• Physics初中電路 http://upload.lsforum.net/users/public/u2711^g72.png?

    (a) I1 = Vs/R1 = 12/75 A = 0.16 A (b) When S2 is open, no current flow through I3. Hence, I3 = 0 A (c) Equivalent resistance of R4 and R5 in parallel = 100 x 150/(100 + 150) ohms = 60 ohms Hence, voltage across R4 = 12 x [60/(20 + 60)] ohms = 9 volts Current I4 = 9/100 A = 0.09 A (d) Equivalent resistance of R3 in series with R4 and R5 in... 顯示更多
    (a) I1 = Vs/R1 = 12/75 A = 0.16 A (b) When S2 is open, no current flow through I3. Hence, I3 = 0 A (c) Equivalent resistance of R4 and R5 in parallel = 100 x 150/(100 + 150) ohms = 60 ohms Hence, voltage across R4 = 12 x [60/(20 + 60)] ohms = 9 volts Current I4 = 9/100 A = 0.09 A (d) Equivalent resistance of R3 in series with R4 and R5 in parallel = (20 + 60) = 80 ohms Hence, equivalent resistance of the circuit = 1/[(1/75) + (1/50) + (1/80)] ohms = 21.81 ohms Therefore, Is = 12/21.81 A = 0.55 A
    1 個解答 · 物理學 · 3 天前
  • To 天同知識長.....?

    最佳解答: You are welcome ! It's my pleasure thast I could share my knowledge and experience in physics with you, and with other candidates in this forum. Your success is attributed to your hard work and willingness to learn. Congratulation to your getting a high score in physics. Wish you would enjoy studying physics at university. 顯示更多
    最佳解答: You are welcome ! It's my pleasure thast I could share my knowledge and experience in physics with you, and with other candidates in this forum. Your success is attributed to your hard work and willingness to learn. Congratulation to your getting a high score in physics. Wish you would enjoy studying physics at university.
    1 個解答 · 物理學 · 2 月前
  • Pressure problem?

    最佳解答: Pressure difference between points at dia = 300mm and dia = 150mm = 1 x (1.25 x 1000)g N/m^2 = 12,263 N/m^2 Using Bernoulli Equation: 12263 = (1/2).(1.25 x 1000).[(V2)^2 - (V1)^2] i.e. (V2)^2 - (V1)^2 = 19.62 (m/s)^2 ------------ (1) where V1 and V2 are the water flow speed respectively at the two specified points. Cross-sectional area at... 顯示更多
    最佳解答: Pressure difference between points at dia = 300mm and dia = 150mm = 1 x (1.25 x 1000)g N/m^2 = 12,263 N/m^2 Using Bernoulli Equation: 12263 = (1/2).(1.25 x 1000).[(V2)^2 - (V1)^2] i.e. (V2)^2 - (V1)^2 = 19.62 (m/s)^2 ------------ (1) where V1 and V2 are the water flow speed respectively at the two specified points. Cross-sectional area at diameter 300mm = (pi).(0.15)^2 m^2 = 0.07069 m^2 Cross-sectional area at diameter 150 mm = (pi).(0.075)^2 m^2 = 0.01767 m^2 Using Continuity Equation at the same two points: q = (V1).(0.07069) = (V2).(0.01767) where q is the theoretical water flow rate i.e. V2 = 4(V1) Hence, equation (1) becomes, 16(V1)^2 - (V1)^2 = 19.62 V1 = 1.144 m/s Thus q = 1.144 x 0.07069 m^3/s = 0.08087 m^3/s Given the actual water flow rate = 0.0037 m^3/s hence the coefficient of discharge = 0.0037/0.08087 = 0.0458
    1 個解答 · 物理學 · 5 月前
  • 我想問紙,如簿,書,要加熱到幾多度才燒著?是不是愈厚所需溫度愈高? 如果是字典,要幾多度才燒著?謝謝!?

    The ignition temperature (also refers as auto-ignition temperature) for paper is commonly taken as 233 degrees Centigrade (or 451' F). Since there are different kinds of paper with various material composition, volume, density and shape, the ignition temperature may varies from around 218'C to 246'C. Reference:... 顯示更多
    The ignition temperature (also refers as auto-ignition temperature) for paper is commonly taken as 233 degrees Centigrade (or 451' F). Since there are different kinds of paper with various material composition, volume, density and shape, the ignition temperature may varies from around 218'C to 246'C. Reference: http://hypertextbook.com/facts/2003/LewisChung.shtml It takes a few minutes for a sheet of paper to burst into flames when it is put at its ignition temperature, but it would take much longer than that for a thick book. The dense material in the center of a book shunts heat away from the outside edges, preventing them from reaching the auto-ignition temperature.
    1 個解答 · 物理學 · 5 月前
  • Physical problems?

    最佳解答: 4. The equations involved in calculating the thrust on an inclined plane are: Thrust on inclined plane Fr = pg(hc).A where p is the density of water (= 1000 kg/m^3), g is the acceleration due to gravity (= 9.81 m/s^2), hc is the vertical depth, from water surface, of the centre of mass of the plane, A is the area of the plane. For a... 顯示更多
    最佳解答: 4. The equations involved in calculating the thrust on an inclined plane are: Thrust on inclined plane Fr = pg(hc).A where p is the density of water (= 1000 kg/m^3), g is the acceleration due to gravity (= 9.81 m/s^2), hc is the vertical depth, from water surface, of the centre of mass of the plane, A is the area of the plane. For a rectangular plane of height H and width W, the thrust acts at a distance (yr), measured in direction along the inclined plane from water surface, is yr = yc + WH^3/(12.WH.yc) where yc is the distance, measured in direction along the inclined plane from water surface, of the centre of mass of the plane, and yc = hc/sin(a) where hc is the vertical depth from water surface of the centre of mass of the plane, and a is the inclined angle of the plane. First consider the thrust acting on the left hand side of the gate, Fr = 1000 x 9.81 x [4 - 1 x sin(60)] x (2 x 1) m = 61,489 N The point of action of Fr is given by, yr = yc + (1 x 2^3)/[12 x (1 x 2)yc] where yc = [4 - 1 x sin(60)]/sin(60) m = 3.619 m Hence, yr = 3.619 + (1 x 2^3)/[12 x (1 x 2) x 3.619] m = 3.711 m Distance along the plane of Fr from the hinge A = yr - [(4 - 2.sin(60))/sin(60)] m = 3.711 - [(4 - 2.sin(60))/sin(60)] m = 1.092 m There the thrust on the left hand side of the gate is 61,489 N and acts at a distance of 1.092 m from the hinge at A. Using the same method, the thrust on the right hand side of the gate Fr' can be found equals to 7,358 N and acts at a distance of 1.423 m from A. Take moment about hinge A, W.cos(60) x 1 + (Fr') x 1.423 = (Fr) x 1.092 i.e. W = (61489 x 1.092 - 7358 x 1.423)/cos(60) N = 113,351 N 5. You could use the same equations given above to solve this problem. Just try it yourself.
    1 個解答 · 物理學 · 5 月前
  • 所算答案有偏差 求解?

    最佳解答: Consider an observer inside the car (which is a non-inertia reference system). The pendulum bob, when in equilibrium inside the car, is subject to the following real and inertia forces: (i) the resolved component of the weight of the bob that is pointing downward parallel to the inclined plane. This equlas to mg.sin(alpha), where m is the mass... 顯示更多
    最佳解答: Consider an observer inside the car (which is a non-inertia reference system). The pendulum bob, when in equilibrium inside the car, is subject to the following real and inertia forces: (i) the resolved component of the weight of the bob that is pointing downward parallel to the inclined plane. This equlas to mg.sin(alpha), where m is the mass of the bob, and g is the acceleration due to gravity; (ii) the resolved component of the weight of the bob that is normal to the inclined plane, which is mg.cos(alpha); (iii) tension in the suspended string of the bob; (iv) the inertia force due to the downward acceleration of the car. This equals to mg.sin(alpha) in direction pointing upward paralllel to the inclined plane. In the above four forces, forces in (i) and (iv) balance one another. The same happens for forces in (ii) and (iii). When the pendulum bob is set into oscillation from its equilibrium position, its period of oscillation T is given by, T = 2.(pi).square-root[L/g'] where g' is the apparent acceleration due to gravity under the situation where the bob is placed. In this case, g' = g.cos(alpha) Hence, T = 2.(pi).square-root[L/(g.cos(alpha))]
    1 個解答 · 體育 · 5 月前
  • Find the time for a wave to travel from one end to the other end of the string.?

    最佳解答: Your calculation seems correct. The right hand side of the equation is the wave velocity expressed in m/s (metres per second). Hence, the time t is in unit of "second"
    最佳解答: Your calculation seems correct. The right hand side of the equation is the wave velocity expressed in m/s (metres per second). Hence, the time t is in unit of "second"
    1 個解答 · 物理學 · 5 月前
  • Plz help me to finish this physics question?

    最佳解答: The equation describes the motion of the proton is, m(dV/dt) = q(V x B) [Note: Capital letters of V and B stand for Vector V and Vector B] i.e. dV/dt = (q/m)(V x B) But (V x B) = [(vox)i + (voy).cos(wt)j - (voy)sin(wt)k] x (bi) where (vox) and (voy) are the initial x and y components of the velocity V; b is the magnitude of the magnetic field... 顯示更多
    最佳解答: The equation describes the motion of the proton is, m(dV/dt) = q(V x B) [Note: Capital letters of V and B stand for Vector V and Vector B] i.e. dV/dt = (q/m)(V x B) But (V x B) = [(vox)i + (voy).cos(wt)j - (voy)sin(wt)k] x (bi) where (vox) and (voy) are the initial x and y components of the velocity V; b is the magnitude of the magnetic field B] After expanding the cross product and simplify, i.e. (V x B) = -b(voy)[sin(wt)j + cos(wt)k] Hence, right-hand-side of the equation = -(qb/m).(voy)[sin(wt)j + cos(wt)k] --------- (1) Now, consider the left-hand side of the equation (dV/dt), dV/dt = d[(vox)i + (voy).cos(wt)j - (voy).sin(wt)k]/dt dV/dt = -(voy)w.sin(wt)j - (voy)w.cos(wt)k i.e. dV/dt = -(voy)w[sin(wt)j + cos(wt)k] Given that w = qb/m, we have, dV/dt = -(qb/m).(voy)[sin(wt)j + cos(wt)k] ----------- (2) By comparing equations (1) and (2), it shows that the given expression for V satisfies the equation dV/dt = (q/m)(V x B). Therefore, it is a possible velocity solution of the equation.
    1 個解答 · 物理學 · 6 月前
  • Plz help me to finish this physics question?

    最佳解答: (a) Given PQ = L, hence PR = L.cos(40), and RQ = L.sin(40) Area of loop = (1/2).(RQ).(PR) = (1/2).(L.sin(40)).(L.cos(40)) = [(L^2)sin(40)cos(40)]/2 Magnetic moment of current loop = current x area = [I.(L^2)sin(40)cos(40)]/2 The direction of magnetic moment is given by the Right-Hand-Grip-Rule, and is pointing into the paper, i.e. in the -ve... 顯示更多
    最佳解答: (a) Given PQ = L, hence PR = L.cos(40), and RQ = L.sin(40) Area of loop = (1/2).(RQ).(PR) = (1/2).(L.sin(40)).(L.cos(40)) = [(L^2)sin(40)cos(40)]/2 Magnetic moment of current loop = current x area = [I.(L^2)sin(40)cos(40)]/2 The direction of magnetic moment is given by the Right-Hand-Grip-Rule, and is pointing into the paper, i.e. in the -ve z-direction. (b) Torque = [B.I.(L^2)sin(40)cos(40)]/2 The direction of torque is in the -ve y-direction.
    1 個解答 · 物理學 · 6 月前
  • Help super urgent!!!!!!!!!!!!! physics about mirror?

    最佳解答: Use the principle that the image formed by a plane mirror is at the same distance behind the mirror as the object in front Therefore, the first image (I1) is formed by object P on mirror 1. The second image ( I') (not shown on the diagram) is formed by object P on mirror 2. The third image (I2) is formed by image (I1) on mirror 2 (Note:... 顯示更多
    最佳解答: Use the principle that the image formed by a plane mirror is at the same distance behind the mirror as the object in front Therefore, the first image (I1) is formed by object P on mirror 1. The second image ( I') (not shown on the diagram) is formed by object P on mirror 2. The third image (I2) is formed by image (I1) on mirror 2 (Note: This image is the same as the one formed by the second image(I') on mirror 1)
    2 個解答 · 物理學 · 6 月前
  • One physics MC about refraction, one MC about transformer, please help?

    最佳解答: 16. How come that the refracted ray be split into two rays after passing the interface? From Snell's law, a given angle of incident only gives one definite angle of refraction. There won't be two refracted rays. 32. A transformer works on electromagnetic induction. The voltage (since a CRO measures voltages) induced in the secondary... 顯示更多
    最佳解答: 16. How come that the refracted ray be split into two rays after passing the interface? From Snell's law, a given angle of incident only gives one definite angle of refraction. There won't be two refracted rays. 32. A transformer works on electromagnetic induction. The voltage (since a CRO measures voltages) induced in the secondary coil follows Faraday's Law of Electromagnetic induction, which states that the induced emf depends on the RATE OF CHANGE of magnetic flux through the coil . The magnetic flux through the secondary coil is, in turn, proportional to the current, and hence voltage, in the primary coil. That said, the induced emf in the secondary coil is proportional to the RATE OF CHANGE of voltage in the primary coil. Given the above understanding, it is not difficult to see that the RATE of change of voltage in the primary coil becomes the highest at the two minimum points of the given wave form. Hence, the induced emf at the secondary coil will show a maximum voltage. Because the turns-ratio of the transformer is 1:2, the induced emf at the secondary coil will be twice as that at the primary coil. When the primary voltage rises from zero to a maximum, the rate of change of the voltage gradually decreases (i.e. the slope of the curve decreases). The induced emf at secondary is thus decreasing accordingly. Likewise, at the two wave crests, the RATE of change of voltage is zero (i.e. the slope of the curve is zero). This gives rise to zero induced emf at the secondary coil. After passing the first crest, the primary voltage decreases. The rate of change of primary voltage now becomes increasing (the slope of the curve increases, but in the -ve manner). This leads to an increase of induced secondary voltage, but in the opposite direction to that before the wave crest. The same induction process repeats itself after the minimum point. The result of the voltage wave form will be that shown in option B.
    1 個解答 · 物理學 · 6 月前
  • Please help, why is the answer D? This is similar to the circuit question I asked before?

    最佳解答: Just treat the voltmeter as a "resistor of very high resistance" and an ammeter as a "resistor of very low resistance". Then the circuit can be regarded as a 2-volt battery connected to a 100-ohm resistance, and then in series with a "high resistance resistor" (represents the voltmeter) and a "low resistance... 顯示更多
    最佳解答: Just treat the voltmeter as a "resistor of very high resistance" and an ammeter as a "resistor of very low resistance". Then the circuit can be regarded as a 2-volt battery connected to a 100-ohm resistance, and then in series with a "high resistance resistor" (represents the voltmeter) and a "low resistance resistor"(represents the ammeter) in parallel. (a) When switch K is opened, the circuit becomes a 2-volt battery connected to a 100-ohm resistance and a "high resistance resistor" in series. Clearly, the "high resistance resistor" will share most the the voltage from the 2-volt battery. Because a voltmeter records the p.d. across its own internal resistance, the voltmeter reading is close to 2 volts. (b) When switch K is closed, the voltmeter and ammeter are in parallel. Remember the fact that when resistors are in parallel, the equivalent resistance must be lower than the smallest resistance in the parallel circuit. Hence, the equivalent resistance of the voltmeter-ammeter parallel arrangement must be smaller than the resistance of the ammeter. Since the resistance of the ammeter is already very small, the equivalent resistance of the parallel arrangement should be much lower than 100-ohm. As such, most of the voltage drop from the 2-volt batter will be taken by the 100-ohm resistor, leaving only a very small potential drop at the parallel arrangement. As said before, the voltmeter records the potential drop across its own internal resistance, the reading shown by the voltmeter is thus close to 0 volt. Note that in the voltmeter-ammeter parallel arrangement, the current flows through the ammeter is larger than that through the voltmeter (due to the small resistance of the ammeter), but the p.d. across both the ammeter and voltmeter are the same. (just compare it with the question you asked before). In the question that you asked before, the voltmeter is more or less directly connected to the two terminals of the battery (if we take the resistance of the light bulb to be small compared with the resistance of the voltmeter, and its presence could somehow be neglected). The voltmeter thus measures the voltage of the battery. But in this problem, one terminal of the voltmeter is NOT connected to the battery. Hence, its reading wouldn't be the battery voltage. Suppose the terminal of the voltmeter that is connected to the right-hand side of the 100-ohm resistor is now re-connected to the left-hand side of that resistor, then the voltmeter will show a reading of 2 volts (the battery voltage) whether switch K is opened or closed.
    2 個解答 · 物理學 · 6 月前
  • Question about moment! please answer and help thanks?

    最佳解答: I am afraid the "model answer" is wrong. You are right that the equation should be, (N1).(0.6+0.9)/cos(5) = 0.9mg In the model answer, only the resolved (vertical) component N1.cos(5) is considered. The other resolved (horizontal) component N1.sin(5) has been neglected. If the component N1.sin(5) is also taken into consideration,... 顯示更多
    最佳解答: I am afraid the "model answer" is wrong. You are right that the equation should be, (N1).(0.6+0.9)/cos(5) = 0.9mg In the model answer, only the resolved (vertical) component N1.cos(5) is considered. The other resolved (horizontal) component N1.sin(5) has been neglected. If the component N1.sin(5) is also taken into consideration, then the clockwise moment becomes N1.cos(5).(0.6+0.9) + N1.sin(5).y where y is the vertical distance between B and the horizontal line at the level of A. From geometry, tan(5) = y/(0.6+0.9), or y = 1.5.tan(5) Hence, the clockwise moment becomes, N1.cos(5).(1.5) + N1.sin(5).(1.5.tan(5)) = 1.5(N1).cos(5) + 1.5(N1)sin(5).sin(5)/cos(5) = 1.5(N1)[(cos^2(5) + sin^2(5)]/cos(5) = 1.5(N1)/cos(5) [because cos^2(5) + sin^2(5) = 1 ] This is just the clockwise moment that you suggested. In fact, in this problem, it is not necessary to resolve the force N1 into vertical and horizontal components. This makes the calculation even more clumsy. Since N1 is already perpendicular to AB, the moment is just simply (N1).(AB)
    1 個解答 · 物理學 · 6 月前
  • Please help. One MC about wave and one about lens?

    最佳解答: 15. Yes. All particles in a stationary wave are moving in phase. If not, then the "crest" (the anti-node) would be moving forward and it is not a stationary wave. 16. I could only say that if you choose option C, the focal length of the convex lens works out to be 12.2 cm, and which is not exactly 12 cm as given in the question. 顯示更多
    最佳解答: 15. Yes. All particles in a stationary wave are moving in phase. If not, then the "crest" (the anti-node) would be moving forward and it is not a stationary wave. 16. I could only say that if you choose option C, the focal length of the convex lens works out to be 12.2 cm, and which is not exactly 12 cm as given in the question.
    1 個解答 · 物理學 · 6 月前
  • 3 physics MC, Please help very urgent?

    最佳解答: 3. The area under the pv graph represents the work-done by the gas. It doesn't not represent the gas internal energy. Internal energy of an ideal gas depends on temperature only. Because the two gases expand under constant temperature, the internal energies do not change. Notice that when an ideal gas expands isothermally (i.e. under... 顯示更多
    最佳解答: 3. The area under the pv graph represents the work-done by the gas. It doesn't not represent the gas internal energy. Internal energy of an ideal gas depends on temperature only. Because the two gases expand under constant temperature, the internal energies do not change. Notice that when an ideal gas expands isothermally (i.e. under constant temperature), there is heat flow from the surroundings towards the gas so as to maintain the temperature of the gas to be constant. 10. Net force acting on the mass = (18 - 6 - 2g.sin(30)) N = 2.19 N (Take g = 9.81 m/s^2) Impulse = force x time = 2.19 x 1.6 Ns = 3.50 Ns Because impulse = change in momentum, hence the change in momentum of the mass = 3.50 kg.m/s 25. Let E be the emf of the cell. Actual current (Ia) through the 6-ohm resistor = E/6 Measured current (Im) using the given ammeter = E/(1 + 6) = E/7 Error = Ia - Im = E/6 - E/7 Percentage error = (Ia - Im)/(Ia) =[(E/6 - E/7)/(E/6)] x 100% = (1 - 6/7) x 100% = 14.3% Note that you should use the actual current as the denominator in the percentage error calculation, as we want to find how much the measured current deviates from the actual current. The actual current should thus be taken as the reference.
    1 個解答 · 物理學 · 6 月前
  • The conical pendulum consists of a bob of mass m attached to the end of an inflexible light?

    (a) Using the symbols given in the diagram, T.sin(theta) = mg T.cos(theta) = mr.w^2 where w is the angular velocity of the bob. Dividing, T.sin(theta)/T.cos(theta) = mg/(mrw^2) i.e. tan(theta) = g/(r.w^2) But from the diagram, tan(theta) = h/r hence, h/r = g/(r.w^2) w^2 = g/h or w = square-root(g/h) (b) 70 rev/min = 70 x 2.pi/60 rad/s = 7(pi)/3... 顯示更多
    (a) Using the symbols given in the diagram, T.sin(theta) = mg T.cos(theta) = mr.w^2 where w is the angular velocity of the bob. Dividing, T.sin(theta)/T.cos(theta) = mg/(mrw^2) i.e. tan(theta) = g/(r.w^2) But from the diagram, tan(theta) = h/r hence, h/r = g/(r.w^2) w^2 = g/h or w = square-root(g/h) (b) 70 rev/min = 70 x 2.pi/60 rad/s = 7(pi)/3 rad/s and 80 rev/min = 80 x 2.pi/60 rad/s = 8(pi)/3 rad/s From (a) above, h = g/w^2 Therefore, at 70 rev/min, ho = g/[7(pi)/3]^2 m = 9g/[49(pi)^2] m At 80 rev/min, h = g/[8(pi)/3]^2 m = 9g/[64(pi)^2] m Difference in height = ho - h = (9g/pi^2).(1/49 - 1/64) m = 0.0428 m (Take g = 9.81 m/s^2) Hence, the level of the bob is raised by 0.0428 m (or 4.28 cm)
    1 個解答 · 物理學 · 6 月前
  • 同上一個問題 == 現在小學老是都這樣出的唷??? 甲、乙兩車作直線運動,其速率分別為50m/s 與30m/s,若兩車以相同之減速度踩煞車而終致停止, 試求煞車距離比? 【Ans:25:9】?

    Use equation of motion: v^2 = u^2 + 2a.s For car 1: 0 = 50^2 + 2a(s1) i.e. s1 = -50^2/2a For car 2: 0 = 30^2 + 2a(s2) i.e. s2 = -30^2/2a where s1 and s2 are the braking distances for car 1 and car 2 respectively, and a is their common deceleration. Hence, s1/s2 = (50/30)^2 = 25/9 or s1:s2 = 25:9
    Use equation of motion: v^2 = u^2 + 2a.s For car 1: 0 = 50^2 + 2a(s1) i.e. s1 = -50^2/2a For car 2: 0 = 30^2 + 2a(s2) i.e. s2 = -30^2/2a where s1 and s2 are the braking distances for car 1 and car 2 respectively, and a is their common deceleration. Hence, s1/s2 = (50/30)^2 = 25/9 or s1:s2 = 25:9
    2 個解答 · 體育 · 6 月前
  • Urgent physics question?

    最佳解答: In the secondary coil of the transformer, current flows from the upper wire to the middle (110 v) or lower wire (220 v) and vice versa (because it is an ac supply). This makes a complete circuit, which is essential for current to flow. If the person touches ANY ONE of the wires in the secondary coil, current cannot return to either one of the... 顯示更多
    最佳解答: In the secondary coil of the transformer, current flows from the upper wire to the middle (110 v) or lower wire (220 v) and vice versa (because it is an ac supply). This makes a complete circuit, which is essential for current to flow. If the person touches ANY ONE of the wires in the secondary coil, current cannot return to either one of the other two wires. Hence, there is no complete circuit, and no current can flow. In (bi), the LIVE wire of the PRIMARY COIL of the transformer carries a potential, and the other wire (the NEUTRAL wire) is earthed. Current thus flows from the LIVE wire to earth. This completes the circuit. When the person touches the LIVE wire, current can flow from the LIVE wire through the person's body and then to earth (this is where the NEUTRAL wire is connected). This also completes a circuit, thus giving an electric shock to that person. The protection given in (bii) is that none of the wires is earthed. Should any one of the wires is earthed, the person would surely get an electric shock when he touches one of the other two wires (except the earthed wire itself, of course). The situation given in (bii) without earthing any wire is similar to the case of a bird staying on a high-tension (i.e. high voltage) power line. The bird will not get any electric shock because no part of its body touches the earth.
    1 個解答 · 物理學 · 7 月前
  • Question about nano meters physics?

    最佳解答: The term [(0.01/1x10^-9) x 4] counts the number of molecules on the top, bottom and the two sides. In such counting, the molecules at the four corners of the square cross-section have been counted twice. Hence, we need to subtract the 4 double-counted molecules.
    最佳解答: The term [(0.01/1x10^-9) x 4] counts the number of molecules on the top, bottom and the two sides. In such counting, the molecules at the four corners of the square cross-section have been counted twice. Hence, we need to subtract the 4 double-counted molecules.
    1 個解答 · 物理學 · 7 月前
  • One MC about circuit concept?

    最佳解答: What you are saying is correct only if both the ammeter and voltmeter are ideal. But the question doesn't tell that the two are ideal meters. Since the two meters are used in an experiment, we cannot take it for granted that they are ideal. Hence, the internal resistances of the meters need to be considered in answering the question. As a... 顯示更多
    最佳解答: What you are saying is correct only if both the ammeter and voltmeter are ideal. But the question doesn't tell that the two are ideal meters. Since the two meters are used in an experiment, we cannot take it for granted that they are ideal. Hence, the internal resistances of the meters need to be considered in answering the question. As a practical voltmeter has a large resistance, the current flows through the bulb and voltmeter branch (you may just take the voltmeter as a resistor of very high resistance) is extremely small as compared with the current that passes through the ammeter branch, because of the low resistance of the ammeter. Therefore, statement (1) is right because the current passing through the bulb-voltmeter branch is small. Statement (2) is right, because a large current flows through the ammeter branch (due to the small resistance of the ammeter). Statement (3) is wrong. The bulb-voltmeter branch is connected to the battery. Thus the potential difference across this branch equals to the battery emf (assume internal resistance of battery is negligible). Because the bulb has resistance much much lower than that of the voltmeter, most of the potential drop will be across the voltmeter (as said above, just imagine the voltmeter as a resistor of high resistance). A voltmeter measures the potential drop across its two terminals (or the potential drop across its own internal resistance), the voltage thus recorded by the voltmeter will be nearer equal to the emf of the battery. Points to note: Do not mistakenly think that a low current must give rise to a low potential drop, and a large current leads to a high potential drop. You need to consider the involved resistances in the circuit too. In this problem, although the current in the bulb-voltmeter branch is small, the potential difference across this branch is not near zero, because of the high resistance of the voltmeter. Similarly, the current in the ammeter branch is large, but the potential difference across this branch equals to that of the other branch (but NOT higher than). Both branches have potential differences equal to the battery emf.
    1 個解答 · 物理學 · 6 月前