I know this is an old post but incase people find it in the future:
rule 1)
logbx + logby = logbxy
rule 2)
x (x + a) = x2 + ax
rule 3)
solving quadratic equations
Yeah, i was taking the radius as the radius of the web.
so I've done this:
nb: s=spider p=person
a) What is the fractional increase in the thread’s length caused by the spider?
Y_s = 4.7x10^9 N/m^2
m = 0.26g = 2.6x10^-^4 kg
r_s = 9.8x10^-^6 m
F = Y(\frac{A_o}{L_o})\Delta L => mg =...
Homework Statement
Q 1. A spider with a mass of 0.26 g hangs vertically by one of its threads. The thread has a Young’s modulus of 4.7 x 10^9 N/m2 and a radius of 9.8 x 10^- ^6 m.
a) What is the fractional increase in the thread’s length caused by the spider?
b) Suppose a 76 kg person...
if you have a pc go through the start menu -> programs -> accessories -> calculator
select view -> scientific for more options
if you have a mac there should just be a pic of a calculator on the desktop
thanks i didn't realized i'd overlooked that information
so:
a) \omega = \frac{v}{r} = \frac{28.8}{2} = \underline{14.4rad/s}
b) f_cp = ma_cp = \frac{mv^{2}}{r} = mr\omega^{2}
= 7.26 x 2 x 14.4^{2}
=3010.8672 = \underline{3.01 x 10^{3}N}
c) F=mg F=f_cp = 3011N, g=9.81m/s^{2}...
On his last throw of the Atlanta Olympics, Lance Deal launched the hammer 81.12m, good enough for a silver medal. The hammer is thrown by rotating the body in a circle, building up rotational speed until releasing it and letting the rotational velocity change to translational velocity. The...
so if i change sin 20 to cos 20 (or sin 110) i get:
\tau1 = Fsin
= mgsin
= 1.1 x -9.81 x cos 20 x 0.15
= -1.521Nm
\tau2 = Fsin
= mgsin
= 20 x -9.81 x cos 20 x 0.3
= -55.310Nm
-3 = 1 + 2
= -1.521 + -55.310
= -56.831Nm
therefore 3 = 56.831Nm
or 56.831Nm in an anticlockwise...
ok so
\tau1 = Fsin\theta
= mgsin\theta
= 1.1 x -9.81 x sin 20 x 0.15
= -0.5536Nm
\tau2 = Fsin\theta
= mgsin\theta
= 20 x -9.81 x sin 20 x 0.3
= -20.131Nm
-\tau3 = \tau1 + \tau2
= -0.5536 + -20.131
= -20.685Nm
therefore \tau3 = 20.685Nm
or 20.685Nm in an anticlockwise direction
sorry i forgot about the 3rd force
\Sigma\tau = 0
\tau1 + \tau2 + \tau3 = 0
\tau1 + \tau2 = -\tau3
but I'm a little unsure about the force of \tau2 (which i have as the object)
is it: -\tau3 = \tau1 + \tau2
= m1gx1 + m2gx2
= 1.1 x 9.81 x 0.15 + 20 x 9.81 0.3
= 60.47865N
\tau3 =...
2)
a) plot a graph on a piece of paper with speed/velocity on the side and time on the bottom (i'd just put 1hr, 2hr, 3hr etc for each of the velocities [unless you got given something])
b) those 3 equations that you have are just different ways of writing the same thing.
anyhow you have two...
would someone be able to tell me if this is right or what I've done wrong? thankyou
Homework Statement
a) What torque does the bicep muscle have to apply to hold the forearm of mass 1.10 kg and length 30 cm horizontally? Assume the arm is a rod of uniform mass density.
b) What torque does...
part 2 retry - electricity
P = VI = \frac{V^{2}}{R} = I^{2}
I_{1+2} = 0.6A
R = 7+5 = 12\Omega
P = 0.6^{2} x 12 = 4.32W
V^{2} = PR = 4.32 x 12 = 51.84V
V = 7.2V
P = I^{2} x 16 = 4.32
I^{2} = \frac{4.32}{16} = 0.2A
I = 0.5A
I = \frac{\epsilon}{R}
\epsilon = IR = 0.5 x 12 = 6V