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 realised 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
The values for the resistors are: R1=7\Omega, R2=5\Omega, R3=4\Omega (see diagram for placement of the resistors)
1) Suppose there is a current of 0.6A going through R1 and that the voltage supplied by the battery is 9V, determine the value of R4
2) Using the information above, determine the...
new pressure
ok so
v = 2.05 x 10^{-3} m^{3}
CSA = 65.2 cm2 = 6.52 x 10^{-3} m^{3}
h = \frac{v}{CSA} =0.3144m
P = \rhogh
= 806 x 9.81 x (14.67 + 0.314)
= 806 x 9.81 x 14.985
= 118 486.05 Pa
= 1.18 x 10^{5} Pa = 118 kPa
thankyou
A cylindrical container with a cross sectional area of 65.2 cm^{2} holds a fluid of density 806kg/m^{3}. At the bottom of the container the pressure is 116 kPa
a) What is the depth of the fluid?
b) Find the pressure at the bottom of the container after an additional 2.05 \times 10^{-3} m^{3}...
Homework Statement
A cylindrical container with a cross sectional area of 65.2 cm^{2} holds a fluid of density 806kg/m^{3}. At the bottom of the container the pressure is 116 kPa
a) What is the depth of the fluid?
b) Find the pressure at the bottom of the container after an additional 2.05...
_{}Homework Statement
If a long jumper at the top of his projectory is moving at 6.5 m/s (horizontally) and his cantre of mass is 1.1m above where it was when he launched into the jump, how fast must he have been moving when he launched?
Homework Equations
v_{x} = v_{0x} + a_{x}t
x =...
Homework Statement
a person sits in a harness hamging in mid-air from a set of scales. the scales display her weight as 550.0 N. she then empties her lungs as much as possible and is completely immersed in water. the scales now give her weight as 21.2 N. what is her average density...
I have a problem from a practice physics exam:
The specifications for design of a playgroud slide say that a child should gain a speed of no more than 15m/s by sliding down the eqipment. How tall can the slide be?
these are the equations that we'll be given in our exam:
v_{x} = v_{0x} +...
thanks that helped heaps, this is what i've got now, just wanted to check that i've done the right thing
a^{2} + b^{2} = c^{2}
b = \sqrt{c^{2} - a^{2}}
= \sqrt{0.72^{2} - 0.55^{2}}
= 0.46m
cos \theta = \frac{A}{H}
cos \theta = \frac{0.55}{0.72}
\theta = cos^{-1}...
my question came with a picture so i get how it's all set out i'm just having a problem with how to calculate the tension of the threads (.72m) leading down to the spider.
I'd be abe to work it out if i had a formula for it, but i've looked through my textbook, my lecture notes and had a quick...
hey i have a question from my physics tute but i'm lost.
a spider is suspended on lines of web between two level branches. each line of web is 72cm long and the branches they are attached to are 1.1m apart
1) given that the spider weighs 5.2g, calculate the tension force in either of the...