• 請問,一個8L(水容量)的鋼瓶裏有30000psi空氣,連接一個沒有空氣的4.3L(水容量)鋼瓶 問: 1)連接後兩個瓶的壓力? 數式? 2)若4.3L的鋼瓶本來已有10000psi空氣,連接後的壓力?數式? 感謝解答?

    (1) Using Boyle's Law, 30000 x 8 = P x (8 + 4.3) where P is the final pressure after the two steel vessels are connected. Hence, P = 30000 x 8/(8 + 4.3) psi = 19,512 psi (2) Using the Ideal Gas Equation, assume the temperature of the air inside the vessel T remains unchanged, the no. of moles of air in each vessel, before they are connected, ... 顯示更多
    (1) Using Boyle's Law, 30000 x 8 = P x (8 + 4.3) where P is the final pressure after the two steel vessels are connected. Hence, P = 30000 x 8/(8 + 4.3) psi = 19,512 psi (2) Using the Ideal Gas Equation, assume the temperature of the air inside the vessel T remains unchanged, the no. of moles of air in each vessel, before they are connected, is given by, For the 8L vessel, no. of moles of air inside n1 = where R is the Universal Gas Constant.8 x 30000/RT For the 4.3L vessel, no. of moles of air inside n2 = 4.3 x 10000/RT Hence, total mass (no. of moles) of air in the two vessles before connected = n1 + n2 = [(8 x 30000) + (4.3 x 10000)]/RT Mass (no. of moles n) of air after the vessels are connected n = (Pf)(8 + 4.3)/RT where Pf is the final pressure after connection. Using Law of Conservation of Mass, no. of moles of air in the two vessels before and after connection are the same. i.e. n1 + n2 = n [(8 x 30000) + (4.3 x 10000)]/RT = (Pf)(8 + 4.3)/RT solve for Pf gives the final pressure = 23,008 psi.
    1 個解答 · 物理學 · 3 星期前
  • About heat transfer (熱傳)?

    (a) Let Q be the amount of heat conducted through unit cross-sectional area of the wall in unit time. Then, Q = -k(dT/dx) -------------- (1) where k is the thermal conductivity of the wall (= 0.77 W/mK) T is the temperature at distance x in the wall. x is the distance from the inside surface of the wall. [The -ve sign indicates that heat flows... 顯示更多
    (a) Let Q be the amount of heat conducted through unit cross-sectional area of the wall in unit time. Then, Q = -k(dT/dx) -------------- (1) where k is the thermal conductivity of the wall (= 0.77 W/mK) T is the temperature at distance x in the wall. x is the distance from the inside surface of the wall. [The -ve sign indicates that heat flows through the wall from the inside to outside, i.e. there is heat loss.] The boundary conditions are: At x = 0 m, T = T1, the inside surface temperature of the wall At x = 0.2 m, T = T2, the outside surface temperature of the wall. (b) From (1), dT = -(Q/k)dx i.e. T = -integral { (Q/k).dx }, with the limits of integration from 0 to x since Q/k is a constant, we have, T = -(Q/k). integral {dx} i.e. T = -(Q/k)x That means the temperature decreases linearly with distance from the inside surface of the wall. (c) On the inside surface, Q = 5.(27 - T1) --------- (2) On the outside surface, Q = 12.(T2 - 8) ---------- (3) Equating (2) and (3), and solve for T2. This gives T2 = 19.25 - 0.4167(T1) ------------- (4) Consider heat conduction through the wall, Q = 0.77(T1 - T2)/0.2 ------------ (5) Equating (3) and (5), 12(T2 - 8) = 0.77(T1 - T2)/0.2 ---------- (6) Substitute T2 from (4) into (6) and solve for T1, we have, T1 = 20 C hence, substitute T1 = 20 C into (4), we have T2 = 10.92 C
    1 個解答 · 物理學 · 1 月前
  • 求基本電學等效電阻?

    最佳解答: Let's start from A to B. In the 1st loop, 40-ohm and 60-ohm resistors in paralle. Equivalent resistance = 40 x 60/(40+60) ohms = 24 ohms In the 2nd loop, 20-ohm and 50-ohm resistors in series.equivalent resistance = (20 + 50) ohms = 70 ohms This 70-ohm resistance is in parallel with the 30-ohm resistor. Equivalent resistance = 70 x 30/(70... 顯示更多
    最佳解答: Let's start from A to B. In the 1st loop, 40-ohm and 60-ohm resistors in paralle. Equivalent resistance = 40 x 60/(40+60) ohms = 24 ohms In the 2nd loop, 20-ohm and 50-ohm resistors in series.equivalent resistance = (20 + 50) ohms = 70 ohms This 70-ohm resistance is in parallel with the 30-ohm resistor. Equivalent resistance = 70 x 30/(70 + 30) ohms = 21 ohms. Because the two loops are in series, hence equivalent resistance of the two loops = (24 + 21) ohms = 45 ohms
    1 個解答 · 物理學 · 2 月前
  • 請問運動力學的問題。 下表有沒有大大可以幫我解釋一下,這分為四個階段,其中包括初始衝擊峰值,二次沖擊峰值,重量接受度 發球drive-ofF。 drive-off 階段是甚麼意思? stance是姿態還是立場?? 第三期重量接受度,我覺得翻得很奇怪。 為什麼第三四期,這麼平穩?

    Sports Mechanics is not my specialty, but I would like to give my view on this problem. Stance I : Initial Impact Peak This happens when the incoming ball is first caught by the racket . Phase II: Secondary Impact Peak When the incoming ball first collides with the racket, it will rebound. The player then swings the racket to catch up with the... 顯示更多
    Sports Mechanics is not my specialty, but I would like to give my view on this problem. Stance I : Initial Impact Peak This happens when the incoming ball is first caught by the racket . Phase II: Secondary Impact Peak When the incoming ball first collides with the racket, it will rebound. The player then swings the racket to catch up with the ball again. This is the second impact. Phase III: Weight Acceptance As far as the player swings the racket, the ball is being pushed forward rapidly, causing it to accelerate At such moment, the force mainly comes from the swinging arm of the player. Hence, the curve on vGRF (vertical ground reaction force) varies not much. Phase IV: Drive-off Here, the racket provides a final strike onto the ball and makes it flying away from the racket .
    1 個解答 · 體育 · 2 月前
  • 一個物理力學問題: 有一個斜面放在水平的卓上,斜面與水平成50度,有一個100N的力,垂直的壓在斜面上,這樣的話, 斜面會向水平方向移動,那麼即是有多少力從水平方向推動斜面?

    If the force is applied in direction as given in your diagram, the force component pushing the wedge to move horizontally will be zero. But I wonder if you have misinterpret the question. It says that the force is "...垂直的壓在斜面上..", it may mean the force is perpendicular to the slant surface of the wedge. In that case, the component of... 顯示更多
    If the force is applied in direction as given in your diagram, the force component pushing the wedge to move horizontally will be zero. But I wonder if you have misinterpret the question. It says that the force is "...垂直的壓在斜面上..", it may mean the force is perpendicular to the slant surface of the wedge. In that case, the component of force pushing the wedge to the right is 100.sin(50) N = 76.6 N
    2 個解答 · 物理學 · 2 月前
  • Physics frequency order http://upload.lsforum.net/users/public/w349222t72.png?

    最佳解答: Frequency of, - 5 GHz radar = 5 x 10^9 Hz - 3 micron infrared radiation = 3 x 10^8/3 x 10^-6 Hz = 10^14 Hz [where 3 x 10^8 m/s is the speed of light ] - 20 kHz sonar = 20 x 10^3 Hz = 2 x 10^4 Hz - 4 kHz of sound = 4 x 10^3 Hz - blue light = 3 x 10^8/450 x 10^-9 Hz = 6.67 x 10^14 Hz [ Take the wavelength of blue light be 450 nm] - 2000m... 顯示更多
    最佳解答: Frequency of, - 5 GHz radar = 5 x 10^9 Hz - 3 micron infrared radiation = 3 x 10^8/3 x 10^-6 Hz = 10^14 Hz [where 3 x 10^8 m/s is the speed of light ] - 20 kHz sonar = 20 x 10^3 Hz = 2 x 10^4 Hz - 4 kHz of sound = 4 x 10^3 Hz - blue light = 3 x 10^8/450 x 10^-9 Hz = 6.67 x 10^14 Hz [ Take the wavelength of blue light be 450 nm] - 2000m radio = 3 x 10^8/2000 Hz = 150 kHz = 1.5 x 10^5 Hz Therefore, the frequencies in ascending order is: 4 kHz sound ( 4 x 10^3 Hz) 20 kHz sonar ( 2 x 10^4 Hz) 2000m radio ( 1.5 x 10^5 Hz) 5 GHz radar ( 5 x 10^9 Hz) 3 micron infrared ( 10^14 Hz) blue light ( 6.67 x 10^14 Hz)
    1 個解答 · 物理學 · 3 月前
  • 請問用細火煲一煲水, 半小時後, 大概幾多ml 既水會被蒸發掉? 是不是 around 600ml?

    最佳解答: "細火" is NOT a scientific term. You need to know how much energy is delivered to the water in a unit time before calculating the amount of water boiled away.
    最佳解答: "細火" is NOT a scientific term. You need to know how much energy is delivered to the water in a unit time before calculating the amount of water boiled away.
    1 個解答 · 物理學 · 3 月前
  • 請問 science 中的right hand grip rule , 如何分辨個方向係clockrise 定anticlock?thanks?

    The Right Hand Grip rule serves TWO purposes. One is to find the direction of magnetic field surrounding a current in a wire. The other is to find the "direction" of the North pole in a current carrying coil. I suppose you ask for the first purpose of using the rule. It says that if the thumb describes the direction of current, then the... 顯示更多
    The Right Hand Grip rule serves TWO purposes. One is to find the direction of magnetic field surrounding a current in a wire. The other is to find the "direction" of the North pole in a current carrying coil. I suppose you ask for the first purpose of using the rule. It says that if the thumb describes the direction of current, then the other fingers describe the direction of the magnetic field. It means that if you are loonking along the direction of the electric current at the end of the wire, the mangetic field will be in the "clockwise direction". On the contrary, if you are looking at the direction against the direction of current flow, then the direction of magnetic field will be in the "anticlockwise direction".
    1 個解答 · 物理學 · 3 月前
  • Physics初中電路 http://upload.lsforum.net/users/public/u2711^g72.png?

    最佳解答: (a) I1 = Vs/R1 = 12/75 A = 0.16 A (b) When S2 is open, no current flow through R4. Hence, I3 = 12/(20+150) A = 0.0706 A [ Thanks for DL081797 for pointing out my mistake in my original calculation. It has been corrected now ]. (c) Equivalent resistance of R4 and R5 in parallel = 100 x 150/(100 + 150) ohms = 60 ohms Hence, voltage across R4 =... 顯示更多
    最佳解答: (a) I1 = Vs/R1 = 12/75 A = 0.16 A (b) When S2 is open, no current flow through R4. Hence, I3 = 12/(20+150) A = 0.0706 A [ Thanks for DL081797 for pointing out my mistake in my original calculation. It has been corrected now ]. (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 · 4 月前
  • 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 · 6 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 · 8 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 · 9 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 · 9 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 · 9 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 · 9 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 · 9 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 · 9 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 · 10 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 · 10 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 · 10 月前