<p>can someone invite me, bandosrock529</p>
<p>I have been staring at this question for a long time. Can anyone help?</p>
<p>A spring scale caliberated in kilograms is used to determine the density of a rock specimen. the reading on the spring scale is .45kg when the specimen is suspended in air and .36kg when the specimen in fully submerged in water. If the density of water is 1000Kg/m^3, the density of the rock specimen is:</p>
<p>(in kg/m^3)</p>
<p>A. 2.0
B. 8.0
c 1.25
d. 4.0
e. 5.0</p>
<p>the answer is e and I am getting everything BUT e.</p>
<p>Help!</p>
<p>2 more questions that im unable to figure out:</p>
<p>Two large, flat, parallel, conducting wires are .04m apart. The lower plate is at a potential f 2 V with respect to the ground and the upper plate is at a potential of 10 v with respect to the ground. Point P is located 0.01 m above the lower plate.</p>
<p>The electric potential at point P is:
A. 10V
B. 8 V
C. 6V
D. 4V
E. 2V</p>
<p>answer is d..how do you do it?</p>
<p>And,</p>
<p>the magnitude of the electric field at point P is:
A. 800V/m
B. 600V/m
c. 400V/m
d. 200V/m
e. 100V/m</p>
<p>answer is again d..but no clue how to get it..Im getting 133.33 which is closest to e!!</p>
<p>For the first question</p>
<p>.45-.36 = .09 is how much is displaced you have the mass but you need the volume, since you know the density of water you set it up the density equation</p>
<p>1000Kg/m^3 = .09kg/? </p>
<p>you will get a volume of 0.00009 or 9.0*10^-5 m^3 </p>
<p>since you know the volume know you can find the density outside of the water with</p>
<p>.45 kg/9*10^-5 m^3 = 5.0 x 10^3 kg/m^3</p>
<p>Ok cool, thanks. I figured out the other questions. I was getting the right answer. Like, 4 for the first electricity one, I just thought the correct answer was 2 and started freaking out even when my sheet clearly said d which is 4. </p>
<p>(I panic a lot)</p>
<p>Thanks dondilx for the help on fluids</p>