Recent content by floridianfisher

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Homework Statement Three 70.0 W, 120 V lightbulbs are connected across a V = 120 V power source, as shown in Figure P18.48. (Assume that the resistance of each bulb is constant even though, in reality, the resistance increases markedly with current.)...

Homework Statement A 1-gram particle with a charge of 1 milliC starts from rest in a uniform electric field of magnitude E = 10 V/m. How far (in m) does the particle travel in 2 seconds? I know the answere is 10 but I don't know how to find it Homework Equations PE/q=-Ex The...

Homework Statement A boy scout starts a fire by using a lens from his eyeglasses to focus sunlight on kindling 5.2 cm from the lens. The boy scout has a near point of 15.2 cm. When the lens is used as a simple magnifier, give values for the following conditions. (Hint: Assume a "normal"...

Please help soon as I do not have much time left to solve this problem. Thanks

So i used the same equation except for Mmax I used 1+ 15.2/7.346= 3.068 but still the answere is not right?

Homework Statement A boy scout starts a fire by using a lens from his eyeglasses to focus sunlight on kindling 5.2 cm from the lens. The boy scout has a near point of 15.2 cm. When the lens is used as a simple magnifier, give values for the following conditions. (Hint: Assume a "normal"...

Thanks for the help!

Homework Statement The object in Figure P23.50 is midway between the lens and the mirror. The mirror's radius of curvature is 20.4 cm, and the lens has a focal length of -16.9 cm. http://www.webassign.net/sf5/p23_54.gif Figure P23.50 A. Considering only the light that leaves the object...

ok that's I got it. Thanks for your help!

I plugged in p= 13.2 +2.1 =15.3 but the answer is wrong?

I used 1/(13.2+x) + 1/x = 1/2.44 for those answers I just tried it the way I believe your talking about which is 1/-(13.2+x) +1/x = 1/2.44 and I got -2.1 and -15.3 I am confused though

Okay so I got this diagram f----------p-------q----lens distance from the lens to f =2.44cm distance from lens to p= 13.2+s distance from lens to q = s then solving for s I get s= 2.88 cm or s= -11.2cm Since q is supposed to be negative for virtual images the answer is 11.2 cm?

Here is the whole question. I got all of a and the second part of b (a) How far is the lens from the slide if the image is real? (Enter 'none' in the second box if there is only one answer.) nearer possible distance = correct check mark cm further possible distance = correct check mark cm (b)...

Homework Statement A transparent photographic slide is placed in front of a converging lens with a focal length of 2.44 cm. The lens forms an image of the slide 13.2 cm from the slide. How far is the lens from the slide if the image is virtual? nearer possible distance = Homework...