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yarga [219]
2 years ago
15

A fixed mass of gas of volume 600cm^3 at a temperature of 30°c is cooled to 0°c. At the same time, the volume is increased by 10

0cm^3. Determine the change in the pressure of the gas if the final pressure of gas is 5.0 × 10^4pa
​
Physics
1 answer:
Lapatulllka [165]2 years ago
7 0
Use Math-way it is very easy
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H e l p.. I have no idea what any of this is, what is the first step of the hydrogen fusion process
klasskru [66]
The answer to this question is the first choice. In the hydrogen fusion process, the first step involves the collision of two protons to emit an antielectron and a neutrino. Only the first choice of
                             1_1H + 1_1H →2_1H + e^+ + v + energy
reflects this process. A hydrogen ion with a mass of one and number protons of 1 means a proton.
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3 years ago
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Fill in the blanks to complete each statement about the heating of Earth's surface.
cricket20 [7]

Answer:

absorption and insolation.

Explanation:

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3 years ago
Carbon-14 has a half-life of 5,700 years. How long will it take for 6.25% of the Carbon-14 to be remaining?
IRISSAK [1]

Answer:

22,800 years

Explanation:

Half life equation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is time,

and T is the half life.

0.0625 = (½)^(t / 5700)

log 0.0625 = (t / 5700) log 0.5

4 = t / 5700

t = 22,800

It takes 22,800 years.

4 0
3 years ago
Electrons do not move unless they are attracted to an electromagnet <br><br> True or false
Tpy6a [65]

The answer is false , because they move without an electromagnetic

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A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangent
Julli [10]

Answer:

The kinetic energy of the merry-go-round is \bf{475.47~J}.

Explanation:

Given:

Weight of the merry-go-round, W_{g} = 826~N

Radius of the merry-go-round, r = 1.17~m

the force on the merry-go-round, F = 57.8~N

Acceleration due to gravity, g= 9.8~m.s^{-2}

Time given, t=3.47~s

Mass of the merry-go-round is given by

m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg

Moment of inertial of the merry-go-round is given by

I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}

Torque on the merry-go-round is given by

\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m

The angular acceleration is given by

\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}

The angular velocity is given by

\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}

The kinetic energy of the merry-go-round is given by

E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J

5 0
3 years ago
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