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3 years ago

If the amount of gas in a sealed container increases, what happens to the pressure in the container

2 answers:

7 0

The pressure of the container will increase, I say this because <span>gases are easily compressed because of the space between the particles in a gas.</span>

lilavasa [31]3 years ago

6 0

The pressure law states that for a constant volume of gas in a sealed container the temperature of the gas is directly proportional to its pressure. This can be easily understood by visualising the particles of gas in the container moving with a greater energy when the temperature is increased.

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If 3.5 paper clips = 1.0 pencils and your paper is 1.5 pencils long, how many paper clips long is your paper?

Andrews [41]

$\bf \begin{array}{ccll}
paperclips&pencils\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
3.5&1.0\\
p&1.5
\end{array}\implies \cfrac{3.5}{p}=\cfrac{1.0}{1.5}\implies \cfrac{3.5\cdot 1.5}{1}=p$

6 0

3 years ago

A reason that one typically does not notice a blind spot in the visual field is that

Sophie [7]

Explanation:

Both of our eyes have a blind spot, in the retina of eye where there are no rods (light vision) or cons (color vision) cells. Its about a size of pinhead. The blind spot is place where the optic nerves exit the eye and connect to our brain. The problem is that we cannot notice this blind as the brain fills in information for us.

3 0

3 years ago

Can someone please tell me how to solve the question below. Thank you!! If 550-nm light is incident normally on a diffraction gr

Likurg_2 [28]

Answer:

The ruling separation must be between 2.2 × 10⁻⁶ m and 3.3 × 10⁻⁶ m

Explanation:

Fringe spacing w = λ·D/s

Diffraction grating, d·sin(θ) = n·λ

n = Number of lines

When θ = 90°, we have, sin(θ) = sin(90°) = 1

Therefore;

d·sin(θ) = d × 1 = d = n·λ

When n = 6 lines and d = 550-nm, we have;

d = 6 × 550-nm = 3,300 nm = 3.3 microns = 3.3 × 10⁻⁶ m

Therefore, the ruling separation must be between 2.2 × 10⁻⁶ m and 3.3 × 10⁻⁶ m.

5 0

3 years ago

What type of circuit is illustrated?A)closed series circuit

RoseWind [281]

I think it’s B open series circuit but I’m not sure

4 0

3 years ago

Read 2 more answers

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated

Lelu [443]

Incomplete question as we have not told to find what quantity.The complete question is here

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated by vacuum and the inner radius of the outer shell is 5.00 cm.calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) the electric field just outside the surface of the inner sphere.

Answer:

(a) $C=16.7pF$

(b) $r_{a} =3.749cm$

Explanation:

Given data

$Q=3.50nC\\V=210V\\r_{b}=5.0cm$

For part (a)

The Capacitance given by:

$C=\frac{Q}{V}\\ C=\frac{3.50*10^{-9} C}{210V}\\C=1.6666*10^{-11}F\\or\\C=16.7pF$

For part (b)

The Capacitance of coordinates is given as

$C=\frac{4\pi e}{\frac{1}{r_{a} }-\frac{1}{r_{b} } }\\ So\\{\frac{1}{r_{a} }-\frac{1}{r_{b} } }=\frac{4\pi *8.85*10^{-12} }{1.666*10^{-11}}=6.672m^{-1} \\ \frac{1}{r_{a} }=6.672+(1 /0.05)\\\frac{1}{r_{a} }=26.672\\r_{a} =1/26.672\\r_{a} =0.0375m\\r_{a} =3.749cm$

For part (c)

The electric field according to Gauss Law is given by:

$EA=\frac{Q}{e}\\ E=\frac{Q}{4\pi er_{a}^{2} }=\frac{kQ}{r_{a}^{2}}\\ E=\frac{9*10^{9}*3.50*10^{-9} }{(0.0375m)^{2} }\\ E=2.24*10^{4} N/C$

7 0

3 years ago