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anonymous
 one year ago
Ballooning Spiders, First in Flight. You are studying Gossamer Spiders in a biology research lab and marvel at their ballooning stunts. These Spiders disperse by spinning strands of silk in open air. The flight of these spiders is electrostatic in nature because everything that moves through air develops static charge and because the glue that coats a spider web strand has electrostatic properties that causes the web to latch onto all charged particles, from pollen to flying insects.2 Suppose a Gossamer Spider has a mass of 5.5 x 104kg. If the spider is suspended from a tree branch by a 1.4m strand of silk, suspended by an angle of 22.00° E of N, what amount of electrostatic force is necessary to keep the spider in equilibrium? Assume the electrostatic force is entirely horizontal. (See reference below)
Knowns:
m = 5.5 x 10^4 kg
length of strand = 1.4m
angle = 22deg E of N
http://arxiv.org/pdf/1309.4731v1.pdf
anonymous
 one year ago
Ballooning Spiders, First in Flight. You are studying Gossamer Spiders in a biology research lab and marvel at their ballooning stunts. These Spiders disperse by spinning strands of silk in open air. The flight of these spiders is electrostatic in nature because everything that moves through air develops static charge and because the glue that coats a spider web strand has electrostatic properties that causes the web to latch onto all charged particles, from pollen to flying insects.2 Suppose a Gossamer Spider has a mass of 5.5 x 104kg. If the spider is suspended from a tree branch by a 1.4m strand of silk, suspended by an angle of 22.00° E of N, what amount of electrostatic force is necessary to keep the spider in equilibrium? Assume the electrostatic force is entirely horizontal. (See reference below) Knowns: m = 5.5 x 10^4 kg length of strand = 1.4m angle = 22deg E of N http://arxiv.org/pdf/1309.4731v1.pdf

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm assuming that the equation they want us to use to calculate the balancing force is \[Q=\frac{ mg }{ E_o }e^{\alpha H_{eq}}\] But I could be wrong. The other equation that they used seems like it's only used when the spider experiences a vertical acceleration.\[Q_{accel} = \frac{ m(a_{net}+g) }{ E_o }\] but I could be wrong.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443080246357:dw

The_Beast_Eli
 one year ago
Best ResponseYou've already chosen the best response.0I think you are right! ;/\

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@The_Beast_Eli The only issue I'm having is that I don't know the values of a or H..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Equilibrium means \[\Sigma F _{x}=0\] so \[\Sigma F _{x}=Tsin \thetaF _{e}=0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\Sigma F _{y}=0\] \[\Sigma F _{y}=Tcos \thetamg=0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\Sigma F _{x}=Tsin \theta=F _{e}\] and \[\Sigma F _{y}=Tcos \theta=mg\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Divide both \[\frac{ Tsin \theta }{ Tcos \theta}=\frac{ F _{e} }{ mg }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\tan \theta=\frac{ F _{e} }{ mg }\] m is mass in kg g is accleration due to gravity F is electric force

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Its pretty simple. Just ignore the article and information at the top.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443157685426:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443157805407:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Electric force should be pointing at those 3 arrows.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443158106792:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wow, I got tricked so hard by my professor for including that article, haha. Thank you!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yea, when I saw the link to the article, I was like "Isnt this a graduate school problem?" I then read the info more carefully and was like "Oh yeah, I know this. Every info here is useless except the one in the bottom."

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No problem! Good luck with the rest.
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