Protons and ____ have electric charge?

Two 800 cm^3 containers hold identical amounts of a monatomic gas at 20Â°C. Container A is rigid. Container B has a 100 cm^2 pis

Vikki [24]

Answer:

1) Final Temperature of the gas in A will be GREATER than the temperature in B

2) Diagram of both processes on a single PV has been uploaded below

3) The Initial pressures in containers A and B is 3039.87 J/liters

4) the final volume of container B is 923.36 cm³

Explanation:

Given that;

Temperature = 20°C = 293 K

mass of piston = 10 kg

Area = 100cm³

Volume V = 800 cm³ = 0.8 L

ideal gas constant R = 8.3 J/K·mol

1)

Final Temperature of the gas in A will b GREATER than the temperature in B

2)

Diagram of both processes on a single PV has been uploaded below,

3)

Initial pressures in containers A and B

PV = nRT

P = RT/V

we substitute

P = (8.3 × 293) / 0.8

P = 2431.9 / 0.8

P = 3039.87 J/liters

Therefore, The Initial pressures in containers A and B is 3039.87 J/liters

4)

Given that;

power = 25 W

time t = 15s

the final volume of container B = ?

we know that;

work done = power × time

work done = 25 × 15 = 375

Also work done = P( V₂ - V₁ )

so we substitute

375 = 3039.87 ( V₂ - 0.8 )

( V₂ - 0.8 ) = 375 / 3039.87

V₂ - 0.8 = 0.12336

V₂ = 0.12336 + 0.8

V₂ = 0.92336 Litres

V₂ = 923.36 cm³

Therefore, the final volume of container B is 923.36 cm³

7 0

2 years ago

In a classroom demonstration, the pressure inside a soft drink can is suddenly reduced to essentially zero. Assuming the can to

umka2103 [35]

Answer:

3141N or 3.1 ×10³N to 2 significant figures. The can experiences this inward force on its outer surface.

Explanation:

The atmospheric pressure acts on the outer surface of the can. In order to calculate this inward force we need to know the total surface area of the can available to the air outside the can. Since the can is a cylinder with a total surface area given by 2πrh + 2πr² =

A = 2πr(r + h)

Where h = height of the can = 12cm

r = radius of the can = 6.5cm/2 = 3.25cm

r = diameter /2

A = 2π×3.25 ×(3.25 + 12) = 311.4cm² = 311.4 ×10-⁴ = 0.031m²

Atmospheric pressure, P = 101325Pa = 101325 N/m²

F = P × A

F = 101325 ×0.031.

F = 3141N. Or 3.1 ×10³ N.

5 0

3 years ago

A pneumatic system consists of several identical 1.5 inch diameter cylinders to lift 450-pound pallets in a warehouse. the gauge

DerKrebs [107]

The applicable equation:

P = F/A

P = pressure
F = Force or weight
A = surface area

Pressure on each cylinder = (W/n)/A
Where n = number of cylinders. Additionally, pressure in the reservoir is equivalent to the pressure in each cylinder.
Net pressure = 75 - 14.7 = 60.3 psi

Therefore,
60.3 = (W/n)/A = (450/n)/(πD^2/4) = (450/n)/(π*1.5^2/4) = (450/n)/(1.7671)
60.3*1.7671 = 450/n
106.03 = 450/n
n = 450/106.3 = 4.244 ≈ 5

The number of cylinders is 5.

3 0

2 years ago

A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless cord passing through a

frosja888 [35]

Answer:

a) Yes

b) 7 rad/s

c) 0.01034 J

Explanation:

a)

Yes the angular momentum of the block is conserved since the net torque on the block is zero.

b)

m = mass of the block = 0.0250 kg

w₀ = initial angular speed before puling the cord = 1.75 rad/s

r₀ = initial radius before puling the cord = 0.3 m

w = final angular speed after puling the cord = ?

r = final radius after puling the cord = 0.15 m

Using conservation of angular momentum

m r₀² w₀ = m r² w

r₀² w₀ = r² w

(0.3)² (1.75) = (0.15)² w

w = 7 rad/s

c)

Change in kinetic energy is given as

ΔKE = (0.5) (m r² w² - m r₀² w₀²)

ΔKE = (0.5) ((0.025) (0.15)² (7)² - (0.025) (0.3)² (1.75)²)

ΔKE = 0.01034 J

6 0

3 years ago

An airtight box has a removable lid of area 1.10 10-2 m2 and negligible weight. The box is taken up a mountain where the air pre

andrezito [222]

Answer:

F= $7.7\times10^2 N$

Explanation:

The magnitude of force required to pull the lid off the box by air pressure.

We know that Pressure, P= Force(F)/Area(A)

Force, F= P×A

Given: A= $1.10\times10^{-2} m^2$

P= $7\times10^{4} Pa$ .

Therefore, F= $7\times10^{4}\times1.10\times10^{-2}$ .

F= $7.7\times10^2 N$

4 0

3 years ago