Have a physics question I can not solve.?
Let at be the tangential acceleration Let ac be the centripetal acceleration Let C be the circumference Let us assume that the tangential acceleration remain constant during the next turn around the track. As the total acceleration is North of West and the car is traveling West, then the center of the circle must be due north of its current location ac = 2.5sin37 ac = 1.5 m/s² at = 2.5cos37 at = 2.0 m/s² ac = v²/R R = v²/ac R = 20²/2.5sin37 R = 265.9 m C = 2πR C = 2π(265.9) C = 1671 m s = s₀ + v₀t + ½at² 1671 = 0 + 20t + ½2.0t² t² + 20t - 1671 = 0 t = (-20 ±√(20² - 4(1)(-1671))) / (2(1)) t = -52 s or t = 32 s ignore the negative time value because it occurred before we begin looking. t = 32 s As a form of exchange, please remember to vote a "Best Answer" from among your results. This is not the same as "like" or "thumbs up", which are nice as well, as points are only awarded for asker's "Best Answer".
Physics questions that i cannot solve?
I have a lengthy summer assingment for physics, and these two questions i just cannot figure out how to solve! Please don't just give the answer, explain to me why it is that 1. A cheetah can run at a maximum speed 96.1 km/h and a gazelle can run at a max- imum speed of 77.3 km/h. If both animals are running at full speed, with the gazelle 92.4m ahead, how long before the cheetah catches its. 2. A river flows with a uniform velocity v. A person in a motorboat travels 0.963 km up- stream, at which time a log is seen floating by. The person continues to travel upstream for 80.4 min at the same speed and then returns downstream to the starting point, where the same log is seen again. Find the flow velocity of the river. Assume the speed of the boat with respect to the water is constant throughout the entire trip. (Hint: The time of travel of the boat after it meets the log equals the time of travel of the log.) Answer in units of m/s.
What should I do if I am not able to solve physics questions even though I have read the theory?
Buddy reading physics and understanding physics are two different things. Applying physics is more difficult than reading it. You all need to do is practice, try to do easy problems then increase your difficulty level. It will take more time to apply than reading it.
I have a physics question! I can't solve it!!?
Ten days after it was launched toward Mars in December 1998, the Mars Climate Orbiter spacecraft (mass 629 kg) was 2.87 106 km from the earth and traveling at 1.20 104 km/h relative to the earth. At this time, what were the following. (a) the spacecraft's kinetic energy relative to the earth I got 4.53e+10 J but that is not correct answer...... please help me
I have a physics question that im not sure how to do?
A cannon fires at a fixed angle θ0=45 degrees and with a muzzle velocity of v0=135 m/s. The target is on a cliff of height h=235 m. From what distance L, should the gunner place the cannon so that the projectile hits the target on its way down? My thought process makes sense to me but i am not getting the right answer. First, i solved for the initial velocity in the y direction. Viy=sin(45)*135=95.5 m/s next i used that velocity to solve for the maximum height of the cannon ball Height=(Viy^(2))/2g=95.5^(2)/2*9.81=46... next took that max height and subtracted the height of the cliff 464.44-235=229.45 m next, i used that height to calculate the velocity it would reach at that point. Knowing that the initial velocity would = 0 because it was at the top of its arc. Vf=sqrt(vi^2+2(g)(y)=sqrt(2*9.81*229.4... m/s now i velocity in the y direction that would be equal to the initial velocity if the target was level. using the fact that Vf=Vi i assumed that would mean the initial velocity out of the cannon in the y direction would equal 67 m/s. using that number, i solved for the initial velocity directly out of the cannon Vi=67/(sin(45))=94.88 finally i used the range equation do find out how far away the cannon would need to be from the cliff. R=(Vi^2/g)*sin2(45)=917 m however that answer was not right. Any ideas?
Please Help!!! I can't solve this physics question?
If the box masses .300 kg, then you can find its weight quite easily. We assume the floor is perfectly horizontal, so the weight will be the same as the normal force from the floor. The model of friction that you're using says that the force of friction will equal the coefficient of friction times the normal force, so that gives you the force of friction. That will be the net force on the box (ignoring air resistance), so dividing that by the mass will give you the acceleration of the box. You have, then, the initial velocity and the acceleration, so this becomes a kinematics problem. You can use those two things to get the time it'll take the box to stop and then use v0t + (1/2) a t^2 to get the distance, or you may have a formula that'll take you to distance directly - your call.
What should I do if I can't solve the physics questions of the JEE?
First of all it is important to realise that if theory is clear to you and you are not able to crack questions it means there is lack of application of concepts.I would suggest in such a case that instead of wasting more time reading theories again you try to do questions and spend some time with it . Think of what u know(theory) and relate it with questions and how can u use it to get nearer to the answer. But please don't waste time spending 15–20 mins on one question instead at this point of time I would suggest u to do questions which are asked in previous years JEE papers and read there solution in case u are not able to figure out the solution behind it. This does not mean that u have to mug up the method but learn how the concept which u knew has been applied in the question and most importantly try to realise where u went wrong in doing the question(was it the concept ,silly mistake ,or the approach)In case this also doesn't work I would suggest you to firstly identify which chapters of physics u like the most and are frequently asked in jee and make them strong firstly instead of wasting time on the whole physics .Since many chapters have very few chances of questions being asked from them.So try to make the most frequently asked chapters your favourite ones because in that way you will be able to score good in physics instead of wasting time on full sylabus..I hope this approach works out well for you.All the best!! Do well!!!
Literally, I cannot solve the even 1 question of physics. What can I do?
Look for the patterns.We human beings follow patterns all the time, even more so in physics.Physics has quantities- stuff like force, moment, distance and so on. The best thing about these quantities is that they are related. So, find the quantities, use the relationship between the quantities and solve.Initially, you will always have one unknown quantity. As you go on, there will be 2 unknowns or more. You have math techniques for solving for unknowns.Also, to understand the relationship between the quantities, you will have to read a lot of content based text in physics. As you read further, you will see that all these quantities are beautifully interrelated.Do not feel stumped. Remember, everyone has the ability to learn.Ps. Walter Lewin Lectures are awesome.This answer is more helpful as well—Mike Wong's answer to How do I excel at physics and solving physics word problems?
I cannot solve this flywheel physics question, please help. i appreciate it.?
A rotating flywheel can be used to store energy. if it is required to store 1.00x10^6 J ( joules) of energy when rotating at 3820 rpm, what is the moment of inertia of the cylindrical wheel? a) 6.25 kg m^2 b) 12.5 kg m^2 c) 25 kg m^2 d) 50 kg m^2 thank you all who attempt
Super super hard physics question i can't even solve!!!!?
you can obtain a rough estimate of the size of a molecule with the following simple experiment: let a droplet of oil spread out on a fairly large but smooth water surface. the resulting "oil slick" that forms on the surface of the water will be approximately one molecule thick. Given an oil droplet with a mass of 8.85x10^-7 kg and a density of 919kg/m^3 that spreads out ot form a circle with a radius of 44.2cm on the water surface, what is the approximate diameter of an oil molecule? Answer in units of m.