Recent content by lynchdemartin

thanks I can finally rest now..

umm this might seem stupid but I think the question is asking what is the work done by the "person" not the "rope" so then it's T 225 N x 8m = 1800J ?

I still get 1690J for work done. Driving me nuts...

That's only if the object is "not moving at constant velocity". In this question it is moving at constant velocity so Fnet must equal 0.

A 50.0 kg crate is dragged at a constant velocity of 1.5 m/s for a distance of 8.0 m. If the force of friction on the crate is 225 N, how much work was done by the person to drag the crate over the 8.0 m distance if the rope used to drag the crate was directed at 20degrees angle above the...

anyways I think the answer is 2.4 magnification & it's a concave virtual larger image. Because like you said I shouldn't have a 2 negative d values. Thanks a lot for the help.

hmm good point.

2 x focal length = radius. so focal length is 10cm/2 = 5cm However the confusing part for me is to find out whether the f is on the positive side or the negative side behind the mirror. Because you don't know what type of mirror right? The only thing I know is that the object is right...

Homework Statement I can't seem to figure out whether I'm dealing with a + or - radius value. Anyways here's the question: An object 5.0 cm tall produces an image that is 7.0 cm behind the mirror. If the radius of the curvature of this mirror is 10.0 cm, what is the magnification of the...