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lukranit [14]
2 years ago
6

A metal object is suspended from a spring scale. The scale reads 920 N when the object is suspended in air, and 750N when the ob

ject is completely submerged in water. Find the volume of the object
Physics
1 answer:
Ostrovityanka [42]2 years ago
3 0

17.3 L.

<h3>Explanation</h3>

The object appears 920 - 750 = 170\;\text{N} lighter in water than in the air. Water has supplied that 170 N of buoyant force.

The size of the buoyant force on an object in water is the same as the weight of water that the object has displaced. The buoyant force on the metal object here is 170 N. The object must have displaced water of the same weight.

g = 9.81 \;\text{N}\cdot\text{kg}^{-1}.

Mass of water displaced:

\text{Mass} = \dfrac{\text{Weight}}{g} = \dfrac{170}{9.81} = 17.3 \;\text{kg}.

Volume of water displaced:

The density of water at room temperature is 1.000\;\text{kg}\cdot\text{dm}^{-3}. Each kilogram of water will occupy a volume of 1 dm³ (one cubic decimeter), which is the same as 1 L (one liter).

V = \dfrac{\text{Mass}}{\text{Density}} =\dfrac{m}{\rho} = \dfrac{17.3\;\text{N}}{1.000\;\text{kg}\cdot\text{dm}^{-3}} = 17.3\;\text{dm}^{-3}=17.3\;\text{L}.

Volume of the object:

The object is completely under water. As a result, the volume of the object will be the same as the volume of water displaced. The volume of the object is also 17.3 L.

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Block 1, with mass m1 and speed 3.6 m/s, slides along an x axis on a frictionless floor and then undergoes a one-dimensional ela
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Answer:

a) The block 1 slides 0.24 m into the rough region.

b) The block 2 slides 2.7 m

Explanation:

Hi there!

First, let´s find the final velocity of each block. With that velocities, we can calculate the kinetic energy of each block. The kinetic energy of the blocks will be equal to the work done by friction to stop them. From the equation of work, we can calculate the distance traveled by the blocks.

Since the collision is elastic, the momentum and kinetic energy of the system composed of the two blocks is constant.

The momentum of the system is calculated as the sum of the momenta of each block:

m1 · v1 + m2 · v2 = m1 · v1´ + m2 · v2´

Where:

m1 and m2 = mass of blocks 1 and 2 respectively.

v1 and v2 = velocity of blocks 1 and 2 respectively.

v1´ and v2´ = final velocity of blocks 1 and 2 respectively.

Using the data we have, we can solve the eqaution for v1´:

m1 · 3.6 m/s + 0.40 m1 · 0 = m1 · v1´ + 0.40 m1 · v2´

3.6 m/s · m1 = m1 · v1´ + 0.40 m1 · v2´

3.6 m/s = v1´ + 0.40 v2´

v1´ = 3.6 m/s - 0.40 v2´

The kinetic energy of the system also remains constant:

1/2 m1 · (v1)² + 1/2 m2 · (v2)² = 1/2 m1 · (v1´)² + 1/2 m2 · (v2´)²

Multiply by 2 both sides of the equation:

m1 · (v1)² + m2 · (v2)² = m1 · (v1´)² + m2 · (v2´)²

Let´s replace with the data:

m1 · (3.6 m/s)² + 0.40 m1 · 0 = m1 · (v1´)² + 0.40 m1 (v2´)²

divide by m1:

(3.6 m/s)² = (v1´)² + 0.40 (v2´)²

Replace v1´ = 3.6 m/s - 0.40 v2´

(3.6 m/s)² = (3.6 m/s - 0.40 v2´)² + 0.40 (v2´)²

Let´s solve for v2´:

(3.6 m/s)² = (3.6 m/s)² - 2.88 v2´ + 0.16 (v2´)² + 0.40 (v2´)²

0 = 0.56 (v2´)² - 2.88 v2´

0 = v2´(0.56 v2´ - 2.88)   v2´ = 0 (the initial velocity)

0 = 0.56 v2´ - 2.88

2.88/0.56 = v2´

v2´ = 5.1 m/s

Now let´s calculate v1´:

v1´ = 3.6 m/s - 0.40 v2´

v1´ = 3.6 m/s - 0.40 (5.1 m/s)

v1´ = 1.56 m/s

Now, let´s calculate the final kinetic energy (KE) of each block:

a) Block 1:

KE = 1/2 · m1 · (1.56 m/s)² = m1 · 1.2 m²/s²

The work done by friction is calculated as follows:

W = Fr · s

Where:

Fr = friction force.

s = traveled distance.

The friction force is calculated as follows:

Fr = N · μ

Where:

N = normal force.

μ = coefficient of friction.

And the normal force is calculated in this case as:

N = m1 · g

Where g is the acceleration due to gravity.

Then, the work done by friction will be:

W = m1 · g · μ · s

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m1 · 1.2 m²/s² = m1 · g · μ · s

Solving for s:

s = m1 · 1.2 m²/s²  / m1 · g · μ

s = 1.2 m²/s²/ 9.8 m/s² · 0.50

s = 0.24 m

The block 1 slides 0.24 m into the rough region.

b) For block 2 the kinetic energy will be the following:

KE = 1/2 · 0.4 · m1 · (5.1 m/s)² = m1 · 5.2 m²/s²

The friction force will be:

Fr = 0.4 m1 · g · μ

And the work done will be:

W = 0.4 m1 · g · μ · s

Since W = ΔKE,

Then:

m1 · 5.2 m²/s² = 0.4 m1 · g · μ · s

Solving for s:

5.2 m²/s²/(0.4 · g · μ) = s

s =  5.2 m²/s²/(0.4 · 9.8 m/s² · 0.50)

s = 2.7 m

The block 2 slides 2.7 m

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