Recent content by Glenboro

That is much better, thanks man

Would you mind if you can briefly explain how "relative" thing works? I need to put those into my head first :smile:

Yeah because gravity would not affect much when it's it orbit. Due to distance At Earth's surface, how far is an object from the centre of the earth? (7.00 x 10^6m) - (Earth's radius)

Gravitational potential energy? PEg = mgh PEg = (1550kg)(7.00 x 10^6m)(g)

Since it is in orbit, yes I calculated part (b) as orbital velocity.

Can you give me a little hint :( My answers keep getting off

My small mistake, Thanks haruspex

So I fully realized my error, I should known it was the radius of orbit. I did re-calculation for all questions. a)PEg = -(6.67x10^-11)(5.98x10^24)(1550)/(7.00 x 10^6) = -8.83 x10^10J b) I didn't had to use Vescape formula since it was in orbit V= root (G)(Me)/R v= root...

I'm sorry if you felt that way, I'm legitimately self-motivated and working hard to achieve decent knowledge in physics. I'm uncertain what made you feel bad and I apologize if you think I only used helpers to check my work which isn't true. Again, I might just leave this forum if you don't seem...

Homework Statement A ball bearing of mass m = 50.0 g, is sitting on a vertical spring whose force constant is 120.0 N/m. The initial position of the spring is at y = 0 m[/B] a) The spring is compressed downward a distance x = 0.200 m. From the compressed position, how high will the ball...

I completely agree with you from beginning to the end. I will remind myself not to confused with standard-form equation.

Yeah, that is a big problem for me. I'm more focused on getting numbers into equation rather than trying to put into my head. I will fix that habit asap. Also, yes hang time is the time when projectile is in the air. Do you recognize any an error from my calculations?

Homework Statement A cannonball is fired with a velocity of 125 m/s at 25.0° above the horizontal a) Determine the horizontal and vertical components of the initial velocity. b) Determine the maximum height the cannonball reaches in its path c) Determine the time it takes to reach maximum...

Oh so to figure out how far it's distance from Earth I had to subtract (6.38 x 10^6m) from 7.00 x 10^6

The radius I used to calculated was (6.38 x 10^6m) The Earth's radius plus 7.00 x 10^6 m from the question. (7.00 x 10^6m)+(6.38 x 10^6) = 1.338 x 10^7 m Is this what you are referring to? Sorry for making mistake, I actually trying my best