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    Solve log(x) + log(x+3) = 1

    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
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    Young's modulus of spider thread

    thanks for all your help
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    Projectile Motion bombarding cancer tumors

    maybe you could try working backwards to see what you've done wrong. i think that you've used some values in the wrong places
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    Forces - Elevator 2 part question

    Do you have any formulae at all? Do you understand the concept of what is happening when the elevator is moving?
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    Young's modulus of spider thread

    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 =...
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    Young's modulus of spider thread

    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...
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    I lost my calculaor Plz calculate this for me

    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
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    Rotational motion

    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}...
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    Rotational motion

    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...
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    Rotational motion of bicep

    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...
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    Rotational motion of bicep

    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
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    Rotational motion of bicep

    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 =...
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    Physics Homework need help for answers/also explain how to do them

    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...
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    Rotational motion of bicep

    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...
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    Electricity prob

    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
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    Electricity prob

    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...
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    Pressure of fluid on bottom of container

    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
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    Pressure of fluid in container

    sorry i didn't mean to finish it yet there is a later post with more in it
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    Pressure of fluid on bottom of container

    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}...
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    Pressure of fluid in container

    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...
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    Another motion prob

    thanks learningphysics cool so it's: vy^{2} = voy^{2} - 2g\Deltay 0^{2} = voy^{2} - 2\times9.81\times1.1 voy^{2} = 21.582 voy = 4.65 m/s vx = vox voy = 6.5 m/s vo = \sqrt{voy^{2} + vox^{2}} = \sqrt{4.65^{2} + 6.5^{2}} = \sqrt{21.582 + 42.25} = \sqrt{63.832} = 7.989 m/s...
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    Another motion prob

    _{}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 =...
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    Density prob

    thanks, that was really helpful
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    Density prob

    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...
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    Motion equations problem

    thanks for that, i didn't even consider energy and those formulae are included too so no memorising once again thanks a bunch
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    Motion equations problem

    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} +...
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    Tension of threads

    ok so that would make the angle 50\circ so T = \frac{mg}{2sin\theta} = \frac{0.0052 \times 9.81}{2 sin 50} = 0.033N = 3.3 x 10^{-2}N
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    Tension of threads

    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}...
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    Tension of threads

    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...
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    Tension of threads

    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...
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