Consider the system shown in the figure below with m1 = 27.0 kg, m2 = 14.6 kg, R = 0.230 m, and the mass of th?
Without a diagram no one is able to offer any advice at all.
Consider a frictionless track as shown in the figure below...?
Answer: 0.112m Work: Well, with an elastic collision both momentum and energy are conserved. So, first thing's first - find the potential energy of the block at its starting point: PE = mgh PE = (5.65)(9.81)(5) = 277.13 From there, we can find the velocity of block of mass m1: KE = 277.13 = (1/2)mv^2 277.13 = (1/2)(5)v^2 v^2 = 110.85 v = 10.53 Now, we set up a system of equations that will allow us to solve for the velocities of the blocks. We know that momentum is conserved, so we write that the momentums of the blocks will have to add up to 10.53. (However, one of the blocks' individual momentums may be greater than the starting momentum because one may have negative momentum, due to directions of the blocks.) (m1)(v1) + (m2)(v2) = (10.53)(5.65) (5.65)(v1) + (8.5)(v2) = 59.5 Also, energy of the blocks is conserved, so we find that: (1/2)(m1)(v1)^2 + (1/2)(m2)(v2)^2 = 277.13 (<---- 277.13 comes from PE earlier in the problem). (2.825)(v1)^2 + (4.25)(v2) = 277.13 Now, if we solve the momentum equation for v2 in terms of v1, we can plug our solution into the energy equation to find a numerical value for v1: (8.5)(v2) = 59.5 - (5.65)(v1) v2 = (59.5 - [5.65][v1])/8.5 (2.825)(v1)^2 + (4.25)[(59.5-[5.65][v1])/8.5]^2 = 277.13 Working out the algebra gives you a value of -1.48084 (or 9.8907, but that value is extraneous) for v1. So, to find how high the block goes, we find the kinetic energy after the collision (using our new v1) and convert it to potential energy. KE = PE (1/2)(m1)(v1)^2 = (m1)(g)(h) (1/2)(5.65)(-1.48084)^2 = (5.65)(9.81)(h) h = 0.111768m ~ 0.112m
Consider the track and carts in the figure below with m1 = 11 kg, m2 = 19 kg, and r = 0.24 m, but the carts?
At the top, we need centripetal acceleration = gravity, or v²/r = g, which leads to v² = rg = 0.24m * 9.8m/s² = 2.352 m²/s² So at the top we need total energy TE = PE + KE = mgh + ½mv² TE = 30kg * 9.8m/s² * 0.48m + ½ * 30kg * 2.352m²/s² = 176 J So at the bottom, post-collision, we need KE = 176 J = ½mv² = ½ * 30kg * v² v = 3.4 m/s (a) Conserving momentum, 11kg * u = 30kg * 3.4m/s u = 9.4 m/s which means the small cart arrives with KE = ½mu² = 481 J so it started at height h = PE / mg = 481J / 11kg*9.8m/s² and h = 4.5 m (b) Conserving momentum, 19kg * u = 30kg * 3.4m/s u = 5.4 m/s which means the large cart arrives with KE = ½mu² = 279 J so it started at height h = PE / mg = 279J / 19kg*9.8m/s² and h = 1.5 m Somewhat of a guess lacking the diagram...
In the figure below, m1 = 9.5 kg and m2 = 3.5 kg.?
Since I can’t see the figure, I will assume that two objects are connected by a string that goes from m1 on the horizontal surface, over a pulley and down to m2. The weight of m2 is causing the blocks to accelerate. The friction force is causing the blocks to decelerate. Weight = 3.5 * 9.8 = 34.3 N Ff = 0.5 * 9.5 * 9.8 = 46.55 N In this problem, there is one thing that you need to remember. That is the fact that the friction force cannot cause an object to accelerate. The friction force can only cause an object to decelerate. Since the static friction force is greater than the weight of m2. The blocks will not move. So, the acceleration is 0 m/s^2. (b) If the system is set in motion with m2 moving downward, what will be the acceleration of the system? m/s2 Since m2 is moving downward, m1 is moving on the across the horizontal surface. The net force on the objects is equal to the difference of the weight of m2 minus the kinetic friction force. Kinetic friction force = 0.3 * 9.5 * 9.8 = 27.93 N Net force = 34.3 – 27.93 = 6.37 N To determine the acceleration, divide by the total mass. a = 6.37 ÷ 13 The acceleration is approximately 0.49 m/s^2. Sorry about my pervious mistake.
Consider the system shown in the figure below with m1 = 30.0 kg?
Consider the system shown in the figure below with m1 = 30.0 kg, Weight = 30 * 9.8 = 294 N m2 = 10.2 kg, Weight = 99.96 N The net force on the system is the difference of these two forces 294 – 99.96 = 194.04 N For pulley: Torque = F * r Torque = I * α F * r = I * α The following equation is used to determine the moment of inertia of the pulley. I = ½ * m * r^2 Since we need to determine the acceleration of the object’s, let’s use the following equation. α = a/r F * r = ½ * m * r^2 * a/r F = ½ * m * a My point is that we use ½ * m for the mass of the pulley in this type problem. Total mass = 30 + 10 * 2.5= 42.5 kg 190.04 = 42.5 * a a = 190.04 ÷ 42.5 This is approximately 4.47 m/s^2. If you need to determine the velocity of m1when it hits the floor, use the following equation. vf^2 = vi^2 + 2 * a * d, vi = 0 vf^2 = 2 * (190.04 ÷ 42.5) * 4.50 vf = √(2 * (190.04 ÷ 42.5) * 4.50) This is approximately 6.34 m/s.
What does CC or litre mean when describing the power of an engine?
The term CC represents the unit “Cubic Centimeter” of Volume. And in case of Engines it is used to represent the Volume of the Internal Combustion Engine Cylinders.The CC can also be represented in Liters, as shown below1000cc = 1000 cm³ = 1 Liter = 1.0L.The above section is the answer in case you are technically aware of what is an Engine Cylinder. If not, please read on.There are many types of Engines used for myriad applications in today’s world. The most common are the Automotive & Marine Internal Combustion Engines which operate on Fossil Fuels (mostly).In these Engines the fuel is burnt in a controlled explosion to generate Mechanical Energy. The controlled explosion happens inside a cavity within the Engine called an “Engine Cylinder.” Based on the type of the Engine & the application the number of Cylinders in the Engine may be just 1 or even as high as 12.Within, these “Engine Cylinders” are the “Engine pistons” that move up and down to create the & maintain the optimum pressure & other conditions needed for the controlled explosion of the fuel in the Engine.For easier understanding of the piston movement just imagine the movement of the plunger (piston) inside a Medical Syringe for an Injection.Every up & down stroke of an “Engine piston” within a cylinder covers a volume lesser that the size of the Cylinder’s Total Volume and is called the “Swept Volume.” Swept Volume simply represents the total Fuel i.e “Air+Fossil Fuel Mixture”Hence, the CC i.e Cubic Centimeters is the quantity of the Total Swept Volume of any Engine.Total Swept Volume, CC = No of Cylinders X Swept Volume of One Cylinder, cc.Now, the importance of this CC is that it represents the Capacity of the Engine. As “CC” is nothing but the amount of Fuel i.e “Air+Fossil Fuel” that can be compressed in one stroke of the piston. More the amount of Fuel that can be compressed, more the power that the Engine can generate.P.S : Sorry for the long post. I believe that “If I can’t explain it to a 6 yr old, I don’t know it myself”. Hence, the detailed post.
The sun attracts the moon with a force twice as large as the attraction of the earth on the moon. Why does the moon not revolve around the sun?
The thing is, the moon just turned around the world? The answer to this question will be clear. The moon revolves around the earth, and the earth moves around the sun together with the moon. What? Is the sun around the sun? Here the moon and earth form a system, which is like a Binary system . If two astrologers rotate around their center of gravity together, then it is called binary system. Earth's mass and size are big so we do not understand the rotation of the earth. But this is not the case even between Jupiter and Sun. Here the sun does not turn around like Jupiter, and the Sun also does not turn around Jupiter! Both of them turn around their center of gravity.Two bodies with an extreme difference in mass orbiting a common Barycenterinternal to one body (similar to the Sun–Earth system)Now the question is why there is such a simple way that the moon travels around the sun in complex ways? What was needed to make the system with the world? Now the answer is given.All the answers in the middle of a term called Hill sphere are hidden! A planet has some impact zones. Any satellites in this region will travel around the center of the satellite. It is called Roche sphere. Needless to say, the satellite also has a heels of the sphere, there is also the sun. If there is a relatively small asteroid in this heel sphere, then it will make the system with that planet.Earth's mass m, the mass of the sun M, the largest distance from the Sun to the Earth, and eccentricity of the orbit e, the radius of the Hill sphere-Formulae for Roche radius where:r = radius of the spherea = semi-major axis of planet's orbit wrt sune = eccentricity of planet's orbitm = mass of planetM = mass of sunAnd its value for the Earth is approximately 1.5 million km, whereas the distance of the Moon from Earth is only 0.384 million km, so the moon is easily trapped in the Hill Sphere.The place between L1, L2, the point is the SphereLastly the Moon's orbit is like thisThe spiral path is for the moon, and the circular way for the earth. All in all, it seems that the moon is just around the earth! But that is not trueNow the question is what happens in this Hill Sphere? The simple answer is that spacetime is more curved here.Therefore, the moon continued to walk on this routeFor more query:Does the Moon Orbit the Sun or the Earth?The Orbit of the Moon around the Sun is Convex!The Cosmic Distance ScaleHow Fast Are You Moving When You Are Sitting Still?
What are the interview questions asked at ISRO?
ISRO INTERVIEW EXPERIENCEI appeared for interview on 20th FEB at DELHI.First of all they checked all the prerequisite documents ..and one by one candidates started going in the interview hall ..When I entered the room i saw a panel of 11 members sitting out there out of which 2 members were ladiesand 1 chairman was sitting in the middle and rest members were encircling him .the very first question the chairman asked to me that you are 2015 pass out and what were u doing since then .. I said sir I was preparing for competitive examination and I have qualified 3 to 4 exams but I didn't join any of the services because my aim is to get into a GROUP A services only , then he said ok and asked other members to follow up and in that panel each member is having a expertise in particular subject so they started asking questions to me one by one1. What is PLL and it's application2. what is maxwell equations ? What is the use of maxwell equations?3. What is figure of merit for an anteena ?4. What is the difference between optical fibre communication and satellite communication? 5.which anteena is used in satellite communication ?6. What is the eccentricity of parabola?7. What is the bandwidth of speech signal ?8. To transmit speech signal digitally sampling rate , no of bits , bandwidth required practically?9 . How do u configure 555 timer for measuring pulse width ?10. What is decimation?11. What is the relationship between carrier frequency and msg bandwidth in optical fibre communication?12. Design a control system to control intensity of light ?13. What is frequency warping?And many more ...I answered only 80 % of the questions ....still I got AIR 10 ..It doesn't matter whether u answer all questions or not what matters is whatever u are answering u have to answer properly you have to influence them with your answers ..I have seen many people talking that they answered all questions but still they are not selected ....so please don't give vague answers if u don't know any questions simply reply them sir I don't have much idea about it .If u have got something in your head then they will take out it and rest of the things are immaterialIf you are thorough with your basics then u will get in easily .I recommend you to please read basics of each subjects before appearing for interview ..Best of luck Vipin sharma( ECE AIR 10 )