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adoni [48]
3 years ago
13

1. When does raising the temperature of a gas increase its pressure? when volume is increased and the number of particles is con

stant when volume is increased and the number of particles is increased when volume and the number of particles are constant
Physics
1 answer:
Neporo4naja [7]3 years ago
3 0

Answer:

when volume and the number of particles are constant

Explanation:

Gay Lussac law states that when the volume of an ideal gas is kept constant, the pressure of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Gay Lussac's law is given by;

PT = K

\frac{P1}{T1} = \frac{P_{2}}{T_{2}}

The ideal gas law is the equation PV = nRT

Where;

P is the pressure.

V is the volume.

n is the number of moles of substance.

R is the ideal gas constant.

T is the temperature.

Generally, raising the temperature of an ideal gas would increase its pressure when volume and the number of particles are constant.

This ultimately implies that, when volume and the number of particles are held constant, there would be a linear relationship between the temperature and pressure of a gas i.e temperature would be directly proportional to the pressure of the gas. Thus, an increase in the temperature of the gas would cause an increase in the pressure of the gas at constant volume and number of particles.

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The strength of the electric field at a certain distance from a point charge is represented by E. What is the strength of the el
LUCKY_DIMON [66]

Answer:

e.)At twice the distance, the strength of the field is E/4.

Explanation:

The strength of the electric field at a certain distance from a point charge is given by:

E=k\frac{Q}{r^2}

where

k is the Coulomb's constant

Q is the charge

r is the distance from the point charge

In this problem, the distance from the point charge is doubled:

r' = 2r

So the new electric field strength is

E'=k\frac{Q}{(2r)^2}=k \frac{Q}{4 r^2}=\frac{1}{4} (k\frac{Q}{r^2})=\frac{E}{4}

so, at twice the distance the strength of the field is E/4.

4 0
3 years ago
series RC circuit is built with a 15 kΩ resistor and a parallel-plate capacitor with 18-cm-diameter electrodes. A 18 V, 36 kHz s
andre [41]

Answer:

d=1.84\ mm

Explanation:

<u>Capacitance</u>

A two parallel-plate capacitor has a capacitance of

\displaystyle C=\frac{\epsilon_o A}{d}

where

\epsilon_o=8.85\cdot 10^{-12}\ F/m

A = area of the plates = \pi r^2

d = separation of the plates

\displaystyle d=\frac{\epsilon_o A}{C}=\frac{\epsilon_o \pi r^2}{C}

We need to compute C. We'll use the circuit parameters for that. The reactance of a capacitor is given by

\displaystyle X_c=\frac{1}{wC}

where w is the angular frequency

w=2\pi f=2\pi \cdot 36000=226194.67\ rad/s

Solving for C

\displaystyle C=\frac{1}{wX_c}

The reactance can be found knowing the total impedance of the circuit:

Z^2=R^2+X_c^2

Where R is the resistance, R=15 K\Omega=15000\Omega. Solving for Xc

X_c^2=Z^2-R^2

The magnitude of the impedance is computed as the ratio of the rms voltage and rms current

\displaystyle Z=\frac{V}{I}

The rms current is the peak current Ip divided by \sqrt{2}, thus

\displaystyle Z=\frac{\sqrt{2}V}{I_p}

I_p=0.65\ mA/1000=0.00065\ A

Now collect formulas

\displaystyle X_c^2=Z^2-R^2=\left(\frac{\sqrt{2}V}{I_p}\right)^2-R^2

Or, equivalently

\displaystyle X_c=\sqrt{\frac{2V^2}{I_p^2}-R^2}

\displaystyle X_c=\sqrt{\frac{2\cdot 18^2}{0.00065^2}-15000^2}

X_c=36176.34\ \Omega

The capacitance is now

\displaystyle C=\frac{1}{226194.67\cdot 36176.34}=1.22\cdot 10^{-10}\ F

The radius of the plates is

r=18\ cm/2=9 \ cm = 0.09 \ m

The separation between the plates is

\displaystyle d=\frac{8.85\cdot 10^{-12} \cdot \pi\cdot 0.09^2}{1.22\cdot 10^{-10}}

d=0.00184\ m

\boxed{d=1.84\ mm}

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3 years ago
What process is mostly responsible for the blue appearance of the sky
malfutka [58]

Answer:reflection by dust particles in air

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How do sea surface temperatures affect evaporation rate?
kotegsom [21]
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8 0
3 years ago
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taurus [48]

Answer:

acceleration

Explanation:

The rate at which velocity changes is the definition of the physical quantity called acceleration, and it is given by the formula: a=\frac{v_f-v_i}{\Delta t}

where \Delta t is the time that took to change from the initial velocity v_i to the final velocity v_f

4 0
3 years ago
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