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3 years ago

In a rigid container, as the temperature of a gas decreases, the pressure of the gas will decrease. t f

2 answers:

6 0

The answer is true: the pressure of a gas will decrease as temperature decreases in a rigid container.

This is one of the central gas laws called the Gay-Lussac law that states for a given gas at a constant volume, the pressure of the gas is directly proportional to its temperature. We also know that as temperature reduces, so too does molecular interaction. Increased temperature results in increased pressure, and decreased temperature therefore results in decreased pressure.

LenaWriter [7]3 years ago

5 0

True please mark brainly

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When an atom releases gamma radiation____________.

wel

Answer:

d. the atomic number remains the same.

Explanation:

First at all it's important to know how to read nuclear information

$_{Z}^{A}X$

X is the atomic symbol, A the mass number and Z the atomic number of the element.

Gamma rays emitted on gamma decay are characterized as $_{0}^{0}\gamma$

If we write the nuclear equation for the decay, we have that:

$_{Z}^{A}X\,\rightarrow{}_{0}^{0}\gamma\,+\,{}_{Z}^{A}Y$

The sum of the mass numbers and atomic numbers on the right side has to be equal to the left side numbers of the equation, that means the mass number and the atomic number remains the same for the resulting atom to preserve the equality.

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2 years ago

After a chemical reaction, the atomic nuclei are _________.

sleet_krkn [62]

After a chemical reaction, the atomic nuclei are unchanged. (C)

The nucleii don't know a thing about the outside world during

physical or chemical processes ... not until NUCLEAR things

happen.

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3 years ago

A mover loads a 100 kg box into the back of a moving truck by

Andrews [41]

Answer:

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Explanation:

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2 years ago

When would the displacement technique of measuring need to be employed?

Tanzania [10]

Answer:

a is the answer is yes I....

Explanation:

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5 0

3 years ago

Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field st

Black_prince [1.1K]

Answer:

Part A:

$E_{midpoint}=0$

Part B:

$E_{center}=2711.7558 N/C$

Explanation:

Part A:

Formula of Electric Field Strength:

$E=\frac{1}{4\pi\epsilon}\frac{xQ}{(x^2+R^2)^{3/2}}$

Where:

x is the distance from the ring

R is the radius of the ring

$\epsilon$ is constant permittivity of free space=8.854*10^-12 farads/meter

Q is the charge

For right Ring E at the midpoint can be calculated as:

x for right plate=25/2=12.5 cm=0.125 m

Radius=R=10/2=5 cm=0.05 m

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{right}=9208.1758 N/C$

For Left Ring E at the midpoint can be calculated as:

Since charge on both plates is +ve and same in magnitude, the electric field will be same for both plates.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{left}=9208.1758 N/C$

Electric Field at midpoint:

Both rings have same magnitude but the direction of fields will be opposite as they have same charge on them.

$E_{midpoint}=E_{left}-E_{right}\\E_{midpoint}=9208.1758-9208.1758\\E_{midpoint}=0$

Part B:

At center of left ring:

Due to left ring Electric field at center is zero because x=0.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0)*(20*10^{-19})}{((0)^2+(0.05)^2)^{3/2}}\\E_{left}=0 N/C$

Due to right ring Electric field at center of left ring:

Now: x=25 cm= o.25 m (To the center of left ring)

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.25)*(20*10^{-19})}{((0.25)^2+(0.05)^2)^{3/2}}\\E_{right}=2711.7558 N/C$

Electric Field Strength at center of left ring is same as that of right ring.

$E_{center}=2711.7558 N/C$

5 0

3 years ago