Hey guys, I'm doing a first year physics subject at uni. I need some help with this question: (this is all the info you get) EDIT: Have another question! Having trouble getting my head around the theory involved in this: A cubical crate with sides of length 1.2 m contains an object that causes the center of mass of the crate and object to be 0.3 m above the crate's geometrical center. As the angle increases on the incline on which the crate is placed, it will either tip or slide. If the coefficient of static friction between the crate and ramp is 0.8, will the crate tip or slide?
first, you need to know the latent heat of ice. I'd assume you'd learn this in class, but its 334 kJ for 1 kg. so 2240g * 334j/g = 748160J. The specific heat of water is 4.2 J/g then set 748,160 = 4.2 (J/g) * m (100 - 0) m = 1781g = 1.78 kg
^ Thank you for the reply. I have another question, it seems really simple but I seem to be getting it wrong, lol. SO BASICALLY, I just thought you make 7+4t=3, work out 't' and then sub that back in. Then work out the magnitude of r. Anyways, that doesn't work since I end up getting t=-1 which makes no sense. The answer is 9.5.
Okay so you have the right idea on this one. t= -1 for sure. So now plug in t =-1 into our function for the y coordinate (j) an we obtain r= 3i+9j and now using this we can obtain the magnitude of the vector by squaring both the x and the y coordinate adding them and then taking the square root as so: magnitude of r = sqrt ( 3^2 + 9^2) magnitude of r= sqrt (9+81) magnitude of r = sqrt(90) magnitude of r= 9.49
Bump! Edited my original post with the new question :S "A cubical crate with sides of length 1.2 m contains an object that causes the center of mass of the crate and object to be 0.3 m above the crate's geometrical center. As the angle increases on the incline on which the crate is placed, it will either tip or slide. If the coefficient of static friction between the crate and ramp is 0.8, will the crate tip or slide?"
