Merry Go Round problem!!!!?
A merry-go-round with a a radius of R = 1.78 m and moment of inertia I = 200 kg-m2 is spinning with an initial angular speed of ω = 1.51 rad/s in the counter clockwise direection when viewed from above. A person with mass m = 64 kg and velocity v = 4.2 m/s runs on a path tangent to the merry-go-round. Once at the merry-go-round the person jumps on and holds on to the rim of the merry-go-round. 1)What is the magnitude of the initial angular momentum of the merry-go-round? 2)What is the magnitude of the angular momentum of the person 2 meters before she jumps on the merry-go-round? 3)What is the magnitude of the angular momentum of the person just before she jumps on to the merry-go-round? 4)What is the angular speed of the merry-go-round after the person jumps on? 5)Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on? 6)Once the person gets half way around, they decide to simply let go of the merry-go-round to exit the ride. What is the linear velocity of the person right as they leave the merry-go-round? [Delayed Feedback] 7)What is the angular speed of the merry-go-round after the person lets go?
Help with a problem?
torque τ = r x F where τ is the torque vector and r is the radius vector and F is the force vector But usually (like in this problem) it is easier to calculate τ = F*d*cosΘ where τ is the magnitude of the torque and F is the magnitude of the force and d is the distance and Θ is the angle from perpendicular at which the force acts. So τ1 = 6.5N * 0.6m * cos0 = 39 N·m τ2 = 2.5N * 0.18m * cos45 = 0.32 N·m τ3 = 12N * 0m * cos30 = 0 N·m τ4 = 15.5N * 0m * cos90 = 0 N·m Total torque, where counterclockwise is positive, is τ = τ2 - τ1 = -38.7 N·m The moment of inertia of a rod about its center is I = mL²/12 = 4.2kg * (1.2m)² / 12 = 0.504 kg·m² τ = I*α -38.7 N·m = 0.504kg·m² * α α = -77 rad/s² ◄ The sign indicates that the rod is spinning clockwise. Hope this helps!
Collision Problem?
We place the weight of the rod at the center of its length or 0.90m from one end. Its Potential Energy, PE at the horizontal position with respect to the particle, is 2.4g*1.40. This energy is converted to a lesser PE at theta=35 degrees. PE of the rod at this point is 2.4g*(1.40-0.90cos35). The particle had initial PE equal to zero, but at 35 degrees from its original position, it gained a PE equal to mg(1.40-1.40cos35). The PE gained by the particle is equal to the dissipated energy during the collision. Based on the principle of conservation of energy: PE of system, initial=PE of system, final a)Substitute the above values: 2.4g*1.4=2.4g(1.40-0.90cos35) +mg(1.40-1.40cos35) g cancels out. 2.4*1.4=2.4*0.67+0.26m m=(3.36-1.60)/0.26 =6.76kg b)Dissipated energy=6.76g*0.26 =17.2 Joules Pls draw a diagram showing the position of the rod and the particle, before and after collision. You'll then find my explanation easier to understand.
How long would it take a human being to count to 1 billion one at a time?
You may think it would take a billion seconds. That's 31.7 years. But the answer is not that simple. Take 999,999,999. Unless you can speak really quickly, you cannot say this number in just one second. That's just one unique number, right? Wrong. 9/10 of all numbers you will count will have the same amount of digits.It takes about 2.5 seconds on average to say a 9 digit number so I will use that. It can then be estimated to take 2.5 billion seconds (79.25 years) to count to a billion. You can be more precise and continue down with 8 digit numbers but this is a good upper bound value.Coincidentally, the life expectancy at birth in the US is around 79 years, so if you somehow knew the numbers the second you were born and started counting then, it would take all of your life to count to a billion. This can't happen however. You wouldn't be able to sleep, work, or eat. If you devoted 8 hours to counting, then it would take 237.25 years so your offspring would have to continue your legacy.As a side note, there is a trick to counting to higher numbers without spending too much time. Count to 100, then restart, but keep track of all the 100s you have counted. You can count a number a second now. With 8 hours of counting a day, and using 31.7 years from earlier, it would take 95.1 years.Overall, you would most likely die before you finished counting to a billion, one way or another.
Can an open cheque be cashed in at the bank counter by anyone who has access to the cheque? If yes, then how?
Open cheque means an uncrossed cheque.Uncrossed cheques can be paid at the bank teller counter and there is no need to present the cheques through account.All cheques are payable to bearer only and in the case of one open bearer cheque, the banker will never demand any proof from the bearer and by getting signature on the backside of the cheque they will make payment to the bearer. The bankers are well protected by NI actHowever, in the case of one open order cheque, the banker has to verify the identify of the bearer and for this purpose they will demand proof of identity namely; driving licence, voter identity card, passport or AATHAAR card etc., before considering payment for the cheque.
How come my VW jetta won't accelerate?
I own a black 2001 jetta automatic 1.8t and recently when I try and accelerate it wont go past 30/40 (RPM) when I floor the gas pedal I can only reach about 27 MPH. To gain speed I have to push on the gas little by little and gradually my speed will pick up to about 60 MPH. What do you think is wrong with my car. (I accidentally shifted into 1st gear going about 30 MPH). Do you think this had any effect? I did not notice this problem until 3 days after that happened.
What is the tension in the cable?
Let the pivot point be at the bottom of the pole. The 120 N force will produce counter clockwise torque. The horizontal component of the tension in the cable will produce clockwise torque. Counter clockwise torque = 120 * 1.8 = 216 N * m Clockwise torque = T * cos 54 * 2.5 T * cos 54 * 2.5 = 216 T = 86.4 ÷ cos 54 This is approximately 147 N.
Which laptop is better, one with 1.8ghz and 8GB ram or one with 2.5ghz and 4GB ram?
There is really no quick answer.Only giving me frequency information is not enough for me to determine the speed of a processor. As the processor architecture is being refined and improved, it is now possible to make new processors run at the same speed in most applications using a frequency much lower than old processors. For example, an Intel Core i3 380UM 1.33GHz (Arrandale) runs at the same speed of an Intel Core 2 Duo T7200 2.0GHz (Merom) under most environments. An Intel Core i5 2467M 2.5GHz (Sandy Bridge) can even run faster than an Intel Core 2 Duo T9600 2.83GHz (Penryn) in nearly all applications. New applications can get extra benefit from new technologies including hardware accelerated AES engine, nested paging, VT-d, Intel Quick Sync Video, and more (similar technologies, except Intel Quick Sync Video, also exists on AMD processors), giving new CPUs a even greater advantage. RAM is important, but in most cases, this is a part which you can upgrade (though some models, including MacBook Air and Dell XPS 13 tend to solder RAM modules onto the motherboard and leave no room for upgrade, which is a suggested but not compulsory requirement for Windows 8 certification), you can usually add more RAM as the need grows.Some other parts also needs to be reviewed, including graphics adapter and storage system. Does the machine's graphics adapter has enough horsepower to meet your needs? Does the machine use rotational media, Solid State Drives or eMMC? Is the main storage media capable of self-encryption? You may also want to investigate parts like WLAN adapter and optical drive (if applicable).Finally, the brand. Is the brand affordable? Is the brand trustworthy? Try to think seriously before buying rather than just comparing some specs which may not help that much.UPDATE: However, if your main focuses are weight and battery life, it would be better to choose newer CPUs with lower frequencies as they use less power and generate less heat.
What is better, a CPU with 4 cores, each having 1.5GHz, or a CPU with 2 cores, with 3GHz? And why?
I’ve answered this a dozen+ times here. The folks at Cray Research, Inc. figured out the answer nearly 30 years ago. So, here’s what they found:2 CPUs =~ 1.6 CPUs4 CPUs =~ 2.8 CPUsThis is in codes that share main memory and are maximally efficient at parallelization. So, the answer is (in general) the 2 CPU system for two reasons:1.6 x 3.0 > 2.8 x 1.5And also, for non parallel codes, 3.0 beats 1.5Some quick notes:It’s the serial (non parallel) codes & their percentages that will drive the speed of the overall programIt’s also the ability to parallelize as well as the setup overheadFinally, cache is critical! The more the internal CPU cache (in general) the lower the chance to suffer a memory stall. This is a RACE condition where one CPU is accessing memory & the other CPU(s) have to wait (STALL) until the access is completed. 2 CPUs have fewer conflicts!Numbers at Cray originally calculated on one of these puppies:Stock Photo used in the advertising poster. (I have the poster framed in my home office.)Cheers!//I added a comment below, but maybe it would also help here with a few slight corrections:Folks forget that the increase in effective processing power in the form of more CPUs/CPU_Cores does not increase in a linear fashion. The problem (in hardware) becomes the resolution of the interconnect and its efficiencies. Even widely distributed memory systems suffer from this.If you look at a multi-core processor, selecting the L1 cache as the example of localized memory, you begin to realize that that cache needs to be loaded from & flushed back to yet another memory system as processing proceeds. The more cores, the harder it gets to keep the central memory working in sync with the cache load & flush cycles.The Cray MPPs [T3D & T3E] tried to solve this, but the interconnect was the most difficult part of the system to design. Today, we have seen the modern supercomputers work massively distributed CPU structures by providing an almost discrete CPU/MBoard/GPU grouping. The problem? It’s assigning the task to this group & then getting the data to & from it while yet again avoiding RACE conditions.PS. Multi-Core counts remind me of the transistors in a radio & jewels in a watch. Many were present only for the bragging rights. A friend had a 7 transistor radio. Upon taking it apart, it was found that three of the transistors were present but not connected to anything!