• Central force question?

最佳解答： The satellite will return to the planet surface when the graviataional pull on it is GREATER than the required centripetal force to keep it in orbit. That is, G(m1)(m2)/(r2)^2 > (m2)(Vb)^2/r2 where G is the Universal Gravitational Constant [Note: when the left term equals to the right term, the satelllite will stay in orbit. If the left... 顯示更多
最佳解答： The satellite will return to the planet surface when the graviataional pull on it is GREATER than the required centripetal force to keep it in orbit. That is, G(m1)(m2)/(r2)^2 > (m2)(Vb)^2/r2 where G is the Universal Gravitational Constant [Note: when the left term equals to the right term, the satelllite will stay in orbit. If the left term is smaller than the right term, the satellite will fly away from point B] thus, (Vb)^2 < G(m1)/r2 or (Vb)^2 < G(m1)/X.r1 ------------------ (1) From conservation of mechanical energy, loss of kinetic energy of satellite = gain in potential energy of satellite i.e. (1/2)(m2)(Va)^2 - (1/2)(m2)(Vb)^2 = G(m1)(m2)(1/r1 - 1/r2) Because r2 = X(r1), we have, after simplification, (Va)^2 - (Vb)^2 = 2G(m1)(X - 1)/(X.r1) i.e. G(m1)/(X.r1) = [(Va)^2 - (Vb)^2]/(2.(X - 1)) Substitute into (1): (Vb)^2 < [(Va)^2 - (Vb)^2]/(2.(X - 1)) Divide both sides by (Vb)^2, 1 < [(Va/Vb)^2 -1]/(2.(X -1)) Inverting the fraction on the right-hand-side, then 2(X -1)/[(Va/Vb)^2 - 1] < 1 --------------- (2) I suppose you have already found the expression for Vb/Va in part (d). Let this be f, say. i.e. Vb/Va = f where f is a function of "alpha" and X Hence, inequality (2) becomes, 2(X -1)/[(1/f)^2 - 1] < 1 The expression on the left-hand-side involves X and "alpha" only. I just leave it for you to complete the derivation, as it involves only algebric manipulation. Hope the above would help.
1 個解答 · 物理學 · 2 月前

最佳解答： In your equation, I suppose the parameter θ is the angle between the rod and the top of the car. In so doing, your equation is actually describing an oscillation about the top of the car (because θ is measured from the top of the car). But the rod is, in fact, oscillating about its equilibrium position, which is at an angle θ with the top of... 顯示更多
最佳解答： In your equation, I suppose the parameter θ is the angle between the rod and the top of the car. In so doing, your equation is actually describing an oscillation about the top of the car (because θ is measured from the top of the car). But the rod is, in fact, oscillating about its equilibrium position, which is at an angle θ with the top of the car. Moreover, according to your equation, the force mg.sin(θ) and the force mA.cos(θ) are acting along the length of the rod, but in opposite directions. Since the two forces pass through the pivot, there should be no torque produced. But you have made a wrong assumption that the two forces are acting perpendicularly to the rod.
5 個解答 · 物理學 · 3 月前
• 現實中有咩不是光學介質(optical medium)?

最佳解答： From your given link: https://en.wikipedia.org/wiki/Optical_medium , an "optical medium" is a substance through which electromagnetic waves propagate. Following this definition, metals (including alloys) are not optical media, as metals are conductors and which electromagnetic waves cannot pass through. ---------------------- Reply... 顯示更多
最佳解答： From your given link: https://en.wikipedia.org/wiki/Optical_medium , an "optical medium" is a substance through which electromagnetic waves propagate. Following this definition, metals (including alloys) are not optical media, as metals are conductors and which electromagnetic waves cannot pass through. ---------------------- Reply to your question in "意見": You are asking, in your question, the material substances that are not optical medium. Copper, clearly, is NOT an optical medium, as electromagnetic waves cannot pass through (see the definition of optical medium in your second link). The reflection of light on polished copper surface indicates that light wave cannot pass through the substance.
2 個解答 · 物理學 · 4 月前
• FInd the the Norton equivalent?

最佳解答： To find the Norton current (In), assume pionts a and b are connected together. Hence, current through the 9-ohm resistor = 18/9 A = 2 A Becasue the 2-ohm resistor is in series with the 3-ohm and 6-ohm resistors in parallel (after a and b are connected), the equivalent resistance (Req) of this branch is, Req = (2 + 3//6) ohms (here the... 顯示更多
最佳解答： To find the Norton current (In), assume pionts a and b are connected together. Hence, current through the 9-ohm resistor = 18/9 A = 2 A Becasue the 2-ohm resistor is in series with the 3-ohm and 6-ohm resistors in parallel (after a and b are connected), the equivalent resistance (Req) of this branch is, Req = (2 + 3//6) ohms (here the symbol 3//6 means "3-ohm resistor connected in parallel with 6-ohm resistor). Thus, Req = [2 + 3x6/(3+6)] ohms = (2 + 2) ohms = 4 ohms Hence, voltage across resistance 3//6 (which equals to 2 ohms) = 18 x [2/(2+2)] v = 9 v or voltage across 3-ohm resistor = 9 v (because 3-ohm and 6-ohm are in parallel) Current through 3-ohm resistor = 9/3 A = 3 A Since the current flows through a and b is the sum of currents from the 9-ohm resistor and 3-ohm resistor, we have, the Norton current In = (2 + 3) A = 5 A ------------------------ To find the Norton resistance (Rn), assume the 18v source is in short-circuit. Here, we have two main branches of resistance. The 1st branch is simply the 9-ohm resistor. The 2nd branch is the 3-ohm resistor in series with the 2-ohm and 6-ohm resistors in parale. First, calculate the resistance of the 2nd branch R2. R2 = (3 + 2//6) ohms = (3 + 2x6/(2+6)) ohms = (3 + 1.5) ohms = 4.5 ohms Therefore, the Norton equivalent resistance Rn = (9//4.5) ohms = 9x4.5/(9 + 4.5) ohms = 3 ohms
1 個解答 · 物理學 · 5 月前

最佳解答： 1. No, water molecules below the surface do move. Water molecules in a "surface wave" move in circular orbits. The diameter of the orbit decreases with distance below the water surface. At a verical distance of one wavelength below the surface, the orbit becomes nearly zero, i.e. water molecules stop to move at such depth and... 顯示更多
最佳解答： 1. No, water molecules below the surface do move. Water molecules in a "surface wave" move in circular orbits. The diameter of the orbit decreases with distance below the water surface. At a verical distance of one wavelength below the surface, the orbit becomes nearly zero, i.e. water molecules stop to move at such depth and below. 2. As said above, water molecules move in circular orbits in a water wave. A circular orbit is indeed a transverse wave and a longitudinal wave superimposed on each other. Hence, water molecules have both vertical and horizontal displacements. Note that these two displacements are 90 degrees (or a quarter of wavelength) out of phase. Because the logitudinal (i.e. horizontal) motion of molecules and the transverse (i.e. vertical) motion of molecules are 90 degrees out of phase. At the wave crest and trough, the vertical displacement is maximum, the horinztonal displace is zero. Hence, at the mid-point between the crest and trough (i.e. the y-displacement, which indicates the vertical displacement, of the wave is zero), the horizontal displacement, which has a phase difference of 90 degrees with the vertical displacement, will be at maximum. Perhaps we can look it on another way using Simple Harmonic Motion (SHM) of water molecules. Just consider the longitudinal motion of the water molecules. These molecules perform Simple Harmonic Motion, the the equilbrium positions are located at the trough and crest of the water wave. Similarly, for the transverse motion of molecules, but the equilibrium positions are at the mid-point between the crest and trough. Because the slope on the shape of the water wave indicates the speed of the water molecules on the transeverse (vertical) motion of molcules, the slope is greatest at the mid-point between the crest and trough. Thus, the vertical speed of molecules is maximum at this point. The horizotnal speed of molecules is thus minimum (i.e. zero) at such point, because the two motions are 90 degrees out of phase. From theroy in Simple Harmonic Motion, a zero velocity implies that the displacement is masximum. The following web-page gives a very clear explanation on water waves. You may take a look on it: http://labman.phys.utk.edu/phys221core/modules/m12/Water_waves.html
1 個解答 · 物理學 · 5 月前
• 物理電路問題 求解?

最佳解答： In the given circuit diagram, the 1.5k and 3k resistances are in parallel and their equivalent resistance R1, say, is in series with the equivalent resistance R2, say, of 2.4k and 1.2 k in parallel. Here, R1 = 1.5 x 3/(1.5 + 3) k-ohm = 1 k-ohm R2 = 2.4 x 1.2/(2.4+1.2) k-ohm = 0.8 k-ohm Hence, equivalent resistance of the branch Re = R1 + R2 =... 顯示更多
最佳解答： In the given circuit diagram, the 1.5k and 3k resistances are in parallel and their equivalent resistance R1, say, is in series with the equivalent resistance R2, say, of 2.4k and 1.2 k in parallel. Here, R1 = 1.5 x 3/(1.5 + 3) k-ohm = 1 k-ohm R2 = 2.4 x 1.2/(2.4+1.2) k-ohm = 0.8 k-ohm Hence, equivalent resistance of the branch Re = R1 + R2 = (1+0.8) k-ohms = 1.8 k-ohms The circuit now reduces to three resistances, namely 1.8k, 1.8k (i.e. Re above) and 3.6k, in paralle. These 3 parallel resistances are connected in series with the 300 ohms resistor. 1. The equivalent resistance RE,say, of the above 3 parallel resistnaces = 1/(1/1.8 + 1/1.8 + 1/3.6) k-ohms = 0.72 k-ohm Hence, Req = (0.3 + 0.72) k-ohms = 1.02 k-ohms 2. Voltage across 1.8k equals volatage acroee RE, since the two are in parallel, Hence, V(1.8k) = 5 x (RE/Req) = 5 x (0.72/1.02) v = 3.529 v 3. Voltage across Re equals V(1.8k) since the two are in parallel, hence, voltage across R2 = 3.529 x [R1/(R1+R2)] = 3.529 x [0.8/(1+0.8)] v = 1.568 v 4. io = 5/Req = 5/1.02 mA = 4.902 mA 5. From (2) above, voltage across 1.8k resistance = 3.529 v Hence, current = 3.529/1.8 mA = 1.96 mA 6. Voltage across 1.2k resistance = voltage across R2(= 0.8k) Hence, V(1.2k) = 1.568/1.2 mA = 1.31 mA
1 個解答 · 物理學 · 5 月前
• 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 · 6 月前

(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 · 7 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 · 7 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 · 7 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 · 7 月前

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 · 8 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 · 8 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 · 8 月前

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 · 9 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 · 11 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 · 1 年前
• 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 · 1 年前
• 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 · 1 年前
• 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 · 1 年前