518. К куску парафиновой свечки массой 4,9 г привязали металлическую шайбу, которая весит в воде 98 мН. Общий вес полностью погруженной в воду системы 78,4 мН. Найдите плотность парафина.

Решение:

Для решения этой задачи нам понадобятся следующие данные:

• Масса парафина: \( m_п = 4.9 \) г \( = 0.0049 \) кг.
• Вес шайбы в воде: \( P_{ш.в} = 98 \) мН \( = 0.098 \) Н.
• Общий вес системы (парафин + шайба) в воде: \( P_{сист.в} = 78.4 \) мН \( = 0.0784 \) Н.
• Ускорение свободного падения: \( g = 9.8 \) м/с².
• Плотность воды: \( \rho_в = 1000 \) кг/м³.

1. Найдем вес шайбы в воздухе:

Вес системы в воде равен разнице веса шайбы в воздухе и выталкивающей силы, действующей на шайбу (так как парафин легче воды и всплывает, а шайба тонет, мы можем считать, что парафин держится на шайбе, и они вместе полностью погружены).

\( P_{сист.в} = P_{ш.в} - F_{а} \)

где \( F_а \) — выталкивающая сила, действующая на шайбу.

2. Найдем выталкивающую силу, действующую на шайбу:

Разница между весом шайбы в воде и общим весом системы в воде — это как раз вес парафина, который выталкивается водой. Однако, здесь парафин сам по себе легче воды. Проблема в том, что не указан вес шайбы в воздухе. Предположим, что 98 мН — это вес шайбы в воздухе, а 78.4 мН — это вес парафина + шайбы в воде.

\( P_{сист.в} = P_{ш} - F_{а} \)

3. Найдем вес шайбы в воздухе (альтернативный подход, если 98 мН - вес шайбы в воздухе):

Let \( P_{ш} \) be the weight of the metal washer in air.

The weight of the system in water is \( P_{system, water} = P_{paraffin} + P_{washer, water} \).

The buoyant force on the washer is \( F_{A} = P_{washer} - P_{washer, water} \).

The weight of the paraffin in water is \( P_{paraffin, water} = P_{paraffin} - F_{A, paraffin} \).

We are given that the weight of the washer in water is 98 mN. This means \( P_{washer, water} = 0.098 \) N. This implies \( P_{washer} - F_A = 0.098 \) N.

The total weight of the system fully submerged is 78.4 mN, which means \( P_{paraffin} + P_{washer, water} = 0.0784 \) N.

Let's assume that the 98 mN is the weight of the washer in air, so \( P_{washer} = 0.098 \) N.

Then the weight of the system in water is \( P_{system, water} = P_{paraffin} + P_{washer} - F_{A, washer} = 0.0784 \) N.

This means \( P_{paraffin} + 0.098 - F_{A, washer} = 0.0784 \) N.

Also, \( F_{A, washer} = \rho_{water} \cdot g \cdot V_{washer} \) and \( P_{washer} = m_{washer} \cdot g \). We don't know the mass or volume of the washer.

Let's re-interpret the problem:

1. Mass of paraffin: \( m_p = 4.9 \) g \( = 0.0049 \) kg.

2. Weight of metal washer in water: \( P_{w,w} = 98 \) mN \( = 0.098 \) N.

3. Total weight of the system (paraffin + washer) fully submerged in water: \( P_{sys,w} = 78.4 \) mN \( = 0.0784 \) N.

The weight of the washer in water is its weight in air minus the buoyant force on it:

\( P_{w,w} = P_w - F_{A,w} \) where \( P_w \) is the weight of the washer in air and \( F_{A,w} \) is the buoyant force on the washer.

The total weight of the system in water is the sum of the weight of the paraffin and the weight of the washer in water (since the paraffin is also submerged and contributes to the total apparent weight).

\( P_{sys,w} = P_p + P_{w,w} \)

However, this would mean \( P_{sys,w} > P_{w,w} \) if \( P_p > 0 \), which is not the case (78.4 mN < 98 mN).

This implies that the 98 mN is NOT the weight of the washer in water. Let's consider another interpretation:

Let \( m_p = 4.9 \) g be the mass of paraffin. Let \( W_{washer, air} \) be the weight of the washer in air.

The weight of the washer in water is \( W_{washer, water} = W_{washer, air} - F_{buoyant, washer} = 98 \) mN \( = 0.098 \) N. This implies \( W_{washer, air} > 0.098 \) N.

The total weight of the system (paraffin + washer) fully submerged in water is \( W_{system, water} = W_{paraffin} + W_{washer, water} \).

But paraffin is less dense than water. If paraffin is fully submerged, it will have an upward buoyant force.

Let's assume the problem means:

1. Mass of paraffin \( m_p = 4.9 \) g \( = 0.0049 \) kg.

2. Weight of the metal washer in air \( P_w = 98 \) mN \( = 0.098 \) N.

3. When the washer is submerged in water, and the paraffin is attached to it and also submerged, the total apparent weight is \( P_{sys,w} = 78.4 \) mN \( = 0.0784 \) N.

The total weight in air is \( P_{sys,air} = P_p + P_w = m_p \cdot g + P_w = 0.0049 \cdot 9.8 + 0.098 = 0.04802 + 0.098 = 0.14602 \) N.

The buoyant force on the system is \( F_{A,sys} = P_{sys,air} - P_{sys,w} = 0.14602 - 0.0784 = 0.06762 \) N.

This buoyant force is the sum of the buoyant force on the washer and the buoyant force on the paraffin: \( F_{A,sys} = F_{A,w} + F_{A,p} \).

\( F_{A,w} = \rho_w \cdot g \cdot V_w \)

\( F_{A,p} = \rho_w \cdot g \cdot V_p \)

We know \( V_p = \frac{m_p}{\rho_p} \), where \( \rho_p \) is the density of paraffin we want to find.

So \( F_{A,sys} = \rho_w \cdot g \cdot V_w + \rho_w \cdot g \cdot \frac{m_p}{\rho_p} \).

We still have \( V_w \) (volume of washer) and \( P_w = m_w \cdot g = 0.098 \) N (weight of washer in air) as unknowns.

Let's try a different interpretation that is more standard for Archimedes' principle problems.

Given:

1. Mass of paraffin \( m_p = 4.9 \) g \( = 0.0049 \) kg.

2. Weight of the metal washer in water \( P_{w,w} = 98 \) mN \( = 0.098 \) N.

3. Total weight of the system (paraffin + washer) fully submerged in water \( P_{sys,w} = 78.4 \) mN \( = 0.0784 \) N.

Let \( P_p \) be the weight of the paraffin in air, and \( P_w \) be the weight of the washer in air.

The weight of the washer in water is \( P_{w,w} = P_w - F_{A,w} = 0.098 \) N, where \( F_{A,w} \) is the buoyant force on the washer.

The total weight of the system in water is the sum of the weight of the paraffin in air and the weight of the washer in water (since the paraffin is also submerged and its weight in water is approximately its weight in air if its density is close to water, or its apparent weight if it's less dense than water). This interpretation is also confusing.

Let's assume the most common phrasing for such problems:

1. Mass of paraffin \( m_p = 4.9 \) g \( = 0.0049 \) kg.

2. Weight of the metal washer in air \( P_w = 98 \) mN \( = 0.098 \) N.

3. When the metal washer is fully submerged in water, its apparent weight is \( P_{w,w} = 78.4 \) mN \( = 0.0784 \) N. (This would mean the washer's weight is actually less than the system's weight in the original problem statement, which doesn't make sense).

Let's go back to the original statement:

518. К куску парафиновой свечки массой 4,9 г привязали металлическую шайбу, которая весит в воде 98 мН. Общий вес полностью погруженной в воду системы 78,4 мН. Найдите плотность парафина.

Given:

1. Mass of paraffin \( m_p = 4.9 \) g \( = 0.0049 \) kg.

2. Weight of the metal washer in water \( P_{w,w} = 98 \) mN \( = 0.098 \) N.

3. Total weight of the system (paraffin + washer) fully submerged in water \( P_{sys,w} = 78.4 \) mN \( = 0.0784 \) N.

This implies that the buoyant force on the *combined system* is such that it reduces the total weight.

Let \( P_p \) be the weight of the paraffin in air.

The total weight in air of the system would be \( P_{sys,air} = P_p + P_w \), where \( P_w \) is the weight of the washer in air.

We are given \( P_{w,w} = P_w - F_{A,w} = 0.098 \) N.

And \( P_{sys,w} = (P_p + P_w) - (F_{A,p} + F_{A,w}) = 0.0784 \) N.

Substituting \( P_w - F_{A,w} = 0.098 \) into the second equation:

\( P_p - F_{A,p} + (P_w - F_{A,w}) = 0.0784 \)

\( P_p - F_{A,p} + 0.098 = 0.0784 \)

\( P_p - F_{A,p} = 0.0784 - 0.098 = -0.0196 \) N.

Since \( P_p = m_p \cdot g = 0.0049 \cdot 9.8 = 0.04802 \) N, this equation becomes:

\( 0.04802 - F_{A,p} = -0.0196 \)

\( F_{A,p} = 0.04802 + 0.0196 = 0.06762 \) N.

This is the buoyant force acting on the paraffin.

The buoyant force is given by \( F_{A,p} = \rho_{water} \cdot g \cdot V_p \).

We know \( V_p = \frac{m_p}{\rho_p} \).

So, \( F_{A,p} = \rho_{water} \cdot g \cdot \frac{m_p}{\rho_p} \).

Rearranging to find \( \rho_p \):

\( \rho_p = \frac{\rho_{water} \cdot g \cdot m_p}{F_{A,p}} \).

Plugging in the values:

\( \rho_p = \frac{1000 \text{ kg/m}^3 \cdot 9.8 \text{ m/s}^2 \cdot 0.0049 \text{ kg}}{0.06762 \text{ N}} \)

\( \rho_p = \frac{48.02}{0.06762} \text{ kg/m}^3 \)

\( \rho_p \approx 710.14 \text{ kg/m}^3 \).

Double check:

If \( \rho_p = 710.14 \) kg/m³, then \( V_p = \frac{0.0049}{710.14} \approx 6.90 \cdot 10^{-6} \) m³.

\( F_{A,p} = 1000 \cdot 9.8 \cdot 6.90 \cdot 10^{-6} = 0.06762 \) N. This matches.

Now, what about the washer's weight in water being 98 mN?

\( P_{w,w} = P_w - F_{A,w} = 0.098 \) N.

And \( P_{sys,w} = P_p - F_{A,p} + P_{w,w} = 0.04802 - 0.06762 + 0.098 = -0.0196 + 0.098 = 0.0784 \) N. This matches.

Therefore, the calculation is consistent.

Final Answer Calculation:

Weight of paraffin in air: \( P_p = m_p \cdot g = 0.0049 \text{ kg} \cdot 9.8 \text{ m/s}^2 = 0.04802 \text{ N} \).

Weight of system in water: \( P_{sys,w} = 78.4 \text{ mN} = 0.0784 \text{ N} \).

Weight of washer in water: \( P_{w,w} = 98 \text{ mN} = 0.098 \text{ N} \).

The total weight of the system in air is \( P_{sys,air} = P_p + P_w \).

The total apparent weight of the system in water is \( P_{sys,w} = (P_p - F_{A,p}) + (P_w - F_{A,w}) \).

We have \( P_{w,w} = P_w - F_{A,w} = 0.098 \) N.

So, \( P_{sys,w} = (P_p - F_{A,p}) + P_{w,w} \).

\( 0.0784 = (P_p - F_{A,p}) + 0.098 \).

\( P_p - F_{A,p} = 0.0784 - 0.098 = -0.0196 \) N.

Since \( P_p = 0.04802 \) N,

\( 0.04802 - F_{A,p} = -0.0196 \).

\( F_{A,p} = 0.04802 + 0.0196 = 0.06762 \) N.

The buoyant force on paraffin is \( F_{A,p} = \rho_{water} \cdot g \cdot V_p \).

The volume of paraffin is \( V_p = \frac{m_p}{\rho_p} \).

\( F_{A,p} = \rho_{water} \cdot g \cdot \frac{m_p}{\rho_p} \).

\( 0.06762 = 1000 \cdot 9.8 \cdot \frac{0.0049}{\rho_p} \).

\( 0.06762 = \frac{48.02}{\rho_p} \).

\( \rho_p = \frac{48.02}{0.06762} \approx 710.14 \text{ kg/m}^3 \).

Ответ: Плотность парафина примерно 710 кг/м³.