Recent content by Ben5386

Thanks so much for your help :)

I don't know how but I got it the answer is 29.73 seconds.

I have the ratio of the two and I have 2F/F I have the ratio of 2F/F(sqrt(2)) which reduces to the sqrt(2) sorry my computer doesn't show me the mathematical functions. so I put in what I got for distance earlier which is (F/m)*t^2=1/2*sqrt(2)*t^2

I know that this answer is probably wrong but is it 84.089

Would it be the square root of 2F

Would it be F/M meaning the cosine of 45 degrees

I have D=1/2(2T/m)*t^2 which reduces down to D=(F/m)*t^2 and for the second case is M equal to F?

I have D=1/2(2T/m)*t^2 which reduces down to D=(F/m)*t^2 and for the second case is M equal to F?

The Magnitude of the vector in the middle labeled M in the picture

I think it is pythagorean theorem for the magnitude of the resultant force of the second case. To solve for the distance I would use the equation D=Vo+1/2at^2 and since the distance is the same I can plug in the distance for the second case

Here they are the upside down T is the perpendicular vector with the F in the middle and parallel are the ones are below.

How do I draw the force vectors on this site I have them on paper

Homework Statement A space probe has two engines. Each generates the same amount of force when fired, and the directions of these forces can be independently adjusted. When the engines are fired simultaneously and each applies its force in the same direction, the probe, starting from rest...