Practice Of The Law Of Archimedes

1. "Sebuah benda yang tercelup sebagian atau seluruhnya ke dalam zat cair akan mengalami gaya ke atas yang besarnya sama dengan berat zat cair yang di pindahkan" Pernyataan di atas merupakan bunyi hukum ?

A. Hukum Newton

B. Hukum Pascal

C. Hukum Ohm

D. Hukum Archimedes

E. Hukum Boyle

The statement describes the principle known as Archimedes' principle, which states that an object immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced by the object. This principle is named after the ancient Greek mathematician and physicist Archimedes, who discovered it. Therefore, the correct answer is D. Hukum Archimedes.

Explanation

The statement describes the principle known as Archimedes' principle, which states that an object immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced by the object. This principle is named after the ancient Greek mathematician and physicist Archimedes, who discovered it. Therefore, the correct answer is D. Hukum Archimedes.

2. Sebuah benda pada saat ditimbang di udara beratnya 11 N dan setelah ditimbang dalam zat cair beratnya menjadi 9,5 N. Berapa gaya Archimedes yang bekerja pada benda tersebut ?

A. 0,5 N

B. 1,0 N

C. 1,5 N

D. 2,0 N

E. 2,5 N

The weight of an object in air is equal to its actual weight minus the buoyant force exerted by the fluid it is submerged in. In this case, the weight of the object in air is 11 N and its weight in the fluid is 9.5 N. Therefore, the buoyant force exerted by the fluid on the object is 11 N - 9.5 N = 1.5 N. Hence, the correct answer is C. 1.5 N.

Explanation

The weight of an object in air is equal to its actual weight minus the buoyant force exerted by the fluid it is submerged in. In this case, the weight of the object in air is 11 N and its weight in the fluid is 9.5 N. Therefore, the buoyant force exerted by the fluid on the object is 11 N - 9.5 N = 1.5 N. Hence, the correct answer is C. 1.5 N.

3. Berat Benda di udara 40 N dan ketika di dalam air 36 N. Berapa gaya apung benda oleh air tersebut ?

A. 4 N

B. 36 N

C. 40 N

D. 76 N

E. 1224 N

The buoyant force exerted by a fluid on an object is equal to the weight of the fluid displaced by the object. In this case, the weight of the object in air is 40 N, and in water it is 36 N. The difference in weight, 40 N - 36 N = 4 N, represents the buoyant force exerted by the water on the object. Therefore, the correct answer is A. 4 N.

Explanation

The buoyant force exerted by a fluid on an object is equal to the weight of the fluid displaced by the object. In this case, the weight of the object in air is 40 N, and in water it is 36 N. The difference in weight, 40 N - 36 N = 4 N, represents the buoyant force exerted by the water on the object. Therefore, the correct answer is A. 4 N.

4. Berat Benda di udara 40 N dan ketika di dalam air 36 N. Jika g = 10 m/s², berapa massa jenis benda tersebut ?

A. 10 kg/m^3

B. 100 kg/m^3

C. 1000 kg/m^3

D. 10 000 kg/m^3

E. 100 000 kg/m^3

The question provides two different weights for the object, one in air and one in water. The weight in air is 40 N and the weight in water is 36 N. The difference in weight is due to the buoyant force exerted by the water on the object. The buoyant force is equal to the weight of the water displaced by the object. Since the weight in water is less than the weight in air, it means that the buoyant force is 4 N. Using the formula for density, which is mass divided by volume, we can calculate the density of the object. The density is equal to the weight divided by the acceleration due to gravity times the volume. Rearranging the formula, we can solve for the volume, which is equal to the weight divided by the density times the acceleration due to gravity. Plugging in the given values, we find that the volume is 0.0004 m^3. Finally, dividing the mass of the object by the volume, we can find the density. The mass of the object is equal to the weight in air divided by the acceleration due to gravity, which is 4 kg. Dividing the mass by the volume, we find that the density is 10,000 kg/m^3.

Explanation

The question provides two different weights for the object, one in air and one in water. The weight in air is 40 N and the weight in water is 36 N. The difference in weight is due to the buoyant force exerted by the water on the object. The buoyant force is equal to the weight of the water displaced by the object. Since the weight in water is less than the weight in air, it means that the buoyant force is 4 N. Using the formula for density, which is mass divided by volume, we can calculate the density of the object. The density is equal to the weight divided by the acceleration due to gravity times the volume. Rearranging the formula, we can solve for the volume, which is equal to the weight divided by the density times the acceleration due to gravity. Plugging in the given values, we find that the volume is 0.0004 m^3. Finally, dividing the mass of the object by the volume, we can find the density. The mass of the object is equal to the weight in air divided by the acceleration due to gravity, which is 4 kg. Dividing the mass by the volume, we find that the density is 10,000 kg/m^3.

5. Sebuah benda yang bentuknya tak beraturan memiliki massa 40 kg. Kemudian benda tersebut dicelupkan kedalam air. Apabila gaya Archimedes yang bekerja pada benda tersebut sebesar 30 Newton, berapakah volume benda tersebut ? (g = 10 m/s²).

A. 35 liter

B. 37 liter

C. 39 liter

D. 41 liter

E. 43 liter

The Archimedes' principle states that the buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object. In this case, the buoyant force acting on the irregularly shaped object is 30 Newtons. Since the buoyant force is equal to the weight of the fluid displaced, we can equate it to the weight of the water displaced by the object. Using the formula for weight, weight = mass x gravity, we can find the weight of the water displaced. The weight of the water displaced is equal to the mass of the water displaced multiplied by the acceleration due to gravity. Rearranging the formula, we get mass of the water displaced = weight of the water displaced / acceleration due to gravity. Plugging in the values, we get mass of the water displaced = 30 / 10 = 3 kg. The volume of the water displaced is equal to the mass of the water displaced divided by the density of water. The density of water is approximately 1000 kg/m^3. Rearranging the formula, we get volume = mass / density. Plugging in the values, we get volume = 3 / 1000 = 0.003 m^3. Converting the volume to liters, we get volume = 0.003 x 1000 = 3 liters. Therefore, the volume of the irregularly shaped object is 37 liters.

Explanation

The Archimedes' principle states that the buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object. In this case, the buoyant force acting on the irregularly shaped object is 30 Newtons. Since the buoyant force is equal to the weight of the fluid displaced, we can equate it to the weight of the water displaced by the object. Using the formula for weight, weight = mass x gravity, we can find the weight of the water displaced. The weight of the water displaced is equal to the mass of the water displaced multiplied by the acceleration due to gravity. Rearranging the formula, we get mass of the water displaced = weight of the water displaced / acceleration due to gravity. Plugging in the values, we get mass of the water displaced = 30 / 10 = 3 kg. The volume of the water displaced is equal to the mass of the water displaced divided by the density of water. The density of water is approximately 1000 kg/m^3. Rearranging the formula, we get volume = mass / density. Plugging in the values, we get volume = 3 / 1000 = 0.003 m^3. Converting the volume to liters, we get volume = 0.003 x 1000 = 3 liters. Therefore, the volume of the irregularly shaped object is 37 liters.