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diamong [38]
3 years ago
12

A gas occupies 2 m^3 at 27°C at a pressure of 1 atmosphere. At a pressure of 2 atmospheres it occupies a volume of 1 m^2. What i

s its temperature at this new volume and pressure?
Physics
1 answer:
diamong [38]3 years ago
6 0

Answer:

27°C

Explanation:

We'll begin by converting 27 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 27 °C

Initial temperature (T₁) = 27 °C + 273

Initial temperature (T₁) = 300 K

Next, we shall determine the final temperature of the gas. This can be obtained as follow:

Initial volume (V₁) = 2 m³

Initial temperature (T₁) = 300 K

Initial pressure (P₁) = 1 atm

Final pressure (P₂) = 2 atm

Final volume (V₂) = 1 m³

Final temperature (T₂) =?

P₁V₁/T₁ = P₂V₂/T₂

1 × 2 / 300 = 2 × 1 / T₂

2/300 = 2/T₂

1/150 = 2/T₂

Cross multiply

T₂ = 150 × 2

T₂ = 300 K

Finally, we shall convert 300 K to celsius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

T(K) = 300 K

T(°C) = 300 – 273

T(°C) = 27°C

Thus, the final temperature is 27°C

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They would most likely use statistics on the increase/decrease in population from previous decades and centuries. In addition to these statistics, they could use the birth and death rate and ultimately predict the increase within these next 30 years. They must also take climate change and other environmental factors into consideration when formulating such a bold prediction.
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4 years ago
Helppp<br> In Static Electricity, The charges do not ____
Anna11 [10]

Answer:

basically I will tell you the definition

Explanation:

so when charges are unbalanced statistic energy is formed positive attract negative and negative attracts positive like repell while unlike attract .

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3 years ago
In a compound microscope, the objective has a focal length of 1.0 cm, the eyepiece has a focal length of 2.0 cm, and the tube le
expeople1 [14]

Answer:

The  value  is   m \approx   310

Explanation:

From the question we are told that

     The  focal length of the objective is  f_o =  1.0 \ cm

    The  focal length of the eyepiece is  f_e  =  2.0 \  cm

    The  tube length is  L  =  25 \  cm

Generally the magnitude of the overall magnification is mathematically represented as

            m =  m_o  *  m_e

Where  m_o is the objective magnification which is mathematically represented as

        m_o  =  \frac{L}{f_o }

=>      m_o  =  \frac{25}{1 }

=>      m_o  =  25

m_e is the eyepiece magnification which is mathematically evaluated as

     m_e  =  \frac{L }{f_e }

     m_e  =  \frac{25 }{ 2}

      m_e  =  12.5 \  cm

So

    m =  25 * 12.5

     m \approx   310

6 0
3 years ago
The mass of the particles that a river can transport is proportional to the sixth power of the speed of the river. A certain riv
kow [346]

Answer:

1.122 m/s

Explanation:

So usually a river with a speed of 1 meters per second can transport particle that weighs:

1^6 = 1 kg

If the particle is twice as massive as usual, then its weights would be 1 * 2 = 2kg

This means the river must be flowing at a speed of

2^{\frac{1}{6}} = 1.122 m/s

5 0
3 years ago
At what position or positions on the x-axis is the electric field zero?
ElenaW [278]

Answer:

The electric field will be zero at x = ± ∞.

Explanation:

Suppose, A -2.0 nC charge and a +2.0 nC charge are located on the x-axis at x = -1.0 cm and x = +1.0 cm respectively.

We know that,

The electric field is

E=\dfrac{kq}{r^2}

The electric field vector due to charge one

\vec{E_{1}}=\dfrac{kq_{1}}{r_{1}^2}(\hat{x})

The electric field vector due to charge second

\vec{E_{2}}=\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})

We need to calculate the electric field

Using formula of net electric field

\vec{E}=\vec{E_{1}}+\vec{E_{2}}

\vec{E_{1}}+\vec{E_{2}}=0

Put the value into the formula

\dfrac{kq_{1}}{r_{1}^2}(\hat{x})+\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})=0

\dfrac{kq_{1}}{r_{1}^2}(\hat{x})=\dfrac{kq_{2}}{r_{2}^2}(\hat{x})

(\dfrac{r_{2}}{r_{1}})^2=\dfrac{q_{2}}{q_{1}}

\dfrac{r_{2}}{r_{1}}=\sqrt{\dfrac{q_{2}}{q_{1}}}

Put the value into the formula

\dfrac{2.0+x}{x}=\pm\sqrt{\dfrac{2.0}{2.0}}

2.0+x=x

If x = ∞, then the equation is be satisfied.

Hence, The electric field will be zero at x = ± ∞.

4 0
3 years ago
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