t = 0.045 s

∴ d = 7 ( 0.045 ) = 0.32 m

Hence the The amount of work done by the head when it ejects the spore = $2.45 * 10^{-7} J$ and The spores will land 0.32 m away from the fungi

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Answer from: nockturnal1993

The work done by the Pilobolus head when it ejects the spore is $2.45 \times 10^{-7} \:\rm J$ .

The fungi will land at a distance of 0.217 m, which is quite negligible distance and approximately equal to 0 m in option (D). Hence, option (D) is correct.

Given data:

The vertical distance of launch is, s = 1 cm = 0.01 m.

The mass of pores is, $m=10^{-8} \;\rm kg$ .

The final velocity is, v = 7.0 m/s.

Time taken for the acceleration is, $t = 2\;\rm \mu s =2 \times 10^{-6} \;\rm s$ .

The gravitational acceleration is, $g = 10 \;\rm m/s^{2}$ .

Apply the work-energy theorem to obtain the work done by the Pilobolus head when it ejects the spore as,

$W = \delta KE\\W = \dfrac{1}{2}mv^{2}-\dfrac{1}{2}mu^{2}\\W = \dfrac{1}{2} \times 10^{-8} \times 7^{2}-(\dfrac{1}{2}m \times 0^{2})\\W = 2.45 \times 10^{-7} \:\rm J$

Thus, the the work done by the Pilobolus head when it ejects the spore is $2.45 \times 10^{-7} \:\rm J$ .

Apply the second kinematic equation of motion to obtain the time taken during the vertical motion as,

$s = ut' +\dfrac{1}{2}gt'^{2} \\0.01=(0 \times t')+\dfrac{1}{2} \times 10 t'^{2}\\0.01=5 t'^{2}\\t' = 0.031 \;\rm s$

Now, horizontal distance covered is,

$v=\dfrac{d}{t'}\\7=\dfrac{d}{0.031}\\d = 0.217 \;\rm m$

Thus, the fungi will land at a distance of 0.217 m, which is quite negligible distance and approximately equal to 0 m in option (D). Hence, option (D) is correct.

Learn more about kinematic of motion here:

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Answer from: Quest

answer: (431,2)

step-by-step explanation: