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Dmitry [639]
3 years ago
6

A body when dropped into a jar ,containing kerosene and glycerine ,sinks below the kerosene level to float in glycerine as infig

.What can we say about the densities of kerosene ,glycerine and the body
Physics
1 answer:
SVETLANKA909090 [29]3 years ago
8 0

Answer:

<em>we can say that that in an increasing order, the body is less dense than the glycerine, and the glycerine is less dense than the kerosene.</em>

<em></em>

Explanation:

If the body sinks below the kerosene level to float in glycerine, then it means that the kerosene is denser than glycerine. This is because bodies will float higher in a denser fluid that a less denser fluid. Also, since the body floats in kerosene and glycerine, then the body is less dense than the kerosene and the glycerine. Finally, <em>we can say that that in an increasing order that the body is less dense than the glycerine, and the glycerine is less dense than the kerosene</em>

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80 grams of iron at 100°C is dropped into 200 of water at 20°C contained in an iron vessel of mass 50 gram find the resulting te
vodka [1.7K]

Answer:

the resulting temperature is 23.37 ⁰C

Explanation:

Given;

mass of the iron, m₁ = 80 g = 0.08 kg

mass of the water, m₂ = 200 g = 0.2 kg

mass of the iron vessel, m₃ = 50 g = 0.05 kg

initial temperature of the iron, t₁ = 100 ⁰C

initial temperature of the water, t₂ = 20 ⁰C

specific heat capacity of iron, c₁ = 462 J/kg⁰C

specific heat capacity of water, c₂ = 4,200 J/kg⁰C

let the temperature of the resulting mixture = T

Apply the principle of conservation of energy;

heat lost by the hot iron = heat gained by the water

m_1c_1 \Delta t_1 = m_2c_2\Delta t_2\\\\m_1c_1 (100 - T) = m_2c_2 (T- 20)\\\\0.08 \times 462 (100-T) = 0.2 \times 4,200 (T-20)\\\\36.96 (100-T) = 840 (T-20) \\\\100 - T = 22.72 (T-20)\\\\100-T = 22.72 T - 454.4 \\\\554.4 = 23.72T\\\\T = \frac{554.4}{23.72} \\\\T = 23.37 \ ^0C

Therefore, the resulting temperature is 23.37 ⁰C

6 0
3 years ago
A rock is thrown vertically upward from some height above the ground. It rises to some maximum height and falls back to the grou
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Answer:

At the highest point the velocity is zero, the acceleration is directed downward.

Explanation:

This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.

I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.

At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.

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The lower the frequency the _____ the pitch sound.
stepan [7]
The lower the frequency the lower the pitch sound.
8 0
3 years ago
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A(n) 10.1 g bullet is fired into a(n) 2.41 kg ballistic pendulum and becomes embedded in it. The acceleration of gravity is 9.8
fomenos

Answer:

v = 186.90\,\frac{m}{s}

Explanation:

The motion of ballistic pendulum is modelled by the appropriate use of the Principle of Energy Conservation:

\frac{1}{2}\cdot (m_{p}+m_{b})\cdot v^{2} = (m_{p}+m_{b})\cdot g \cdot h

The final velocity of the system formed by the ballistic pendulum and the bullet is:

v = \sqrt{2\cdot g\cdot h}

v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.031\,m)}

v\approx 0.78\,\frac{m}{s}

Initial velocity of the bullet can be calculated from the expression derived of the Principle of Momentum:

(0.0101\,kg)\cdot v = (2.41\,kg + 0.0101\,kg)\cdot (0.78\,\frac{m}{s} )

v = 186.90\,\frac{m}{s}

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