Recent content by jenador

Homework Statement Two blocks are sliding to the right across a horizontal surface, as the drawing shows. In Case A the masses of both blocks are 2.2 kg. In Case B the mass of block 1, the block behind, is 7.0 kg, and the mass of block 2 is 3.5 kg. No frictional force acts on block 1 in either...

I am having a lot of problems with the tension/force questions on my homework. Here is the last question I don't understand in my homework set: The drawing shows a circus clown who weighs 860 N. The coefficient of static friction between the clown's feet and the ground is 0.41. He pulls...

Oh! Therefore there are two strings that both have 81.5 N of tensions. I assumed 81.5 was the whole amount of tension in both strings. thank you SO much for clearing that up.

here's the question: Part a of the drawing shows a bucket of water suspended from the pulley of a well; the tension in the rope is 81.5 N. Part b shows the same bucket of water being pulled up from the well at a constant velocity. What is the tension in the rope in part b? picture...

i did that too but the computer is telling me that's the wrong answer as well. is it possible that 65 is the wrong angle to use ?

Homework Statement Interactive Solution 4.11 offers help in modeling this problem. Two forces, vector F 1 and vector F 2, act on the 5.00 kg block shown in the drawing. The magnitudes of the forces are F1 = 30.5 N and F2 = 47.5 N. What is the horizontal acceleration (magnitude and direction)...

by the way i have to write it in a way solving for just v_y

i used the equation LaTeX Code: v_y^2=(v_0)^2_y-2ay and rearranged it so it would be sqtroot(v(initial)^2 + 2ay). i think i got like -8.9 m/s, but the website told me i was wrong.

Homework Statement A golfer hits a shot to a green that is elevated 4.0 m above the point where the ball is struck. The ball leaves the club at a speed of 15.0 m/s at an angle of 40.0° above the horizontal, which is the +x axis. It rises in the +y direction to its maximum height and then falls...

omg i was not using the quad equation. thank you for pointing that out to me! since someone said it is the quad equation, i just plugged in using a=-4.9, b=11.8, and c=840. i got t =14.35 s and 187.9 m. and the computer said that was the right answer. thank you all so much! i feel really dumb...

using Y= (Viy)t + (-4.9t^2) when i plug in numbers, i get: -840 = 11.8*t - 4.9 t^2 is this correct? and initial velocity is not zero, like belliot said right?

thank you for responding belliott4488! i did what you said and took v initial in the y direction to not be zero. i found this out by using sin42*17.6 m/s = 11.8 m/s. i found time to be 1.9 s using d=v(initial)*t+1/2*a*t^2 in the y direction. then i 1.9s*13.1m/s = 24.9 m. but the website where i...

so i am having issues with finding the distance for this question. here is the question and what i tried to do to solve it. Homework Statement An airplane with a speed of 17.6 m/s is climbing upward at an angle of 42° counterclockwise from the positive x axis. When the plane's altitude is...