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sashaice [31]
1 year ago
7

9. A balloon is filled with 1000 cm3 of a gas weighing 1000 g. Will it rise or fall when it is released?

Physics
1 answer:
torisob [31]1 year ago
3 0

We can calculate the density of the balloon as follows:

\rho=\frac{mass}{volume}=\frac{1000g}{1000cm^3}=\frac{1g}{cm^3}

Therefore, the balloon will fall

Since the density of air is about 0.00123 g/cm^3 , the balloon is much more dense than the surrounding air. As a result, the balloon weighs more than the air that it displaces so the balloon will fall.

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Dimensional formula of Amplitude ?​
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[ M⁰L¹T⁰ ]

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T = Time

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The amplitude of wave = Maximum displacement

Since, the dimensional formula of displacement = [ M⁰L¹T⁰ ]

Therefore, the amplitude of a wave is dimensionally represented as [ M⁰L¹T⁰ ]

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A spring scale hung from the ceiling stretches by 6.1cm when a 2.0kg mass is hung from it. The 2.0kg mass is removed and replace
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Why is it important to have a balance of gases in the atmosphere?
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The greenhouse gases absorb some of this energy and radiate much of it back towards the surface whilst the rest is radiated out to space. This plays an important role in keeping the Earth's surface warm and able to sustain life.
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3 years ago
An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10
Alexeev081 [22]

Answer:

a) c=1822.3214\ J.kg^{-1}.K^{-1}

b) This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c) The material is peat, possibly.

d) The material cannot be ice because ice doesn't exists at a temperature of 100°C.

Explanation:

Given:

  • mass of aluminium, m_a=0.1\ kg
  • mass of water, m_w=0.25\ kg
  • initial temperature of the system, T_i=10^{\circ}C
  • mass of copper block, m_c=0.1\ kg
  • temperature of copper block, T_c=50^{\circ}C
  • mass of the other block, m=0.07\ kg
  • temperature of the other block, T=100^{\circ}C
  • final equilibrium temperature, T_f=20^{\circ}C

We have,

specific heat of aluminium, c_a=910\ J.kg^{-1}.K^{-1}

specific heat of copper, c_c=390\ J.kg^{-1}.K^{-1}

specific heat of water, c_w=4186\ J.kg^{-1}.K^{-1}

Using the heat energy conservation equation.

The heat absorbed by the system of the calorie-meter to reach the final temperature.

Q_{in}=m_a.c_a.(T_f-T_i)+m_w.c_w.(T_f-T_i)

Q_{in}=0.1\times 910\times (20-10)+0.25\times 4186\times (20-10)

Q_{in}=11375\ J

The heat released by the blocks when dipped into water:

Q_{out}=m_c.c_c.(T_c-T_f)+m.c.(T-T_f)

where

c= specific heat of the unknown material

For the conservation of energy : Q_{in}=Q_{out}

so,

11375=0.1\times 390\times (50-20)+0.07\times c\times (100-20)

c=1822.3214\ J.kg^{-1}.K^{-1}

b)

This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c)

The material is peat, possibly.

d)

The material cannot be ice because ice doesn't exists at a temperature of 100°C.

7 0
3 years ago
A force of 20 N is executed to raise a rock weighing 30 N. What is the actual mechanical advantage?
xxMikexx [17]
I am absolutely sure its 1.5
5 0
3 years ago
Read 2 more answers
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