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## Homework Statement

A chunk of carbon steel with density, ρ = 7.84 g/cm3 , is completely submerged in fresh water. The chuck of steel weighs 30 N more in air than in water

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- Thread starter jdg
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- #1

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A chunk of carbon steel with density, ρ = 7.84 g/cm3 , is completely submerged in fresh water. The chuck of steel weighs 30 N more in air than in water

- #2

jgens

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- #3

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- #4

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1. Find the buoyant force acting on the chunk of steel

2. Find the the volume of the chunk of steel

- #5

jgens

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The weight of the carbon steel chunk in water should be: W

And: W

Edit: Thanks for the questions! If you can solve this first one, you can solve the second question really simply!

- #6

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Yea, my problem is, how do I find the mass? If I assume 1 m3 for the volume, it is 7840 kg

- #7

jgens

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Hint: To do this problem, you don't need to make any assumptions! :)

- #8

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Wair - Wwater

Ok, so Wair - (Wair-30N)?

Ok, so Wair - (Wair-30N)?

- #9

jgens

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Yes! So this means that: W_{air} - (W_{air} - 30 N) = ???

- #10

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mg-30?

- #11

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wait, so the Wair's cancel and you get -30N?

- #12

jgens

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Hint: Your expression should include the buoyant force.

Edit: Yes, the W

- #13

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- #14

jgens

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Yes! *F*_{b} = 30 N.

That's what I get for the Volume as well, or equivalently*V = 390 cm*^{3}.

That's what I get for the Volume as well, or equivalently

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- #16

jgens

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Edit: Oops! Used the wrong density in calculating the volume, remember that the buoyant force equals the weight of the water displaced! I'm terribly sorry about that.

- #17

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thanks a lot! you have time for one or 2 more?

- #18

jgens

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- #19

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ok, so for weight in air it would be 30 N + 30 N= 60 N?

I don't think you can find the weight in air. It's juts "30N" heavier than in the weight water.

Following from jgens, the volume should be about 3.058e-3

- #20

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m = dV = (7840)(3.901e-4) =3.06 kg

V = Fb/dg = (30 N) / (7840*9.81) = 3.901 m3

Next question: A garden hose with internal diameter of 13.5 mm lies flat on a sidewalk while water is flowing in it at a speed of 6 m/s. A person happens to step on it at the very edge of the opening of the hose and decreases its internal diameter by a factor of 9

So D (1) = 0.0135m

r (1) = 0.00675m

D (2) = 0.0135/9 = 0.0015m

r (2) = 0.00075m

A (1) = pi*r^2 = (3.14...)(0.00675^2) = 1.4134...e-4

A (2) = pi*r^2 = (3.14...)(0.00075^2) = 1.767...e-6

1. What was the water flow rate in the hose prior to the person stepping on it?

- I got this part: J(1) = A(1)V(1) = 8.59 m3/s

2. What is the flow rate of water after the person steps on it?

3. What is the speed of the water just as it exits the hose after the person steps on it?

- #21

jgens

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Ok, I got: Fair= mg = 30 N

m = dV = (7840)(3.901e-4) =3.06 kg

V = Fb/dg = (30 N) / (7840*9.81) = 3.901 m3

Not sure I agree with this. Start with

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ok, I got V = 0.003058...

m = 23.975...

m = 23.975...

- #23

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So V2 = V1*(A1/A2) = 486 m/s

Is this right?

And for part 3 I did

J = (A2)(V2) = 8.59e-4 m3/s

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