Recent content by gjh

St. Louis Arch: The St. Louis arch was constructed using the hyperbolic cosine function. The equation used for construction was 693.8597 – 68.7672 cosh .0100333x. -299.2239<=x<=299.2239. (1) Cross sections of the arch are equilateral triangles, and (x,y) traces the path of the centers of mass...

(a) How high above the ground is the center of the highest triangle? (At ground level, y=0) Substituting, x=0 in equation (1), I got 693.8597-68.7672 cosh (0) = 625.0925 (b) What is the height of the arch? (Hint: For an equilateral triangle, A = sqrt(3) c^2, where c is one half the base of...

Apply Principle of Archimedes - upward force is equal to the weight of the volume of water displaced. Density (boat) = 120 kg/(3x1x.24)m(3) = 500/3 kg/m(3) Density of water = 1000kg/m(3). Density (water)/Density of boat = 1000 kg/m(3)/(500/3) kg/m(3) = 6. So density of water is 6 times the...

We know x = R =max range (28m) on level ground. Need to find v()^2. Subbing y=0 into (1) above, get v(0)^2 = (gR^2/)/(2*cos (theta)^2 * tan (theta). ... (2) This didn't seem right, since this means v(0)^2 is a negative number ... maybe my orientation or algebra wrong? Anyway, didn't see...

Yes, plus I made a stupid computational error. .22j/300 = 7.33*(10)^4 not 8.7jx(10)-4. Thus, my answer would be a (average) =(-4.3i+7.3j)10^-4 in component form. mag a then is sqrt [(-4.3)^2+(7o.3)^2(10^-4)] = 8.9*10^-4 m/s^2. This is very close to book answer of -4.4i+7.6j)*10^-4 with mag...

Started over. Back to vectors. v =.26[cos 60i+sin 60j) = .13i+.22j (this is instantaneous velocity). avg a = delta v/delta t = (.13i+.22j-.26i)/300 = (-.13i+.22j)/300 = (-4.3i+8.7j)10^-4 - this is closer, but magnitude of a would be 9.7(10)^-4 ... which is way off from 9(10)^-4. Almost there...

I grappled with this problem also: my approach was simply to set coordinate system at the center of the wheel. The initial position vector would be -75j. After 5 minutes, the wheel will have completed (5/30)*360 degrees or 60 degrees. The final position vector (after 5 mins) would be R = r...

Not to beat a dead horse ... well, maybe a little since I spent many hours a/analyzing this problem, I checked my answer by finding the range at 16.7 m/s as suggested. Using x = (v(0)^2)sin (2sin (theta)*cos (theta)/g -> x = 28 m. But we know the range was 39 m - conclusion is he was speeding...

I think this line of reasoning works. Here's how I applied it: Assume 39 m is the max range and it was achieved at launch angle of 45 degrees. Use the range formula X=2v(0)^2 sin (theta)*cos (theta)/g, where X=39 m and theta = 45 degrees and solve for v(0) -> v(0) = 19.6 m/s, which is larger...

I assumed the cyclist was going 60-kph. Then worked backwards as shown above to calculate the deceleration in the x direction and then the time to decelerate to a stop at 39 m. Then used this time to calculate the initial speed if cyclist thrown 39 m, which was 8.4 m/s. But if he was going...

Hello - new member here. Am retired 75-year-old doing course work in various areas of mathematics and physics to pass the time while the COVID situation resolves itself. Found I needed help with some of the physics problems, so I joined the Physics Forum as an aid. Cheers!