A particle is sliding in a helical path, using cylindrical coordinates where P = 4 , ø = 21 and z = 0.51* so the magnitude of the acceleration of the particle is :
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![A particle is sliding in a helical path, using cylindrical coordinates where P = 4, ¢ = 21 and z
so the magnitude of the acceleration of the particle is :
0.51 ?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa929ea7c-fcd0-4f7e-8027-72aca0bcd877%2Fa91733c1-eb04-422c-8d99-978ea23b8fc8%2Ft6ebzpc_processed.png&w=3840&q=75)
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- A marble slides down on a curved path defined by the parabola y = 0.4x2. When it is at point “A” (xA = 2m; yA = 1.6m), the velocity of the marble v = 8m/s and the tangental acceleration due to gravity is dv / dt = 4ms2. Calculate the normal component and the total magnitude of the accelarion of the mable in that instant.You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this statement, consider a ball thrown off the top of a building with a velocity v at an angle 0 with respect to the horizontal. Let the building be 54.0 m tall, the initial horizontal velocity be 9.10 m/s, and the initial vertical velocity be 10.5 m/s. Choose your coordinates such that the positive y-axis is upward, and the x-axis is to the right, and the origin is at the point where the ball is released. (a) With these choices, find the ball's maximum height above the ground and the time it takes to reach the maximum height. maximum height above ground time to reach maximum height (b) Repeat your calculations choosing the origin at the base of the building. maximum height above ground time to reach maximum heightA particle moves along a circular path having a radius of 1.1 m. At an instant when the speed of the particle is equal to 3.22 m/s and changing at the rate of 6.46 m/s², what is the magnitude of the total acceleration of the particle?
- A particle confined to the xy-plane is in uniform circular motion around the origin. The x- and y-coordinates of the particle vary with time as follows: x(t) = r sin(cot) y(t) = r cos(cot). where >= 2.60 m and c = 2.15 s-¹. What are the x- and y-components of the particle's acceleration at an instant when x = 1.00 m and y = 2.40 m. Give your answers by entering numbers into the empty boxes below. ax || || X x² ms ms-2.The position r of a particle moving in an xy plane is given by ř seconds. In unit-vector notation, calculate (a) 7, (b) V , and (c) a for t = 3.00 s. (d) What is the angle between the positive direction of the x axis and a line tangent to the particle's path at t = 3.00 s? Give your answer in the range of (-180°; 180°). (4.00r3 – 1.00t)î + (5.00 – 1.00r4)j with 7 in meters and t in (a) Number i i Units (b) Number ît i Units i (c) Number i i Units (d) Number i Units