All categories

3 years ago

Calculate the ratio of change in the mass of the molecules of a gas to the initial mass, if its

Engineering

1 answer:

kirill115 [55]3 years ago

3 0

<h2>Answer:</h2>

<h3>Required Answer is as follows :-</h3>

$\sf \dfrac{P _{i}}{ P_{f}} = \dfrac{3}{4}$

3/4 = (⅓ × m₁/Volume × V²)/(⅓ × m₂/Volume) × (V/2)²

¾ = m₁/Volume × (V)²/m₂/Volume × (V/2)²

¾ = m₁ × (V)²/m₂ × (V²/4)

¾ = m₁/m₂ × 4

m₂ = 4 × 4/3 = 16/3m₁

$\bold{ }$

∆m = m₂ - m₁
∆m = 16m₂/3 - m₁ = 13m₁/3
Ratio = (13m₁/3)/ m₁
Ratio = 13/3

Ratio = 13:3

$\bold{ }$

Mass => It is used to measure it's resistance to its acceleration. SI unit of mass is Kg. $\bold{ }$

You might be interested in

QUESTION:

svet-max [94.6K]

OA bloom is smaller than a bar

6 0

3 years ago

The clepsydra, or water clock, was a device that the ancient Egyptians, Greeks, Romans, and Chinese used to measure the passage

Nookie1986 [14]

Suppose a tank is made of glass and has the shape of a right-circular cylinder of radius 1 ft. Assume that h(0) = 2 ft corresponds to water filled to the top of the tank, a hole in the bottom is circular with radius in., g = 32 ft/s2, and c = 0.6. Use the differential equation in Problem 12 to find the height h(t) of the water.

Answer:

Height of the water = √(128)/147456 ft

Explanation:

Given

Radius, r = 1 ft

Height, h = 2 ft

Radius of hole = 1/32in

Acceleration of gravity, g = 32ft/s²

c = 0.6

Area of the hold = πr²

A = π(1/32)² ---- Convert to feet

A = π(1/32 * 1/12)²

A = π/147456 ft²

Area of water = πr²

A = π 1²

A = π

The differential equation is;

dh/dt = -A1/A2 √2gh where A1 = Area of the hole and A2 = Area of water

A1 = π/147456, A2 = π

dh/dt = (π/147456)/π √(2*32*2)

dh/dt = 1/147456 * √128

dh/dt = √128/147456 ft

Height of the water = √(128)/147456 ft

3 0

4 years ago

A particular cloud-to-ground lightning strike lasts 500 ÂµÂµsec and delivers 30 kA across a potential difference of 100 MV. Assu

sasho [114]

Answer:

a) 15 C charge was delivered by the lightening bolt

b) the lightning delivered 3.0 × 10¹² W of power

- total energy delivered by the lightening strike in J is 1500 × 10⁶ J

- total energy delivered by the lightening strike in Wh is 416666.67 Wh

the residential retail value of the energy delivered by the strike is $ 40.83

a total of 26 lightening strikes would be required to power an average US home for a year.

Explanation:

Given that;

the lighting strike lasted for t ( time ) = 500 μsecs = 500×10⁻⁶ s

Current I = 30 kA = 30×10³ A

voltage V = 100 mV = 100×10⁶ v

we know that; I = Q/t

so Q = I × t

we substitute

Q = 30×10³ × 500×10⁻⁶

Q = 15 C

Therefore 15 C charge was delivered by the lightening bolt

Power P = V × I

we substitute

Power P = 100×10⁶ × 30×10³

P = 3.0 × 10¹² W

Therefore, the lightning delivered 3.0 × 10¹² W of power

we know that; Power = Energy / Time

Energy = Power × Time

we substitute

E = 3.0 × 10¹² × 500×10⁻⁶

E = 1500 × 10⁶ J

- total energy delivered by the lightening strike in J is 1500 × 10⁶ J

- total energy delivered by the lightening strike in Wh is;

⇒ 1500×10⁶ / 3600 Wh

= 416666.67 Wh

given that; 1 KWh → $ 0.098

energy delivered by the strike = 416666.67 Wh = 416.66667 KWh

so the residential retail value of the energy delivered by the strike will be;

416.66667 KWh × $ 0.098

= $ 40.83

∴ the residential retail value of the energy delivered by the strike is $ 40.83

Given that; average monthly residential energy consumption is 900 kWh.

for a year; energy consumption = 12 × 900 kWh = 10,800 KWh = 10800000 Wh

Now

1 lightening strike ⇒ 416666.67 Wh

x lightening strike ⇒ 10800000 Wh

x = 10800000 / 416666.67

x = 25.9199 ≈ 26

Therefore; a total of 26 lightening strikes would be required to power an average US home for a year.

7 0

3 years ago

Batteries typically have ratings stated in ampere-hours, that let you estimate the length of time any particular current level c

DerKrebs [107]

Answer: attached

Explanation:

6 0

3 years ago

Diodes can be used to protect electronic components from certain flowing in another Direction describe another application outsi

Anvisha [2.4K]

Answer:

Plumbing using a one-way check valve to stop water flowing back on a pump when the pump shuts off.

Explanation:

Diodes are like check valves, keeping current from flowing both ways. Used to create d.c. out of a.c by rectification. Also to block flow if d.c. power like a battery is hooked up in reverse polarity.

4 0

3 years ago