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3 years ago

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A cylinder measuring wide and high is filled with gas. The piston is pushed down with a steady force measured to be . Calculate

the pressure of the gas inside the cylinder. Write your answer in units of kilopascals. Be sure your answer has the correct number of significant digits

1 answer:

jekas [21]3 years ago

7 0

Explanation:

Let us assume that the given cylinder is 2.6 cm wide and its height is 3.1 cm. And, when piston is pushed down then the steady force is equal to 15 N.

Now, radius of the cylinder will be as follows.

r = $\frac{diameter}{2}$

= $\frac{2.6 cm}{2}$

= 1.3 cm

or, = 0.013 m (as 1 m = 100 cm)

As, area of cylinder = $\pi \times r^{2}$

= $3.414 \times (0.013 m)^{2}$

= $5.77 \times 10^{-4} m^{2}$

Relation between pressure and force is as follows.

Pressure = $\frac{Force}{Area}$

= $\frac{15 N}{5.77 \times 10^{-4} m^{2}}$

= 25996 $N/m^{2}$

Since, 1 $N/m^{2}$ = 1 Pa (as 1 kPa = 1000 Pa)

Therefore, P = 25996 $N/m^{2}$

= 25.99 kPa

= 26 kPa (approx)

Thus, we can conclude that pressure of the gas inside the cylinder is 26 kPa.

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What do the length of the cord and gravity determine for a pendulum?

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Content:

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Besides ethical considerations, what is another reason why Milgram’s experiment may be difficult to duplicate?

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3 years ago

A uniform meter stick is pivoted at the 50.00 cm mark on the meter stick. A 400.0 gram object is hung at the 20.0 cm mark on the

babunello [35]

Answer:C

Explanation:

Given

mass $m_1=400\ gm$ is at $x=20\ cm$ mark

mass $m_2=320\ gm$ is at $x=75\ cm$ mark

Scale is Pivoted at $x=50\ cm mark$

For scale to be in equilibrium net torque must be equal to zero

Taking ACW as positive thus

$T_{net}=0.4\times g\times (0.5-0.2)-0.32\times g\times(0.75-0.50)$

$T_{net}=0.12g-0.08g=0.04g$

Therefore a net torque of 0.04 g is required in CW sense which a mass $400\ gm$ can provide at a distance of $x_o$ from pivot

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6 0

3 years ago

Two point charges of +20.0 μC and -8.00 μC are separated by a distance of 20.0 cm. What is the intensity of electric field E mid

algol13

Answer:

The intensity of the net electric field will:

$E_{net}=E_{1}+E_{2}=2.52*10^{7}\: N/C$

Explanation:

Here we need first find the electric field due to the first charge at the midway point.

The electric field equation is given by:

$|E_{1}|=k\frac{q_{1}}{d^{2}}$

Where:

k is Coulomb's constant
q(1) is 20.00 μC or 20*10⁻⁶ C
d is the distance from q1 to the midpoint (d=10.0 cm)

So, we will have:

$|E_{1}|=(9*10^{9})\frac{20*10^{-6}}{0.1^{2}}$

$|E_{1}|=1.8*10^{7}\: N/C$

The direction of E1 is to the right of the midpoint.

Now, the second electric field is:

$|E_{2}|=k\frac{q_{2}}{d^{2}}$

$|E_{2}|=(9*10^{9})\frac{8*10^{-6}}{0.1^{2}}$

$|E_{2}|=7.2*10^{6}\: N/C$

The direction of E2 is to the right of the midpoint because the second charge is negative.

Finally, the intensity of the net electric field will:

$E_{net}=E_{1}+E_{2}=2.52*10^{7}\: N/C$

I hope it helps you!

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3 years ago