A) The amount of work done by the head when it ejects the spore ;[tex]2.45 * 10^{-7} J[/tex]
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 = [tex]1/2mv^2 - 1/2 mu^2[/tex]
where ; m = 10⁻⁸⁸ kg
v = 7.0 m/s
u = 0 m/s
∴ W = [tex]1/2 * (10^{-8})(7)^2 - 1/2(10^{-8})(0)^2 = 2.45 * 10^{-7} J[/tex]
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 = [tex]\sqrt{\frac{2s}{g} } = \sqrt{\frac{2(0.01)}{10} }[/tex] = 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 = [tex]2.45 * 10^{-7} J[/tex] and The spores will land 0.32 m away from the fungi
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The work done by the Pilobolus head when it ejects the spore is [tex]2.45 \times 10^{-7} \:\rm J[/tex].
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, [tex]m=10^{-8} \;\rm kg[/tex].
The final velocity is, v = 7.0 m/s.
Time taken for the acceleration is, [tex]t = 2\;\rm \mu s =2 \times 10^{-6} \;\rm s[/tex].
The gravitational acceleration is, [tex]g = 10 \;\rm m/s^{2}[/tex].
Apply the work-energy theorem to obtain the work done by the Pilobolus head when it ejects the spore as,
[tex]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[/tex]
Thus, the the work done by the Pilobolus head when it ejects the spore is [tex]2.45 \times 10^{-7} \:\rm J[/tex].
Apply the second kinematic equation of motion to obtain the time taken during the vertical motion as,
[tex]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[/tex]
Now, horizontal distance covered is,
[tex]v=\dfrac{d}{t'}\\7=\dfrac{d}{0.031}\\d = 0.217 \;\rm m[/tex]
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.
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