/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 79 You are hiking through a lush fo... [FREE SOLUTION] | 91影视

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You are hiking through a lush forest with some of your friends when you come to a large river that seems impossible to cross. However, one of your friends notices an old metal barrel sitting on the shore. The barrel is shaped like a cylinder and is \(1.20 \mathrm{m}\) high and \(0.76 \mathrm{m}\) in diameter. One of the circular ends of the barrel is open and the barrel is empty. When you put the barrel in the water with the open end facing up, you find that the barrel floats with \(33 \%\) of it under water. You decide that you can use the barrel as a boat to cross the river, as long as you leave about $30 \mathrm{cm}$ sticking above the water. How much extra mass can you put in this barrel to use it as a boat?

Short Answer

Expert verified
The barrel has a diameter of 0.76 m and a height of 1.2 m, and it is currently floating with 33% of it underwater. Answer: Approximately 229.71 kg of extra mass can be added to the barrel and still have 30 cm sticking above the water level.

Step by step solution

01

Find the submerged volume of the barrel

First, we need to find the total volume of the barrel, as well as the current submerged volume which is 33% of its total volume. To find the total volume of a cylinder, use the formula \( V = 蟺r^2h \), where \(V\) is the volume, \(r\) is the radius, and \(h\) is the height. Given the barrel's diameter is 0.76 m, the radius is 0.38 m. The height of the barrel is 1.20 m. So, the total volume of the barrel is: $$ V = 蟺(0.38)^2(1.20) $$ $$ V \approx 0.5467 \mathrm{m}^3 $$ Now calculate the submerged volume when 33% of the barrel is underwater: $$ V_{submerged} = 0.33 * V $$ $$ V_{submerged} \approx 0.1804 \mathrm{m}^3 $$
02

Calculate the current buoyant force acting on the barrel

To calculate the buoyant force, the following equation should be used, where \(B\) is the buoyant force, \(蟻_{water}\) is the water density (1000 kg/m鲁), \(V_{submerged}\) is the submerged volume, and \(g\) is the acceleration due to gravity (9.81 m/s虏): $$ B = 蟻_{water} * V_{submerged} * g $$ Using the submerged volume calculated in Step 1, find the buoyant force: $$ B = 1000 * 0.1804 * 9.81 $$ $$ B \approx 1768.84 \mathrm{N} $$
03

Calculate the initial mass of the barrel without any external weight

We can calculate the initial mass (\(m_{initial}\)) of the barrel using the buoyant force and acceleration due to gravity: $$ B = m_{initial} * g $$ $$ m_{initial} = \frac{B}{g} $$ Calculate the initial mass of the barrel: $$ m_{initial} = \frac{1768.84}{9.81} $$ $$ m_{initial} \approx 180.21 \mathrm{kg} $$
04

Find the submerged volume when there's 30 cm above the water level

Now, we are given that we need to leave 30 cm (0.30 m) above the water level. Therefore, the new submerged height is: $$ h_{new} = 1.20 - 0.30 = 0.90 \mathrm{m} $$ The new submerged volume would be: $$ V_{new} = 蟺(0.38)^2(0.90) $$ $$ V_{new} \approx 0.4110 \mathrm{m}^3 $$
05

Calculate the new buoyant force and mass

Now find the new buoyant force using the new submerged volume: $$ B_{new} = 蟻_{water} * V_{new} * g $$ $$ B_{new} = 1000 * 0.4110 * 9.81 $$ $$ B_{new} \approx 4020.71 \mathrm{N} $$ Calculate the new total mass (\(m_{new}\)) with the new buoyant force: $$ m_{new} = \frac{B_{new}}{g} $$ $$ m_{new} = \frac{4020.71}{9.81} $$ $$ m_{new} \approx 409.92 \mathrm{kg} $$
06

Find the extra mass that can be added to the barrel

Finally, subtract the initial mass of the barrel from the new total mass to find the additional mass that can be added: $$ m_{extra} = m_{new} - m_{initial} $$ $$ m_{extra} = 409.92 - 180.21 $$ $$ m_{extra} \approx 229.71 \mathrm{kg} $$ So, you can add approximately 229.71 kg of extra mass to the barrel and still have 30 cm sticking above the water level.

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