Physics: Vector's Problem?
a = dv/dt = (6.0 – 8.0 t ) i + 0 j + 0 k. (a) a(1.0) = (6.0 – 8.0 × 1.0 ) i + 0 j + 0 k = –2.0 i + 0 j + 0 k. (b) a = 0 when 6.0 – 8.0 t = 0 t = 6.0 / 8.0 = 0.75 s. (c) Never. Speed along the y-axis is a constant 9.0 m/s, and irrespective of the value vx may assume, these 9.0 m/s cannot be cancelled out. (d) | v |² = (6.0 t – 4.0 t²)² + 9.0² = 10² (6.0 t – 4.0 t²)² = 100 – 81 = 19 6.0 t – 4.0 t² = ±√19 4.0 t² – 6.0 t ±√19 = 0 Using +√19 yields complex roots. Not useful. Using –√19 yields real roots; one of them is negative. Since t > 0, there's only one solution left: t = 2.03539 s.
Physics Vector Problem?
During the rodeo, a clown runs 11 m north, turns 23◦ east of north, and runs 1.8 m. Then, after waiting for the bull to come near, the clown turns due east and runs 8.1 m to exit the arena. a) What is the magnitude of the clown’s total displacement? Answer in units of m. Can somebody run me through this? Thanks
Vector physics problem please help!!!?
hello my teacher didnt explain how to do this problem Vector R has x-component = +14.0 and y-component = +3.00 find the lenght there is no example like this on the book please help me understand how to do it thanks
Physics vector problems help!?
I really don't know why I struggle like hell with vector problems but someone please help me with this. And can you guys please answer the questions on scratchpad or write it down on paper and take a picture {good quality :) } It is hard for me to understand what you mean if it is typed out in the comments. Who ever does this with correct answers gets an easy 5 star best answer award. 1) Vector A is 5.5cm long and points along the x-axis. Vector B is 7.5cm long and points at +30 degrees above the negative x axis. a. Determine the x and y components of vector A and vector B b. Determine the sum of these two vectors in terms of components. c. Determine the sum of these two vectors in terms of magnitude and direction 2) The magnitude of a vector is 3.1 units and its x component is 0 units. What are possible y components and respective angles with the x-axis of the vector?
Physics need help with vectors? 2 problems?
I'll try to help you 2....this is physics 30? grade 12 physics? This is my 3rd year in mech engineering...so i kinda forgot most of these stuff, but i'll try my best! I could be wrong... 1. UPSTREAM Vb = Va + Vb/a <-- remember this vector formula assuming Vb= 10km/h velocity of the boat that it can move (doesn't mean its speed) Va=5 km/h velocity of stream Vb/a=10km/h beat can move relative to the stream Vb= 5km/h + (-10km/h); Vb= -5km/h <---5km/h upstream Vb= 5km/h + (+10km/h); Vb= 15km/h<---15km/h downstream 2. wow this question is kinda tricky. What is water surface? Did you copy the question right? because part b) what is his velocity walking to the rail? It said 4m/s (the same rate to the rail) I think the question is what is his velocity walking to the rail relative to the water surface (whatever water surface is)... But I think i know what the question is asking....Picture the ship travelling left and a guy walking to the rear (to the right). Find the velocity of his speed relative to the water..This is how u do it. convert 18km/h to m/s => 5m/s Vb= Va + Vb/a => Vb= 5m/s speed of ship relative to the water (left) Va=4m/s speed of man walking relative to the deck of the ship(right) Vb/a speed of the man walking relative to the water Vb/a= Vb - Va => 4m/s - 5m/s = -1m/s
Physics vector problem --- Kinematics...help?
In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.050 s, during which time it experiences an acceleration of 360 m/s2. The ball is launched at an angle of 47° above the ground. Determine the horizontal and vertical components of the launch velocity.
HELP ME PLZ! i need physics help, vectors problem?
Turn them into vectors (x,y), with x being east (negatives being west), y being north (negatives being south) So: 1 is (70, 0) 2 is (0, -210) 3 is (-86.6, -50) - use trigonometry to calculate 4 is (-90, 155.9) - as above And add together: (-106.6, -104.1) Use trigonometry again to measure back to distance and angle (drawing a diagram may help) Gives 149.0 meters at 44.3° south of west
Can someone help me with this physics vector problem?
First you want to get the actual displacement of te plane on the map. notice that the 2cm east at the end will cancel out two of the cm travelled west in the beginning. So in total the plane wil travel 11cm west and 9 cm north. If you drew this, the displacement of the plane (let's call it d) would be the hypotenuse of a right angle triangle with small sides length 11 and 9 so using the pythagorean theorem, 11^2 + 9^2 = d^2 121 + 81 = d^2 202 = d^2 d = sqrt(202) = 14.2cm But we know that 1 "map cm" = 20 "real km" so 14.2 cm on the map is the same as 14.2(20) = 284 km
Physics Vector Addition Problem!!! help anyone?
A > is in (x, -y) quardant A = 2080[cos 32.5 (i) + sin 32.5 (-j)] where (+i) and (+j) are unit vectors in east and north B > is along south B = 3160 (-j) ============= resultant (R) R = A+B R = 2080[cos 32.5 (i) + sin 32.5 (-j)] + 3160 (-j) R = 1754.254 (i) - 1117.583 (j) - 3160 (j) R = 1754.254 (i) - 4277.583 (j) |R| = [1754.254^2 + 4277.583^2]^1/2 = 4623.324 N direction tan (p) = - 4277.583 / 1754.254 = - 2.4384 p = 292.3 degree angle(p) made by Ris measured from (+x) or east counter-clockwise with is (-67.74 degree south of east clockwise)
What should I do if I am facing problem in vectors physics portion in class 11 even after regular practice?
Well, I too had faced the same problem in class 11th. The problem is that the Practical part (class 11th) has been taught before the basic part (class 12th).So we end up struggling in Physics.I suggest you to solve Vectors first from class 12th NCERT as it will help clear your basics. (Even if you are a PCB) student.Also try solving Calculus (Integration and differentiation) , Basic trigonometry of class 12th. Try completing it in your vacations else you'll end up struggling in Physics entire class 11th.Seek a teacher's help and guidance.Hope it Helps!!Good luck!!