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jolli1 [7]
3 years ago
14

A gas has an initial volume of 212 cm3 at a temperature of 293 K and a pressure of 0.98 atm. What is the final pressure of the g

as if the volume decreases to 196 cm3 and the temperature of the gas increases to 308 K? 0.86 atm 0.95 atm 1.0 atm 1.1 atm
Physics
2 answers:
PIT_PIT [208]3 years ago
8 0

Answer: 1.1 atm

Explanation:

Using ideal gas equation:

\frac{P_1V_1}{T_1}={P_2V_2}{T_2}

P_1 = initial pressure = 0.98 atm

V_1 = initial volume = 212cm^3

T_1 = initial temperature= 293 K

P_2 = final pressure = ?

V_2 = final volume= 196cm^3

T_2 = final temperature=308 K

\frac{0.98\times 212}{293}={P_2\times 196}{308}

P_2= 1.1atm

Thus final pressure of the gas is 1.1 atm.

Serggg [28]3 years ago
7 0
If the ga has 212 cm3 the. Temperature was at a 2
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mars1129 [50]
<h3><u>Answer;</u></h3>

= 20.436 seconds

<h3><u>Explanation;</u></h3>

Speed = Distance × time

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Time = Distance/speed

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Time = 7.50/0.367

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7 0
3 years ago
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3 years ago
Water flows through a 2.5cm diameter pipe at a rate of
My name is Ann [436]

Answer:

From the Bernoulli energy principle,

ΔP + 1/2ρΔv² = 0 -------------------------- eqn 1

where

ΔP = pressure drop = P2 - P1 = (1 - 0.25)x10⁵ N/m =7.5 x 10⁴N/m

Δv²= velocity change = v₂² - v₁²

ρ = water density = 1kg/m3

Recall volumetric flow rate, Q=A v = constant

A = cross sectional area = πr²=πd²/4

d=pipe diameter at point 2 = 2.5cm = 0.025m and Q =0.20m³/min = 0.00333m³/s

So A= 0.000491m²

we can get v2 = Q/A = 6.79m/s

From eqn 1, v₁² = 2(P2 - P1)/ρ + v₂²

v₁² =  (2 x 7.5 x 10⁴)/1000 + 6.79²

v₁² = 196

v₁ = 14m/s

we can now get the area of the constriction point 2, A₁ = Q/v₁

A₁ = 0.000238m² and the diameter now will be d₁

d₁² = 4 x A₁ / π

     = 4 x 0.000238/3.14 = 0.000303m²

d₁ = √0.000303 = 0.0174m

Therefore, the diameter of  a constriction in the pipe  at the new pressure = 0.0174m = 1.74cm

6 0
2 years ago
Guys No one's answering my question so sad! Once again I'm asking the same question –Here
nexus9112 [7]

For the front glass of the car to get wet, V_c \geq 10 \ m/s.

The given parameters:

  • <em>Speed of the car, = Vc</em>
  • <em>Speed of the rain, = 10 m/s</em>

The relative velocity of the car with respect to the falling rain is calculated as;

V_{C/R} = V_C- V_R

  • If the speed of the car equals the speed of the rain, the rain will fall behind the car.
  • If the speed of the rain is greater than speed of the car, the rain will fall far in front of the car.
  • If the speed of the car is greater than speed of the rain, the rain will fall on the car.

Thus, for the front glass of the car to get wet, V_c \geq 10 \ m/s.

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8 0
2 years ago
You are hungry and decide to go to your favorite neighborhood fast-food restaurant. You leave your apartment and take the elevat
Viktor [21]

Answer:

A)  R = (200 i ^ + 100 j ^ + 30k ^) m , B)    L = 223.61 m , C)   R = 225.61 m

Explanation:

Part A

This is a vector summing exercise, let's take a Reference System where the z axis corresponds to the height (flights), the x axis is the East - West and the y axis corresponds to the North - South.

Let's write the displacements

Descending from the apartment

10 flights of 3 m each, the total descent is 30 m

                Z = 30 k ^ m

Offset at street level

            L1 = 0.2 i ^ km

            L2 = 0.1 j ^ km

Let's reduce everything to the SI system

          L1 = 0.2 * 1000 = 200 i ^ m

          L2 = 100 j ^ m

The distance traveled is

          R = (200 i ^ + 100 j ^ + 30k ^) m

Part B

The horizontal distance traveled can be found with the Pythagorean theorem for the coordinates in the plane

                L² = x² + y²

                L = √ (200² + 100²)

                L = 223.61 m

Part C

The magnitude of travel, let's use the Pythagorean theorem for the sum

             R² = x² + y² + z²

              R = √ (30² + 200² + 100²)

             R = 225.61 m

7 0
3 years ago
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