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2 years ago

7

With the temperature held constant, the piston of a cylinder containing a gas is pulled out so that the volume increases from 0.

3 m3 to 0.9 m3. If the final pressure of the gas is 80 kPa, what was the initial pressure

1 answer:

klio [65]2 years ago

3 0

Answer:

P1 = 240 kPa.

Explanation:

Given the following data;

Initial volume = 0.3 m³

Final volume, V2 = 0.9 m³

Final pressure, P2 = 80 kPa

To find the initial pressure, we would use Boyle's law;

Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.

Mathematically, Boyles law is given by the formula;

$PV = K$

$P_{1}V_{1} = P_{2}V_{2}$

Substituting into the formula, we have;

$P_{1} * 0.3 = 80 * 0.9$

$0.3P_{1} = 72$

$P_{1} = \frac {72}{0.3}$

$P_{1} = 240$

Therefore, the initial pressure of the gas is 240 kPa

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Which is the most common direction of motion in the solar system, both for orbital revolution and axial rotation?.

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Answer:

Clockwise

Explanation:

All of the planets rotate the same way around the sun.

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2 years ago

monochromatic light from a distant source is incident on a slit 0.75 mm wide. on screen 2 m away, the distance from the central

hjlf

Displacement from the center line for minimum intensity is 1.35 mm , width of the slit is 0.75 so Wavelength of the light is 506.25.

<h3>How to find Wavelength of the light?</h3>

When a wave is bent by an obstruction whose dimensions are similar to the wavelength, diffraction is observed. We can disregard the effects of extremes because the Fraunhofer diffraction is the most straightforward scenario and the obstacle is a long, narrow slit.

This is a straightforward situation in which we can apply the

Fraunhofer single slit diffraction equation:

y = mλD/a

Where:

y = Displacement from the center line for minimum intensity = 1.35 mm

λ = wavelength of the light.

D = distance

a = width of the slit = 0.75

m = order number = 1

Solving for λ

λ = y + a/ mD

Changing the information that the issue has provided:

λ = 1.35 * 10^-3 + 0.75 * 10^-3 / 1*2

=5.0625 *10^-7 = 506.25

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Wavelength of the light 506.25.

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1 year ago

A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci

ollegr [7]

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity ( $v$ ), measured in meters per second, by using this kinematic equation:

$v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s}$ (1)

Where:

$a$ - Acceleration, measured in meters per square second.

$\Delta s$ - Travelled distance, measured in meters.

$v_{o}$ - Initial velocity, measured in meters per second.

If we know that $v_{o} = 0\,\frac{m}{s}$ , $a = 2.35\,\frac{m}{s^{2}}$ and $\Delta s = 595\,m$ , the final velocity of the rocket is:

$v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}$

$v\approx 52.882\,\frac{m}{s}$

The time associated with this launch ( $t$ ), measured in seconds, is:

$t = \frac{v-v_{o}}{a}$

$t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }$

$t = 22.503\,s$

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

$v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o})$ (2)

Where:

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

If we know that $v_{o} = 52.882\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ , $a = -9.807\,\frac{m}{s^{2}}$ and $s_{o} = 595\,m$ , then the maximum height reached by the rocket is:

$v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})$

$s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}$

$s = 737.577\,m$

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

$s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}$ (2)

Where:

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

$v_{o}$ - Initial speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

If we know that $s_{o} = 595\,m$ , $v_{o} = 52.882\,\frac{m}{s}$ , $s = 0\,m$ and $a = -9.807\,\frac{m}{s^{2}}$ , then the time needed by the rocket is:

$0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}$

$-4.904\cdot t^{2}+52.882\cdot t +595 = 0$

Then, we solve this polynomial by Quadratic Formula:

$t_{1}\approx 17.655\,s$ , $t_{2} \approx -6.872\,s$

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

7 0

2 years ago

Why the greater the depth, the greater the pressure?

Tatiana [17]

Well, the reason behind this is because water depth (I assume you're talking about water/liquid) increases the pressure because of the weight and volume of the water above. I hope this helps! ~Mia

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3 years ago

A spherical shell of radius 9.0 cm carries a uniform surface charge density σ= 9.0 nC/m2. The electric field at r= 9.1 cm is app

maria [59]

Answer:

995.12 N/C

Explanation:

R = 9 cm = 0.09 m

σ = 9 nC/m^2 = 9 x 10^-9 C/m^2

r = 9.1 cm = 0.091 m

q = σ x 4π R² = 9 x 10^-9 x 4 x 3.14 x 0.09 x 0.09 = 9.156 x 10^-10 C

E = kq / r^2

E = ( 9 x 10^9 x 9.156 x 10^-10) / (0.091 x 0.091)

E = 995.12 N/C

8 0

2 years ago