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Block $A$ in Figure 13.45 hangs by a cord from spring balance $D$ and is submerged in a liquid $C$ contained in beaker $B$ . The mass of the beaker is $1.00 \mathrm{kg} ;$ the mass of the liquid is 1.80 $\mathrm{kg} .$ Balance $D$ reads 3.50 $\mathrm{kg}$ and balance $E$ reads 7.50 $\mathrm{kg} .$ The volume of block $A$ is $3.80 \times 10^{-3} \mathrm{m}^{3}$ . (a) What is the density of the liquid? (b) What will each balance read if block $A$ is pulled up out of the liquid?

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b) The mass of the block is obtained from $E q \cdot 1 : 3.50 g+B=m_{A} g(E q .1)$$$m_{A}=\frac{3.50 g+B}{g}=\frac{3.50(9.80)+46.06}{9.80}=8.20 \mathrm{kg}(\text {Reading in } D)$$Scale $E$ will Read the mass of the beaker and the mass of the liquid: $1.00+1.80=2.80 \mathrm{kg}$

Physics 101 Mechanics

Chapter 13

Fluid Mechanics

Temperature and Heat

Cornell University

Rutgers, The State University of New Jersey

University of Winnipeg

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03:45

In physics, a fluid is a substance that continually deforms (flows) under an applied shear stress. Fluids are a subset of the phases of matter and include liquids, gases, plasmas and, to some extent, plastic solids.

09:49

A fluid is a substance that continually deforms (flows) under an applied shear stress. Fluids are a subset of the phases of matter and include liquids, gases and plasmas. Fluids display properties such as flow, pressure, and tension, which can be described with a fluid model. For example, liquids form a surface which exerts a force on other objects in contact with it, and is the basis for the forces of capillarity and cohesion. Fluids are a continuum (or "continuous" in some sense) which means that they cannot be strictly separated into separate pieces. However, there are theoretical limits to the divisibility of fluids. Fluids are in contrast to solids, which are able to sustain a shear stress with no tendency to continue deforming.

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all right. So the starting observation for this problem is that the scale that the beaker is sitting on is going to read a value that's higher than on the mass of the beaker, plus the mass of the liquid. And the reason why that value is higher is because the liquid is exerting a buoyant force on the block and by Newton's second law than the blockhouse to be exerting unequal in opposite force down on the liquid in the beaker. So that adds to the reading on the scale. And we can use thiss fact tio deduce what the density of the liquid is. So what we wanted Teo. It is set the force that the bottom scale reads, which is 7.5 kilograms times the gravitational acceleration chief. That's going to be equal to the way of the beaker, plus the weight of the liquid, plus the buoyant force. So we divide both sides of this equation by CI Weaken, draw off the cheese and on the right here we have F buoyant, divided by H E and F. Buoyant, divided by G, is also equal to buy our committee's principal equal to the density of the liquid times the volume of the block and the volume of the block was given to us A 3.8 ten 10 sends the negative three meters Cute. So we solve this first equation for F buoyant over G. Yeah, that's going to be 7.5 minus the massive beaker, which is one minus the massive illiquid, which is 1.8. Then, after we have Chef Bui over G, we consent that equal to roll liquid times v block and solve this for rolling quid. And it's just that previous value 7.5 minus one minus one point A divided by the volume of the block, which is 3.8 times 10 to the negative through. Okay, now in part B, we want to know what the two scales with Reed if the block is removed from the liquid. And in that case, there's no buoyant force adding to the scale. What the bottom scale will read is just the sum of the mass of the liquid and the mass of the beaker, which comes out to 2.8 kilograms. And to figure out what the spring scale read, we just need to know the mass of the of the block. So, um to find not we know that what What the scale reads is the mass. Sorry is the way of here. Let me take a step back. The scale reads a mass, which is the force on the scale divided by the gravitational acceleration. So then, if we want to know the total force on the spring scale, we've multiply again by G So this would be in the parentheses here, the reading and this is the reading when the block is still submerged. That's what we want to find and block that's going to be equal to the way of the block, minus the buoyant force on it. And so once again, we can divide both sides by G and replace a buoyant with point over G and then and lock equals FT over. Ji close F buoyant over. And she where f buoyant over ji is thiss value here 7.5 minus one minus 1.8 and F d over CI is the mass that the spring scale initially reads. Then, after the block is submerged from the water, the scale will just read on block, which is the sum of these two

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