Recent content by justinh8

Geometry Question? A triangle has vertices, A(1,-1) , B(0,5) and C(-3,0). Find the length of the altitude from A to BC. You must find the POI of the altitude and BC: that's the hint So basically i have no idea how to do this... i got 4.3... but its wrong

Homework Statement A 5.00 x10^2-kg satellite is in circular orbit 2.00 x 10^2 km above Earth’s surface. Determine the gravitational potential energy of the satellite Determine kinetic energy Determine The Binding Energy The Attempt at a Solution I solved for gravitational potential...

and also, does binding energy = -0.5(Gmm/r) or 0.5(Gmm/r)?

So when it asks for just the gravitational potential energy, i use Eg = -GMm/r ?

Ok, so the formula of -0.5(GMm/re + altitude) is for the total energy?

Homework Statement What is the change in gravitational potential energy of a 64.5-kg astronaut, lifted from Earth’s surface into a circular orbit of altitude 4.40 3 102 km? The Attempt at a Solution Ok, i know to find this i have to find the gravitational energy on Earth using...

Soo.. what should I go with than, you mean find the y component of the second situation and don't subtract it from the first?

Ok i understand, what i was wondering was since i found the new downward component using pythogorean with tan (angle) = downward component / 9.64N shouldn't I do the same for the first sequence by doing tan(50) = downward component / 9.64N? or is 15sin(50) the same thing? EDIT: i didnt realize i...

Ok so than new downward component would be: tan 30 = downward component/9.64 so downward component would equal 5.56 Newtons and therefore the difference would be 11.49N - 5.56N = 5.924N. But when i find the distance do i not have to find the same downward component using the angle 50 degrees...

Ok but can you explain to me why the equation mg = 15sin50 - 15sin30 is correct? instead of -mg = 15sin50 - 15sin 30?

But wouldn't that be saying 15sin 30 = 15sin50 -mg which is saying the second situation = the first situation subtract the force of the bird whereas the force of the bird would have to be added to the first situation to equal the second situation and also why would you only consider the y...

I did the exact same thing as stanc, however, i think that since the force of the wind stays the same for both situations, the force in the x would always 9.64N and the force in the y would always equal 11.49N You would than have to say to find hypotenuse in situation 2 it is tan 30 = opp/adj =...

So basically, the tension in sequence number 2 is not 15 Newtons? Ya but in the question it states that the force exerted by the wind is the same both times doesn't it?

But in this situation doesn't Jane go down and Josh go up because of the forces Janes force = 539.5 Newtons Josh's Force = 466 Newtons Since Janes force is greater than Joshs force, won't Jane go down and Josh go up??

[b]1. Homework Statement Jane, mass of 55kg holds one end of a vertical rope that passes up over smooth rock and down along a 30 degrees sloping hill as shown. But Josh, mass of 95kg tightly holds the other end of the rope. Assume no friction. Calculate the acceleration. The picture is included...