Recent content by faoltaem

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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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