• If the voltage is constant throughout a region of space, the electric field in that region is zero. True or false? and WHY?

    If you limit yourself to electrostatics, that's true because the electric field is then simply equal to (the opposite of) the gradient of the voltage. If the voltage is constant, that gradient is zero. See first link below. However, that's not true for dynamic fields because (the opposite of) the electric field is then equal equal to the... show more
    If you limit yourself to electrostatics, that's true because the electric field is then simply equal to (the opposite of) the gradient of the voltage. If the voltage is constant, that gradient is zero. See first link below. However, that's not true for dynamic fields because (the opposite of) the electric field is then equal equal to the sum of two terms. One is the aforementioned gradient. The other is the time-derivative of the vector potential (usually denoted A). The latter term is not necessarily zero. See second link. For example, the usual way to describe a planar electromagnetic wave involves zero (constant) voltage everywhere. Yet, the electric field isn't zero. See third link.
    4 answers · Physics · 3 years ago
  • Instrument string 80.0cm long, mass 8.77g. Played in room w/ speed of sound=334 m/s. Find TENSION so that second overtone wavelength=0.775m?

    From the speed of sound in the air, we deduce the frequency from the wavelength (in air). f = (334 m/s) / (0.775 m) = 430.96774... Hz Properly rounded (see first link below) this is: f = 431 Hz For a string of length L to vibrate at that frequency in the second overtone, the speed V of propagation of the string wave (UNRELATED to the speed of... show more
    From the speed of sound in the air, we deduce the frequency from the wavelength (in air). f = (334 m/s) / (0.775 m) = 430.96774... Hz Properly rounded (see first link below) this is: f = 431 Hz For a string of length L to vibrate at that frequency in the second overtone, the speed V of propagation of the string wave (UNRELATED to the speed of sound in the air) must verify the relation: f = 2 ( V / 2L ) = V / L [in the second overtone one full wave matches exactly the string length] Thus V = L f = (0.8 m) (431 Hz) = 344.8 m/s More precisely, before rounding, V = (0.8 / 0.775) 334 m/s = 344.774... m/s Now, the speed V is related to the mass m of the string and its tension T by the following formula V^2 = T / (m/L) [ see second link ] Therefore: T = m V^2 / L = ( 0.00877 kg ) (344.774... m/s)^2 / (0.8 m) = 1303 N Note that the rounding to 4 significant digits is appropriate because it implies only 1.48 times better relative precision than the data (which gives 8.77 to 3 significant digits). So, this isn't really misleading. On the other hand, quoting the answer as "1300 N" would be very inappropriate because the 2 significant digits thus given represent a precision which is far too corase (67 times worse than the data). See first link (again).
    1 answer · Physics · 3 years ago
  • Physics hw question?

    Best answer: Let R = 7.05 e-4 m be the radius of the (long) wire. (That's half of 1.41 mm.) Let mu_0 = 4 pi 10^-7 (N/A^2) denote the permeability of the vacuum.(see first link below) If a total current I is flowing in the wire, the magnetic field B at the surface is given by Ampere's law, using the cylindrical symmetry of the system (see... show more
    Best answer: Let R = 7.05 e-4 m be the radius of the (long) wire. (That's half of 1.41 mm.) Let mu_0 = 4 pi 10^-7 (N/A^2) denote the permeability of the vacuum.(see first link below) If a total current I is flowing in the wire, the magnetic field B at the surface is given by Ampere's law, using the cylindrical symmetry of the system (see second link): B = (I/R) mu_0 / (2 pi) = (I/R) 2e-7 This is below 0.1075 T when the current I is less than the following value: I = B R / (2 e-7) = (0.1075) ( 7.05 e-4) / (2 e-7) = 378.9375 A We round this "exact" result to match the precision of the data (see last link). The maximum current compatible with superconductivity is thus I = 379 A
    1 answer · Physics · 3 years ago
  • What is the magnitude of the electric field?

    Best answer: Simply add the fields due to each particle (BTW the microcoulomb is abbreviated uC, not uc). The position described is 26 cm from the +25uC charge and 15 cem from the -15uC charge. So, the total field (taking left-to-right as the positive direction) is equal to: E = k (+25 e-6 C) / (0.026 m)^2 + k (-15 e-6 C) / (0.015 m)^2 where k is the... show more
    Best answer: Simply add the fields due to each particle (BTW the microcoulomb is abbreviated uC, not uc). The position described is 26 cm from the +25uC charge and 15 cem from the -15uC charge. So, the total field (taking left-to-right as the positive direction) is equal to: E = k (+25 e-6 C) / (0.026 m)^2 + k (-15 e-6 C) / (0.015 m)^2 where k is the electric constant 1/(4 pi epsilon_0) EXACTLY equal to k = 8.9875517873681764 e+9 N.m^2 / C^2 (see first link below) so that: E = -0.26679... e+9 N/C (Note that 1 N/C = 1 V/m) Properly rounded to two significant digits (see second link): E = -0.27 e+9 N/C = -270 N/uC The negative sign indicates that the field points to the left.
    1 answer · Physics · 3 years ago
  • Why does a photon have no mass?

    Best answer: Well, actually, the energy of a particle of mass m at speed v is E = m c^2 / sqrt (1-v^2/c^2) This implies two complementary aspects of the same thing: 1) A particle with nonzero mass MUST move at a speed less than c (or else it would have infinite energy). 2) A particle moving at speed c MUST have zero mass. In the latter case, the... show more
    Best answer: Well, actually, the energy of a particle of mass m at speed v is E = m c^2 / sqrt (1-v^2/c^2) This implies two complementary aspects of the same thing: 1) A particle with nonzero mass MUST move at a speed less than c (or else it would have infinite energy). 2) A particle moving at speed c MUST have zero mass. In the latter case, the above formula cannot be used to obtain the energy of the particle (the expression 0/0 is undefined) which must come from other considerations (i.e., the formula E=hf). Formerly, the mass m of a particle used to be called "rest mass" and the symbol "m0" was typically used to denote it (those notations were used by Einstein himself but are now deprecated). The quantity E/c^2 was once called "relativistic mass" (a term which is now unused except, unfortunately, at the introductory level) and renamed "m" to FORCE the famous equation E=mc^2 to hold in any frame of reference! Not a good idea, pedagogically speaking, as it turned out. To avoid all confusions, it's best to state that the equation E=mc^2 only holds in the rest frame of a particle. A particle that moves at speed c has NO REST FRAME (it's moving at speed c in every possible reference frame) and no mass. The first link below puts things in perspective for particles of nonzero masses. The second link applies to particles of either zero or nonzero masses (it explains E=hf). (Full disclosure: I'm the author of both.)
    5 answers · Physics · 4 years ago
  • A towing van of mass 6.0 * 10^3 kg used in towing a car of mass m moved with a speed of 4ms^-1 just before the?

    Best answer: With the usual simplifying assumptions, the situation amounts to a "shock" of two masses which end up moving at the same speed (when the rope is "taut"). When the rope is still loose, the van of mass M = 6000 kg moves at speed V = 4 m/s and the car of mass m is motionless (at speed 0 m/s). When the rope becomes taut,... show more
    Best answer: With the usual simplifying assumptions, the situation amounts to a "shock" of two masses which end up moving at the same speed (when the rope is "taut"). When the rope is still loose, the van of mass M = 6000 kg moves at speed V = 4 m/s and the car of mass m is motionless (at speed 0 m/s). When the rope becomes taut, both vehicles move at the same speed v = 2.5 m/s (i) Conservation of momentum holds (as discussed in the footnote below) so: M V + 0 = (M+m) v Solving for m, we obtain: m = M ( V/v - 1 ) = (6000 kg) ( 4/2.5 - 1 ) = 3600 kg = 3.6 10^3 kg (ii) The loss E in kinetic energy boils down to: E = M V^2 / 2 - (M+m) v^2 / 2 = (M/2) [ V^2 - (V/v) v^2 ] = M V (V-v) / 2 E = (6000 kg) (4 m/s) (1.5 m/s) / 2 = 18000 J (iii) The impulse P transferred by the rope from the van to the car is either the change in the car's momentum or the opposite of the change in the van's momentum: P = m v = (3600 kg) (2.5 m/s) = 9000 kg.m/s or (same thing): P = M (V-v) = (6000 kg) (1.5 m/s) = 9000 kg.m/s = 9000 N.s FOOTNOTE: Horizontal interactions with the road (friction and/or traction from the van's engine) involve only forces of limited magnitude (see second link). The resulting change in total momentum can thus be neglected if the duration of the "shock" is brief. Our assumption of momentum conservation is good for a fairly rigid rope (it could be questionable for a rubber rope). The maximum error due to this assumption is proportional to the duration of the "shock".
    2 answers · Physics · 4 years ago
  • Physics: Rotational motion?

    Best answer: This setup is traditionally called a "conical pendulum" I refer to the first link below, which includes a picture. PRECISE SOLUTION: Apply the theorem of Pythagoras to the right triangle shown there, whose hypotenuse is F/m (where F is the tension, of at most 100 N and m = 200 g is the mass of the bob). The vertical side is g... show more
    Best answer: This setup is traditionally called a "conical pendulum" I refer to the first link below, which includes a picture. PRECISE SOLUTION: Apply the theorem of Pythagoras to the right triangle shown there, whose hypotenuse is F/m (where F is the tension, of at most 100 N and m = 200 g is the mass of the bob). The vertical side is g (acceleration of gravity) and the horizontal side is v^2/R (the centripetal acceleration along the circular trajectory of radius R). (F/m)^2 = g^2 + (v^2/R)^2 Solving for v^4 at the maximum allowed value of F, we obtain: v^4 = R^2 [ (F/m)^2 - g^2 ] = (2.5)^2 [ (100/0.2)^2 - (9.80665)^2 ] = 1561900 Raising this to the power of 1/4, we obtain: v = 35.35194 m/s APPROXIMATIVE SOLUTION: If we had neglected the acceleration of gravity we'd have obtained: v^2 = 2.5 (100/0.2) = 1250 v = 35.35534 m/s At the precision of two digits implied by the data (see last link below) the rounded result is the same with either "method": v = 35 m/s WHAT THE QUESTION MIGHT HAVE BEEN: The latter (approximative) approach is only applicable when the maximum allowed tension is much greater than the weight of the bob itself. For example: 1) If the maximum allowed tension had been 8 N instead of 100 N, the simplistic approach would have given v = 10 m/s which already differs substantially from the correct result of v = 9.846223 m/s (properly rounded to 9.8 m/s). 2) If the maximum tension is only 2 N (which is barely above the weight of the bob) the simplistic method is completely wrong... It would give v = 5 m/s instead of the correct result of v = 2.211866 m/s (properly rounded to 2.2 m/s). That would have been a relative error of about 126% !
    2 answers · Physics · 4 years ago
  • Physics honors - electric fields and forces.?

    Best answer: The equation to use is simply F = q E. See first link. E = F / q = 2.4 E10 N/C = 2.4 E10 V/m Now, the data is SILLY since the charge of a proton is 1.602 E-19 C You can only observe multiples of that. NEVER 1.25 E-19 C (Quarks may have 1/3 or 2/3 the elementary charge. Even if they could be isolated in a dryer--which isn't the... show more
    Best answer: The equation to use is simply F = q E. See first link. E = F / q = 2.4 E10 N/C = 2.4 E10 V/m Now, the data is SILLY since the charge of a proton is 1.602 E-19 C You can only observe multiples of that. NEVER 1.25 E-19 C (Quarks may have 1/3 or 2/3 the elementary charge. Even if they could be isolated in a dryer--which isn't the case--that wouldn't do either.) Likewise, an electric field can't be this big in this mundane context: Air breaks down under a much smaller electric field (about 10000 times less is enough, see second link). Feel free to forward this to your teacher, or whomever. I wish generations of students (and/or teachers) wouldn't be led to believe that physics is all about applying formulas mindlessly. Please, don't be one of them; figure out what the numbers mean! Best regards. (N.B. I'm the author of the first page quoted below.)
    2 answers · Physics · 4 years ago
  • Why doesn't Hawking Radiation just get pulled back into a black hole?

    Anything traveling at the speed of light won't fall into the black hole unless it happens to travel inside the so-called "event horizon". Anything traveling outward at or "near" the speed of light won't cross the event horizon, if it's just a little bit outside of it to begin with. Such is the case of Hawking... show more
    Anything traveling at the speed of light won't fall into the black hole unless it happens to travel inside the so-called "event horizon". Anything traveling outward at or "near" the speed of light won't cross the event horizon, if it's just a little bit outside of it to begin with. Such is the case of Hawking radiation (whether it consists of massless or massive particles). Let's be more precise... According to the most popular explanation (more sophisticated explanations being compatible with this) Hawking radiation is produced near the event horizon when a pair of virtual particles is spontaneously created but one of them falls into the black hole and the other doesn't. It then looks like the non-absorbed particle was emitted from a black hole (which lost mass by virtue of "absorbing" a virtual particle of negative energy). Now, not every pair production results in Hawking emission. If both particles recombine (inside or outside the black hole, makes no difference) it's as if nothing happened. An "Hawking particle" (bad name, I'm just being provocative) that would fall back into the black hole would, so-to-speak, find its former partner there and everything would look as if nothing happened (in fact, in a deep sense, nothing did). Hawking radiation consists, by definition, of what was emitted without being reabsorbed. The statement "without being reabsorbed" is not even logically needed: If it was reabsorbed, it wasn't emitted, strictly speaking. The wonderful thing is that some stuff does escape the vicinity of the black hole so we may call it "radiation". Some stuff doesn't escape and we call it... "nothing".
    4 answers · Physics · 4 years ago
  • Sum of alternating series?

    This is a special case of the exponential series: e^z = SUM ( n = 0 to infinity, z^n / n! ) with z = -1/3 Therefore, the sum is e^(-1/3) which is equal to: 0.716531310573789250425604096925... Properly rounded to 4 decimal places, this is: 0.7165 If you wanted to compute that in an elementary way. notice that it's an alternating series of... show more
    This is a special case of the exponential series: e^z = SUM ( n = 0 to infinity, z^n / n! ) with z = -1/3 Therefore, the sum is e^(-1/3) which is equal to: 0.716531310573789250425604096925... Properly rounded to 4 decimal places, this is: 0.7165 If you wanted to compute that in an elementary way. notice that it's an alternating series of decreasing terms. Therefore, the error we make by using a partial sum to approximate the whole sum is less than the magnitude of the first discarded term. The sum of the first terms for n = 0,1,2,3,4 is 0.716563786... The next term (n=5) is -0.00003429... This isn't quite good enough to stop at n=4 since all we would know so far is that the whole sum is at least 0.71652949 (n=5, rounded down) at most 0.71656379 (n=4, rounded up) Those two are different to a 4-digit precision (0.7165 and 0.7166) so we must add terms up to n=5 and check that the next term (+0.0000019 for n=6) is small enough to validate the approximation to 0.7165. More precisely, we can say that the result is at least 0.71652949 (n=5, rounded down) at most 0.71653140 (n=6, rounded up). This does establish both the correctness of 0.71653 at a 5-digit precision and that of 0.7165 at a 4-digit precision.
    3 answers · Mathematics · 4 years ago
  • Is this parametric curve a knot?

    The ONLY thing to remark is that this curve is drawn on the surface of a sphere (of radius 1, centered at (2,2,0). Knotted curves can't exist on a sphere or a plane.
    The ONLY thing to remark is that this curve is drawn on the surface of a sphere (of radius 1, centered at (2,2,0). Knotted curves can't exist on a sphere or a plane.
    5 answers · Mathematics · 4 years ago
  • A proton accelerates from rest to 3.00 x 106 m/s in 1.00 x 10-6 s in a uniform electric field E.?

    Best answer: Because the speed reached is only 1% of the speed of light, a nonrelativistic approach is sufficient, at the 3-digit precision needed. The force applied to the proton is qE. The uniform acceleration it undergoes is: q E / M = (3.00 e+6 m/s) / (1.00 e-6 m/s) = 3.00 e+12 m/s^2 Solving for E with q = 1.60 e-19 C [charge of the... show more
    Best answer: Because the speed reached is only 1% of the speed of light, a nonrelativistic approach is sufficient, at the 3-digit precision needed. The force applied to the proton is qE. The uniform acceleration it undergoes is: q E / M = (3.00 e+6 m/s) / (1.00 e-6 m/s) = 3.00 e+12 m/s^2 Solving for E with q = 1.60 e-19 C [charge of the proton] and M = 1.67 e-27 kg , we obtain: E = (1.67 e-27 kg)(3.00 e+12 m/s^2) / (1.60 e-19 C) E = 31300 N/C = 31300 V/m = 31.3 kV/m
    1 answer · Physics · 5 years ago
  • A charge is moving at an angle of 30.0 degrees to a magnetic field and experiences a force of 3.95mN.?

    Best answer: The vectorial force applied to a charge q moving at velocity v in a magnetic field B is q times the cross product of v and B. F = q v x B If A is the angle between v and B, this yields: || F || = q ||v|| . ||B|| . | sin A | Therefore, the force is proportional to the sine of the angle. (It's maximum when the velocity is... show more
    Best answer: The vectorial force applied to a charge q moving at velocity v in a magnetic field B is q times the cross product of v and B. F = q v x B If A is the angle between v and B, this yields: || F || = q ||v|| . ||B|| . | sin A | Therefore, the force is proportional to the sine of the angle. (It's maximum when the velocity is perpendicular to the field.) If the angle were A' = 60 degrees, the magnitude of the force would be: || F' || = || F || . | sin A' / sin A | = (3.95 mN) ( 0.8660254... / 0.5 ) So, the magnetic force experienced by the charge would have a magnitude of (1.73205...) (3.95 mN) = 6.8416... mN We round that result to the 3-digit precision of the data to state that the force is 6.84 mN.
    1 answer · Physics · 4 years ago
  • Can you give an example of an infinite set such that the set of its parts is countable?

    Best answer: No, there is no such thing. Every infinite set E contains a set N equipollent to the natural integers. The set of the parts of E (also called the powerset of E and often denoted 2^E) contains the set of the parts of N (2^N) which is not countable (by Cantor's theorem, see link below). Therefore, 2^E is not countable. show more
    Best answer: No, there is no such thing. Every infinite set E contains a set N equipollent to the natural integers. The set of the parts of E (also called the powerset of E and often denoted 2^E) contains the set of the parts of N (2^N) which is not countable (by Cantor's theorem, see link below). Therefore, 2^E is not countable.
    4 answers · Mathematics · 5 years ago
  • What is the ratio C1/C2?

    Best answer: Charge is conserved. On the other hand energy may not be conserved (some of it can be dissipated by the violent currents and/or sparks that occur when the conductors at different potentials come in contact). When a capacity C has a charge q, its voltage is such that q = CU. Its energy is: 0.5 C U^2 = 0.5 ^2 / C Before contact, the... show more
    Best answer: Charge is conserved. On the other hand energy may not be conserved (some of it can be dissipated by the violent currents and/or sparks that occur when the conductors at different potentials come in contact). When a capacity C has a charge q, its voltage is such that q = CU. Its energy is: 0.5 C U^2 = 0.5 ^2 / C Before contact, the energy of the first capacitor is thus W = 0.5 q^2 / C1 After contact, the voltage is the same on both capacitor so their respective charges q1 and q2 verify the relation q1/C1 = q2/C2 with q1+q2 = q. Plug the value q2 = q1 C2/C1 from the first equation into the second one to obtain: q = q1 (1 + C2/C1) The energy of the first capacity is then W1 = 0.5 q1^2 / C1 = W (q1/q)^2 = W / (1 + C2/C1)^2 Since we're told that W1/W = 1/2 we have: 2 = (1 + C2/C1)^2 = (1 + 1/x)^2 with x = C1/C2 Solving for x we obtain: 1+1/x = sqrt(2) x = 1 / (sqrt(2)-1) = sqrt(2) + 1
    1 answer · Physics · 3 years ago
  • Electricity and Magnetism?

    Let q = 6,40 e-19 C (more precisely, 4 times the elementary charge = 6.4087... e-19 C) Let R = 4.68 e-3 m be the radius of the trajectory. Let B = 1.65 T by the magnetic induction (= "magnetic field"). Let v be the speed of the particle (assumed to be nonrelativistic). Let m be the mass of the particle; its (nonrelativistic) momentum is p... show more
    Let q = 6,40 e-19 C (more precisely, 4 times the elementary charge = 6.4087... e-19 C) Let R = 4.68 e-3 m be the radius of the trajectory. Let B = 1.65 T by the magnetic induction (= "magnetic field"). Let v be the speed of the particle (assumed to be nonrelativistic). Let m be the mass of the particle; its (nonrelativistic) momentum is p = mv). The centripetal (Lorentz) force exerted by the magnetic field has magnitude F = q v B (see first link below) F is also equal to m times the ventripetal acceleration v^2 / R (see second link). Therefore, F = m v^2 / R = q v B which boils down to: m v = q B R = (6.40 e-19 C) (1. 65 T) ( 4.68 e-3 m) = 4.95 e-21 kg.m/s This is the answer to question (a) properly rounded (see third link). The orbital angular momentum (question b) is simply R m v (see fourth link). Numerically, this is the above multiplied into R: L = (m v) R = 2.32 e-23 kg.m^2/s For "extra credit", this last result can also be given in the correct theoretical units for angular momentun (rarely used, see last link). L = 2.32 J.s/rad
    1 answer · Physics · 3 years ago
  • Has liquid helium ever been used to attempt to extend the nuclear life of super heavy man made elements?

    No. The lifetime of a radioactive nucleus is "almost" independent of the surrounding temperature. I am saying "almost" because there's a tiny relativistic effect which makes the "internal clock" of nuclei slow down at high temperature (because they move faster). When you cool radioactive nuclei (initially at... show more
    No. The lifetime of a radioactive nucleus is "almost" independent of the surrounding temperature. I am saying "almost" because there's a tiny relativistic effect which makes the "internal clock" of nuclei slow down at high temperature (because they move faster). When you cool radioactive nuclei (initially at room temperature) you make their lifetime DECREASE (very, very, very slightly).
    3 answers · Chemistry · 4 years ago
  • Physics question on rotational motion?

    Best answer: J = 0.12 kg.m^2 is the rotational inertia of the disk of radius R = 0.08 m m = 10 kg is the mass of the load. g = 9.80665 N/kg is the acceleration of gravity. T = 9.0 N.m is the torque applied by the crank to the disk. Let F be the tension of the string. If A is the (upward) acceleration of the mass, the angular acceleration of the disk... show more
    Best answer: J = 0.12 kg.m^2 is the rotational inertia of the disk of radius R = 0.08 m m = 10 kg is the mass of the load. g = 9.80665 N/kg is the acceleration of gravity. T = 9.0 N.m is the torque applied by the crank to the disk. Let F be the tension of the string. If A is the (upward) acceleration of the mass, the angular acceleration of the disk is A/R. We may apply Newton's law to the linear motion of the mass and to the rotational motion of the disk (first and second links below): F - mg = m A T - RF = J A/R This is just a linear system of two equations in two unknowns (A and F). We eliminate F by plugging into the latter equation the value obtained from the former: T - R (A+g) m = J A/R We may then solve this for A to obtain: T - m g R = A (m R + J/R) A = (T-mgR) / (mR+J/R) = (9.0-10*9.80665*0.08) / (10*0.08+0.12/0.08) A = 0.50203478... m/s^2 Properly rounding this result to the precision of the data (see last link below) we state the answer as: A = 0.50 m/s^2 The same result can also be obtained directly (without introducing the string tension F) by equating the total external torque (T-mgR) to the rate of change of the total angular momentum (mRA+JA/R).
    2 answers · Physics · 4 years ago
  • Solenoid Magnetic Field - Physics Question?

    Best answer: Not too close to its extremities and not too close to the wires themselves, the magnetic field inside a LONG solenoid is constant and equal to what it would be inside an infinite ideal solenoid around which would circulate a uniform current NI per unit of length (in practice, I is the actual current in the wire and N is the number of... show more
    Best answer: Not too close to its extremities and not too close to the wires themselves, the magnetic field inside a LONG solenoid is constant and equal to what it would be inside an infinite ideal solenoid around which would circulate a uniform current NI per unit of length (in practice, I is the actual current in the wire and N is the number of turns of wire per unit of length). The formula is: B = (mu0) N I where mu0 is the magnetic constant (second link below). Here N = 1350 / 0.64 (assuming uniform winding) and I = 6.0 A. So, B = (4 pi 10^-7) (1350 / 0.64 ) 6.0 = 0.0159 T The formula for an infinite solenoid can be obtained as a straightforward application of Ampere's law, since the cylindrical symmetry of an infinite solenoid implies that the field is constant both inside the wiring and outside of it (where it's zero). Details are available at the first link quoted below.
    1 answer · Physics · 4 years ago
  • A 5-m uniform board with a mass of 25 kg is supported by a fulcrum 2 m from one end. A 12-kg mass ........?

    Best answer: The weight of the board is applied at its center, 0.5 m from the fulcrum. Equilibrium is realized with a mass M on the same side (3 m from the fulcrum) and the 12 kg mass (2 m on the other side of the fulcrum) if the torques balance: (25 kg) (0.5 m) + M (3 m) = (12 kg)( 2 m) We solve this for M: M = [ ( 12 * 2 - 25 * 0.5) / 3 ]... show more
    Best answer: The weight of the board is applied at its center, 0.5 m from the fulcrum. Equilibrium is realized with a mass M on the same side (3 m from the fulcrum) and the 12 kg mass (2 m on the other side of the fulcrum) if the torques balance: (25 kg) (0.5 m) + M (3 m) = (12 kg)( 2 m) We solve this for M: M = [ ( 12 * 2 - 25 * 0.5) / 3 ] (kg) = 3.83333... kg Properly rounding to 2 significant digits (see last link below) we obtain: M = 3.8 kg
    1 answer · Physics · 4 years ago