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KatRina [158]
3 years ago
5

What would the final volume of 40 L of gas at 80 pascals be if the pressure increases to 130 pascals?

Physics
1 answer:
melomori [17]3 years ago
5 0

Answer:

Final volume, V2 = 24.62 L

Explanation:

Given the following data;

Initial volume = 40 L

Initial pressure = 80 Pa

Final pressure = 130 Pa

To find the final volume V2, we would use Boyles' 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;

PV = K

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

Substituting into the equation, we have;

80 * 40 = 130V_{2}

3200 = 130V_{2}

V_{2} = \frac {3200}{130}

V_{2} = 24.62

Final volume, V2 = 24.62 Liters

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Curve of space time

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Part 1 :
tatuchka [14]

1) 9.57 N

We have two forces applied on the apple:

- The force of gravity, in the downward direction:

W = 9.42 N

- The force exerted by the wind, in the horizontal direction (to the right):

Fw = 1.68 N

The two forces are perpendicular to each other, so we can find the magnitude of the net force by using Pythagorean's theorem.

Therefore, we have:

F=\sqrt{W^2+F_w^2}=\sqrt{(9.42)^2+(1.68)^2}=9.57 N

2) 10^{\circ}

The direction of the net external force, measured from the downward vertical, can be measured using the following formula:

\theta = tan^{-1}(\frac{F_x}{F_y})

where

F_x is the force in the horizontal direction

F_y is the force in the vertical direction

In this problem,

F_x = F_w = 1.68 N

F_y = W = 9.42 N

and so we find:

\theta = tan^{-1}(\frac{1.68}{9.42})=10^{\circ}

4 0
3 years ago
Help with this one please! Please give correct answers!
Verdich [7]

the answer is C because the air balloon is  going down meaning that negative


5 0
3 years ago
Point charges of 28.0 µC and 42.0 µC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el
olga_2 [115]

Answer: a) electric field will be zero at zero meters apart

b) for smaller charge q, E = 6.048X10^6N/m towards away from the charge,

for bigger charge Q, E = 4.032X10^6N/m

Explanation:

Detailed explanation and calculation is shown in the image below.

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A charge of 1.0 × 10-6 μC is located inside a sphere, 1.25 cm from its center. What is the electric flux through the sphere due
Korvikt [17]

Answer:

\Phi_E=0.11\frac{N\cdot m^2}{C}

Explanation:

According to Gauss's Law, the electric flux of a charged sphere is the electric field multiplied by the area of ​​the spherical surface:

\Phi_E=EA\\\Phi_E=E(4\pi R^2)\\\Phi_E=\frac{q}{4\pi \epsilon_0 R^2}(4\pi R^2)\\\Phi_E=\frac{q}{\epsilon_0}

This is identical to the electric flux of a point charge located in the center of the sphere.

\Phi_E=\frac{1*10^{-12}C}{8.85*10^{-12} \frac{C^2}{N\cdot m^2}}\\\Phi_E=0.11\frac{N\cdot m^2}{C}

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