Home Page Textbooks Physics: Principles and Problems Page 358: Section Review

Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

Textbook solutions

All Solutions

Page 358: Section Review

Exercise 32

Step 1

1 of 1

A full soda pop can either sink or float in water depending on their content. It matters whether the drink is diet or not because the sweeteners used for them vary in density. The can that contains sugar sinks because it is denser that water. The can that contain artificial sweeteners (diet soda) float because it is less dense that water.

Exercise 33

Solution 1

Solution 2

Step 1

1 of 6

**Known:**
$rho_text{water} = 1.0 times 10^3 frac{text{kg}}{text{m}^3}$
$frac{V_text{submerged}}{V_text{cork}} = 0.10$

**Unknown:**
$rho_text{cork} = ?$

Step 2

2 of 6

**Calculation:**
Since the buoyant force $F_text{buoyant}$ equalizes the weight of the cork $F_text{g}$, we can express their relationship as follows

$$ tag{1}
F_text{g} = F_text{buoyant}
$$

Step 3

3 of 6

The weight of the cork is given by the equation

$$
F_text{g} = m_text{cork}g
$$

The buoyant force on the submerged part of the cork is given by the equation

$$
F_text{buoyant} = rho_text{water}V_text{submerged}g
$$

Thus, we can express Equation (1) as follows

$$ tag{2}
m_text{cork}g = rho_text{water}V_text{submerged}g
$$

Step 4

4 of 6

The density of the cork is given by the equation

$$
rho_text{cork} = frac{m_text{cork}}{V_text{cork}}
$$

Isolating $m_text{cork}$ one the left side of the equation

$$
m_text{cork} = rho_text{cork}V_text{cork}
$$

We can now express Equation (2) as follows

$$
rho_text{cork}V_text{cork}g = rho_text{water}V_text{submerged}g
$$

Step 5

5 of 6

Isolating $rho_text{cork}$ on the left side of Equation (2)

$$
rho_text{cork} = rho_text{water}frac{V_text{submerged}}{V_text{cork}}
$$

Plugging in the given values, we have

$$
rho_text{cork} = (1.0 times 10^3) cdot (0.10)
$$

$$
boxed{rho_{cork} = 100 frac{text{kg}}{text{m}^3}}
$$

Result

6 of 6

$$
rho_text{cork} = 100 frac{text{kg}}{text{m}^3}
$$

Step 1

1 of 2

The buoyant force and force of gravitation are the same:

$F_{buoyant} = F_{g}$

$rho_{water} V_{floated} g = rho_{cork} V_{total} g$

cancel g:

$rho_{water} V_{floated} = rho_{cork} V_{total}$

$rho_{cork} = rho_{water} (dfrac{V_{floated}}{V_{total}})$

$rho_{cork} = (1000) * (dfrac{1}{10})$

$$
rho_{cork} = 100 kg/m^3
$$

Result

2 of 2

$$
100 kg/m^3
$$

Exercise 34

Solution 1

Solution 2

Step 1

1 of 5

**Known:**
$rho_text{helium} = 0.18 frac{text{kg}}{text{m}^3}$
$rho_text{air} = 1.3 frac{text{kg}}{text{m}^3}$
$F_text{brick} = 10 text{N}$

**Unknown:**
$V = ?$

Step 2

2 of 5

**Calculation:**
Since the buoyant force of the helium balloon must be greater than or equal to the weight of the brick in order to lift it, we can express their relationship as follows

$$ tag{1}
F_text{brick} = F_text{buoyant}
$$

Step 3

3 of 5

The buoyant force $F_text{buoyant}$ is given by the equation

$$
F_text{buoyant} = (rho_text{air}-rho_text{helium})Vg
$$

Plugging this into Equation (1), we have

$$ tag{2}
F_text{brick} = (rho_text{air}-rho_text{helium})Vg
$$

Step 4

4 of 5

Isolating $V$ on the left side of Equation (2)

$$
V = frac{F_text{brick}}{(rho_text{air}-rho_text{helium})g}
$$

Plugging in the given values, we have

$$
V = frac{(10)}{(1.3 – 0.18) cdot (9.8)}
$$

$$
boxed{V = 0.91 text{m}^3}
$$

Result

5 of 5

$$
V = 0.91 text{m}^3
$$

Step 1

1 of 2

$F_{brick} = (rho_{air} – rho_{helium}) V g$

Solve for V:

$V = dfrac{F_{brick}}{(rho_{air} – rho_{helium}) g}$

$V = dfrac{10}{(1.3 – 0.18) * (9.80)}$

$$
V = 0.91 m^3
$$

Result

2 of 2

$$
0.91 m^3
$$

Exercise 35

Solution 1

Solution 2

Step 1

1 of 4

**Known:**
$F_1 = 150 text{N}$
$A_1 = 2.5 times 10^{-3} text{m}^2$
$A_2 = 4.0 times 10^{-4} text{m}^2$

**Unknown:**
$F_2 = ?$

Step 2

2 of 4

**Calculation:**
To solve for the force exerted by the hydraulic lift, we use the equation below

$$
F_2 = frac{F_1A_2}{A_1}
$$

Step 3

3 of 4

Plugging in the given values, we have

$$
F_2 = frac{(150) cdot (4.0 times 10^{-4})}{(2.5 times 10^{-3})}
$$

$$
boxed{F_2 = 24 text{N}}
$$

Result

4 of 4

$$
F_2 = 24 text{N}
$$

Step 1

1 of 2

We should use $F_{2}=dfrac{F_{1}A_{2}}{A_{1}}$ equation ;

So;

$$
F_{2}=dfrac{F_{1}A_{2}}{A_{1}} = dfrac{(150N)(4.0 times 10^{-4} m^{2})}{2.5 times 10^{-3}m^{2}} = 24 N
$$

Result

2 of 2

$$
textit{color{#c34632} 24N }
$$

Exercise 36

Solution 1

Solution 2

Step 1

1 of 6

**Known:**
$F_1 = 2.3 times 10^4 text{N}$
$A_1 = 0.15 text{m}^2$
$A_2 = 0.0082 text{m}^2$

**Unknown:**
a. $P = ?$
b. $F_2 = ?$

Step 2

2 of 6

**Calculation:**
(a)
To calculate the pressure in the hydraulic cylinder, we use the equation below

$$
P = frac{F_1}{A_1}
$$

Step 3

3 of 6

Plugging in the given values, we have

$$
P = frac{2.3 times 10^4}{0.15}
$$

$$
boxed{P = 1.5 times 10^5 text{Pa}}
$$

Step 4

4 of 6

(b)
To calculate the force exerted on the small cylinder to lift the automobile, we use the equation below

$$
F_2 = frac{F_1A_2}{A_1}
$$

Step 5

5 of 6

Plugging in the given values, we have

$$
F_2 = frac{(2.3 times 10^4) cdot (0.0082)}{(0.15)}
$$

$$
boxed{F_2 = 1300 text{N}}
$$

Result

6 of 6

**a.** $P = 1.5 times 10^5 text{Pa}$
**b.** $F_2 = 1300 text{N}$

Step 1

1 of 2

$P = dfrac{F}{A} = dfrac{2.3 times 10^4}{0.15} = 1.5 times 10^5 Pa$

The pressure is the same:

$P_1 = P_2$

$dfrac{F_1}{A_1} = dfrac{F_2}{A_2}$

$$
F_1 = dfrac{F_2 A_1}{A_2} = dfrac{(2.3e4)*(0.0082)}{0.15} = 1300 N
$$

Result

2 of 2

a) $1.5 times 10^5 Pa$

b) $1300 N$

Exercise 37

Step 1

1 of 2

The density of the block of lead is greater than aluminum, therefore the volume of the aluminum is greater and displaces more water.

The volume of the blocks is the same and therefore they displace equal amount of water when they are placed in water.

Result

2 of 2

a) aluminum

b) same

Exercise 38

Step 1

1 of 3

Tornado increases the velocity of the surrounding air. According to Bernoulli’s principle, as the velocity of the air increases, its pressure decreases. This confirms the fact stated in Problem 4 that tornado causes the pressure outside to drop by 15% below the atmospheric pressure.

Step 2

2 of 3

Since the pressure outside the house drops and the pressure inside the house stays equal to the atmospheric pressure, there will be a pressure difference between them. This will exert a force exerted outward the house that can cause the doors and windows to explode outward.

Step 3

3 of 3

To reduce the danger of a door or window from exploding outward, the following things can be done:
1. Reinforce the doors and the windows so that it can withstand the force exerted on them
2. From the equation $F = PA$, we can reduce the force by making the surface area of the doors and windows smaller.

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Page 358: Section Review

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