Recent content by hobbes1235
-
[Grade 12 Physics] Gravitational/circular motion #2
I actually didn't know that the apparent weight was Fn but it turned out to be correct. What even is "apparent weight" I thought it was the weight when moving.- hobbes1235
- Post #3
- Forum: Introductory Physics Homework Help
-
[Grade 12 Physics] Gravitational/circular motion #2
Diagram for question 2: Please check if my work is correct. Main equation: Fc = Fg - Fn Fn = Fg - Fc I assume that: Fn = Fg when stationary Fg = mg I divide by 5 because the apparent weight must be one fifth of her weight when stationary Fg = mg / 5 Fn = Fg Fn = mg / 5 Now to substitute...- hobbes1235
- Thread
- Grade 12 physics Motion Physics
- Replies: 5
- Forum: Introductory Physics Homework Help
-
[Grade 12 Physics] Circular/gravitational motion
I think I finished the first question, I got 2.06 m/s^2 for my gravitation strength and was wondering if you could confirm this.- hobbes1235
- Post #20
- Forum: Introductory Physics Homework Help
-
[Grade 12 Physics] Circular/gravitational motion
EP1 = EP2 + EK2 mgh - mg2r = 1/2(mV^2)mgh - mg2r = Change in EP Change in EP = 4.03g Change in EP = Kinetic energy 4.03g = 1/2(mv^2)- hobbes1235
- Post #18
- Forum: Introductory Physics Homework Help
-
[Grade 12 Physics] Circular/gravitational motion
so the I found that the potential energy lost was 4.03g's. Using the law of conservation of energy, that should equal the kinetic energy: 4.03g = 1/2(mv^2) which turned out to be: 6.2g = v^2 (I'm not going to square root this because we need v^2 in the next equation) mV^2 / r --> (1.3)(6.2g)...- hobbes1235
- Post #16
- Forum: Introductory Physics Homework Help
-
[Grade 12 Physics] Circular/gravitational motion
How am I supposed to calculate the loss of potential energy when I don't have the mass of the planet or the gravitational field strength? On the equation above, I canceled out the mass of the planet but I didn't know if that worked.- hobbes1235
- Post #14
- Forum: Introductory Physics Homework Help
-
[Grade 12 Physics] Circular/gravitational motion
EP1 = EP2 + EK2 ((-6.67x10^-11)(1.3) / 4.50) = ((-6.67x10^-11)(1.3) / 1.40) + 1/2(v^2) V = 9.237621478x10^-6 Fc = (1.3)(9.237621478x10^-6) / 0.70 Fc = 1.715558277x10^-5 Fc - 21 = Fg 1.715558277x10^-5 - 21 = Fg Fg = 16.2m/s^2- hobbes1235
- Post #12
- Forum: Introductory Physics Homework Help
-
[Grade 12 Physics] Circular/gravitational motion
I tried that but I'll see if I made a mistake, I probably did. For the other question, do you get 16.2m/s^2 for the gravitational strength? I'm just checking because it seems very big.- hobbes1235
- Post #10
- Forum: Introductory Physics Homework Help
-
[Grade 12 Physics] Circular/gravitational motion
Could you also help me with Question 2. ?- hobbes1235
- Post #7
- Forum: Introductory Physics Homework Help
-
[Grade 12 Physics] Circular/gravitational motion
Oh wait, I was thinking of the law of conservation of energy wrong. I think I get it now. On one of my other questions, there was a force of friction but this problem is frictionless. So, the GPE1 is for the whole system. thanks :)- hobbes1235
- Post #6
- Forum: Introductory Physics Homework Help
-
[Grade 12 Physics] Circular/gravitational motion
But what would the height value for the GPE2 be? Like the r value- hobbes1235
- Post #5
- Forum: Introductory Physics Homework Help
-
[Grade 12 Physics] Circular/gravitational motion
So GPE1 = GPE2 + KE2 ?- hobbes1235
- Post #3
- Forum: Introductory Physics Homework Help
-
[Grade 12 Physics] Circular/gravitational motion
Diagram for question 1: I know the mass, I need Fg. My work: Main equation: g = Fg/m I need to find Fg. Fg= Fc - Fn [Fn = 21 N Fc = ?] {I need to find Fc.} Fc = ma --> Fc = (mV^2)/ r [Mass = 1.3kg V = ? r = 0.70] {Now I need the velocity at that point where Fn = 21 N (the top of the...- hobbes1235
- Thread
- Circular motion Grade 12 physics Motion Physics
- Replies: 21
- Forum: Introductory Physics Homework Help