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bonufazy [111]
3 years ago
10

According to Archimedes’ principle, the mass of a floating object equals the mass of the fluid displaced by the object. Use this

principle to solve the following problems. (a) A wooden cylinder 30.0 cm high floats vertically in a tub of water (density 1:00 g/cm3 ). The top of the cylinder is 13.5 cm above the surface of the liquid. What is the density of the wood? (b) The same cylinder floats vertically in a liquid of unknown density. The top of the cylinder is 18.9 cm above the surface of the liquid. What is the liquid density? (c) Explain why knowing the length and width of the wooden objects is unnecessary in solving Parts (a) and (b).
Physics
1 answer:
Andrew [12]2 years ago
3 0

Answer:

Part a)

\rho = 0.55 g/cm^3

Part b)

\rho_L = 1.49 g/cm^3

Part c)

Since we know that the base area will remain same always

so here the length and width of the object is not necessary to obtain the above data in such type of questions

Explanation:

Part a)

As we know that when cylinder float in the water then weight of the cylinder is counter balanced by the buoyancy force

So here we know

buoyancy force is given as

F_b = \rho_w V_{sub} g

F_b = (1 g/cm^3) (30 - 13.5) Ag

F_b = 16.5 Ag

Now we know that the weight of the cylinder is given as

W = \rho (30 cm)A g

now we have

\rho (30 cm) A g = 16.5 A g

\rho = 0.55 g/cm^3

Part b)

When the same cylinder is floating in other liquid then we will have

F_b = \rho_L (30 - 18.9 )A g

so we have

\rho_L (11.1) Ag = 0.55(30) Ag

\rho_L = 1.49 g/cm^3

Part c)

Since we know that the base area will remain same always

so here the length and width of the object is not necessary to obtain the above data in such type of questions

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Answer:

1 is 90, 2 is 200 and 3 is 5

Explanation:

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3 years ago
A short-wave radio antenna is supported by two guy wires, 155 ft and 175 ft long. Each wire is attached to the top of the antenn
Vladimir79 [104]

Answer:

163.8 ft

Explanation:

In triangle ABD

AB = 155 ft

Cos63 = \frac{BD}{AB} = \frac{BD}{155}\\BD = 155 Cos63 \\BD = 70.4 ft

Sin63 = \frac{AD}{AB} = \frac{AD}{155} \\AD = 166 Sin63\\AD = 148 ft

Using Pythagorean theorem in triangle ADC

AC^{2} = AD^{2} + DC^{2} \\175^{2} = 148^{2} + DC^{2} \\DC = 93.4 ft

d = distance between the anchor points

distance between the anchor points is given as

d = BD + CD = 70.4 + 93.4\\d = 163.8 ft

5 0
3 years ago
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From a set of graphed data the slope of the best fit line is found to be 1.35 m/s and the slope of the worst fit line is 1.29m/s
Svetradugi [14.3K]

Solution:

Let the slope of the best fit line be represented by 'm_{best}'

and the slope of the worst fit line be represented by 'm_{worst}'

Given that:

m_{best} = 1.35 m/s

m_{worst} = 1.29 m/s

Then the uncertainity in the slope of the line is given by the formula:

\Delta m = \frac{m_{best}-m_{worst}}{2}               (1)

Substituting values in eqn (1), we get

\Delta m = \frac{1.35 - 1.29}{2} = 0.03 m/s

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3 years ago
A 4.0 cm × 4.2 cm rectangle lies in the xy-plane. You may want to review (Pages 664 - 668) . Part A What is the electric flux th
padilas [110]

Answer:

(A). The flux is 0.336 N.m²/C

(B). The flux is zero.

Explanation:

Given that,

Length = 4.2 cm

Width = 4.0 cm

Electric field E=(150 i-200 k)\ N/C

Area vector is perpendicular to xy plane

(A). We need to calculate the flux

Using formula of flux

\phi=E\cdot A

Where, E = electric field

A = area

Put the value into the formula

\phi=(150 i-200 k)\times(4.2\times10^{-2}\times4.0\times10^{-2})k

\phi=-200\times4.2\times10^{-2}\times4.0\times10^{-2}

\phi=-0.336\ N.m^2/C

(B). Given electric field

E=(150i-200j)\ N/C

We need to calculate the flux

Using formula of flux

\phi=E\cdot A

Put the value into the formula

\phi=(150 i-200 j)\times(4.2\times10^{-2}\times4.0\times10^{-2})k

Here, The component of k is not given

So, the flux is

\phi=0

Hence, (A). The flux is -0.336 N.m²/C

(B). The flux is zero.

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