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A 16.6 kg silver mass rests on the bottom ofa pool. (The densityof silver is 10.5 ✕ 103 kg/m3 andthe density of wateris 1.00 ✕ 103 kg/m3.)HINT: Archimed...

Question

A 16.6 kg silver mass rests on the bottom ofa pool. (The densityof silver is 10.5 ✕ 103 kg/m3 andthe density of wateris 1.00 ✕ 103 kg/m3.)HINT: Archimedes' principle states, "Any object completely orpartially submerged in a fluid is buoyed up by a force withmagnitude equal to the weight of the fluid displaced by theobject."(a)What is the volume of the silver (inm3)? ______m3(b)What buoyant force acts on the silver (in N)? (Enterthe magnitude.) _____N(c)Find the silver's w

A 16.6 kg silver mass rests on the bottom of a pool. (The density of silver is 10.5 ✕ 103 kg/m3 and the density of water is 1.00 ✕ 103 kg/m3.) HINT: Archimedes' principle states, "Any object completely or partially submerged in a fluid is buoyed up by a force with magnitude equal to the weight of the fluid displaced by the object." (a) What is the volume of the silver (in m3)? ______m3 (b) What buoyant force acts on the silver (in N)? (Enter the magnitude.) _____N (c) Find the silver's weight (in N). (Enter the magnitude.) _____ N (d) What is the normal force acting on the silver (in N)? (Enter the magnitude.) _____N



Answers

The density of silver is 10.5 $\mathrm{g} / \mathrm{cm}^{3} .$ What is the mass (in kilograms) of a cube of silver that measures 0.62 $\mathrm{m}$ on each side?

In this problem of centering measurement, we have given that a graduated cylinder was filled with water 2 15 ml. So this is 15 ml. You see the volume mark and wait on a balance. Its mouth was. So this will be the volume and the mass was yeah, 27.35 g. So this is 27.35 g. An object made of silver was placed in the cylinder and completely submerged in the water. The water level rose to, so this was a new water level rose to this is 18.3 mm. When we wait the cylinder, the water and the silver object has a total mass of. So this is a total mass of this is 62 0.0 grams. We have to calculate the density of silver and now we have to find the density of silver. In this is a gram per centimeter cube. So this would be ground percent in metric. You. So, first we will find the massive silver. So this will be ma so this would be um book now this is a mass of silver So this would be equals two M two minus m now divided. So this would be this was subtracted. So this will be 62.0 g minus 27.35g is equals to 34 65 g. And now we have to find the volume of silver. So this will be volume of silver. This is silver, silver. So this would be volume to minus volume one. So this will be volume to minus volume one. So this is equal to 18 point three ml -15.0 mm. inc was 3.3 mm. So this is 3.3 emulated and we know that one millimeter is equal to one centimeter too. So this will be also equal to 3.3 centimetre go. And now we want the density of silver. So density is given by mass of silver. So this will be a mass of silver and this would be a volume of silver. Now put the value. So this is 34.65 g Divided with 3.3 centimeter cube. And now also we have to keep only two significant figures also here and now we have to They were 34.65 with 3.3. So this is equal to 10 points here 10.5. So this is here then point same Graham's first centimeter cube. And now we have to round off a year. So zero is even and if Before five this is a zero even number we have to increase by one. So this would be equal to 11 g per centimeter cube. So this is the density of silver. Sure.

Well said number 74. The density off world is given as G is equal though 19 three 00 Yeah, g for me that you that then city off silver. So as is given as 5 10,001 bid here, do you but me that you the mass or that Katie in a is given us three G but And the muscle of the ground in water is given us through. Born 75 Do you get now? Let must off. They told me I m g and the marks off. That's it Would be, um as for air begin by the mass off gold less the mosque off silver is equal Toe the mass off the around in here So OMG um as is equal to three jg in water. The Byron Force is the quality. It'll roll ward and G into total volume off that crowd. And this is equal to that difference in that two words in air and water. So three minus 2.75 in the nine points. Yeah. One. This can be detainers. Rhoda blue G. The total volume of the crown is made up of the volume of gold and the volume off silver so we can die Marcel Gold upon density or role Yes, Mars off silver upon density or silver isn't for your school Fine four by now substituting that given values the density of or terrorist 1000 kg permitted in 29.81 into OMG upon 19 1000 feet and a less the mine s nd upon 5 10,001 day is to 2.45 When we saw this, we will get zero buying 50 a. M d. That is muscle gold less a little bored. 934 times three minus Must off board is equal to 2.4 by this will give us the muscle bold as zero point it to six. You ready? So the marks of gold is it would be £6 and mars off. Silver is three miners, zero point a blue states is equal. Good point 174 You ready

As we all know that the expression for the buoyant force is given by Abby is equal to. Hello, liquid multiplication, be multiplication T. Two. I can write the value of the buoyant force acting on the cube when it immersed in water. F B water is equal to peru water multiplication pt. So according to the event problem, we have B is equal to 3.9 centimeter multiplication 10 to depart. Managed to meter. I'm just changing the unit per centimeter whole cube which is equal to zero point oh dear. Oh dear. Oh dear. Oh dear. Oh 59 32 m cube. Now going forward and putting the value in the apart equation. So A B is equal. Do 1000 kg per meter cubed multiplication, zero point 0000 5932 m Q. Multiplication, 9.8 m per second. The square. So finally I get 0.6 newton to option B is correct. Answer for this problem.

And this problem. We have a submarine that is 255 meters, but low the surface of the water. The seawater has a density of 1.3 grand per cubic centimeter. The cross sectional area of the hall of the submarine. It's 2200 meters where we wanted him out. How much weight of the water is on top of the submarine so you could draw a picture of our submarine here. It's a distance deep below the surface and were given that the the density of sea water actually have used 1024 year. So we know that the pressure at it given depth is the density of the fluid above it times gravity times the distance to the point we're looking for. We know that the weight above the submarine is just the pressure times, the area of the submarine. That's then the density of the water, times gravity and the area times the distance below the below the water. So this tournament brackets is actually just the volume of the water above the sub fucking all our numbers in we have 1024 kilogram for me to cube 9.8 meters per second squared play 200 meters squared and 255 meters and we wind up with 5.66 times 10 to the ninth Newton's. They were asked if a diver we're at this depth in the water so we could find the pressure and that's 1024 kilograms from Eater's cubed 9.8 meters per second squared times 255 meters and then we convert the atmosphere. We have one atmosphere is 1.13 times 10 to the fifth past cows fucking all that into the calculator and we wind up with 25.4 atmospheres.


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