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3 years ago

An object increases its velocity from 22 m/s to 36 m/s in 5.0 s. What is the average velocity of the object?

Physics

1 answer:

Luden [163]3 years ago

7 0

Answer:

$a=2.8\ m/s^2$

Explanation:

Given that,

Initial velocity of an object, u = 22 m/s

Final velocity of an object, v = 36 m/s

Time, t = 5 s

It can be assumed to find the average acceleration of the object instead of average velocity.

The change in velocity per unit time is equal to average acceleration of an object. It can be given by :

$a=\dfrac{v-u}{t}\\\\a=\dfrac{36\ m/s-22\ m/s}{5}\\\\\a=\dfrac{14}{5}\ m/s^2\\\\a=2.8\ m/s^2$

So, the acceleration of the object is $2.8\ m/s^2$ .

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If we assume that the metallic plates are perfect conductors, the electric field in their interiors must vanish. given that the

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here the charge density of metal plate is given as

$charge density = \sigma$

now the electric field is given Gauss law

$\int E. dA = \frac{q}{\epsilon_0}$

now here E = constant

so we will have

$E. \int dA = \frac{q}{\epsilon_0}$

Since total area on both sides of plate will be double and becomes 2A

$E. 2A = \frac{q}{\epsilon_0}$

$E = \frac{q/A}{2\epsilon_0}$

$E = \frac{\sigma}{2\epsilon_0}$

Now if we will find the electric field inside the metal plate

Then as we know that total charge inside the plate will always be zero

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2 years ago

Air enters the compressor of an ideal air-standard Braytoncycle at 100 kPa, 300 K, with a volumetric flow rate of 5 m3/s.The tur

Harrizon [31]

Answer:

Explanation:

Given that

Air Inlet Pressure, P1 = 100 KPa

Air Inlet temperature, T1 = 300 K

Volume flow rate, Q = 5 m³/s

Turbine inlet temperature, T₃ = 1400 K

Compressor pressure ratio, r = 6, 8, 12

Heat capacity ratio or air = 1.4

γ= 1.4

Specific heat constant pressure of air, cp = 1.005 KJ/kg.k

At r = 6,

For Brayton cycle,

T2/T1 = r ^ (γ - 1)/γ

T3/T4 = r ^ (γ - 1)/γ

Now by putting the values

T2/300 = 6 ^ (1.4 - 1)/1.4

1400/T4 = 6 ^ (1.4 - 1)/1.4

T₂ = 1.67 × 300

= 500 K

T₄ = 1400/1.67

= 839.07 K

Efficiency, η = 1 - ((T4 - T1)/(T3 - T2)

Inputting values,

= 1 - ((839.07 - 300)/(1400 - 500))

= 0.40

= 40%

Bwr = Wcomp/Wturb

Where,

Wcomp = workdone by compressor

Wturb = workdone by turbine

= ((T2 - T1)/(T3 - T4))

= ((500 - 300)/(1400 - 839.07))

= 0.36

Net work = Net heat

Net heat = Qa - Qr

Qr = Cp ( T₄-T₁)

Qa = Cp ( T₃-T₂)

Imputting values,

Net heat, Qnet = 1.005 (1400 - 500 - 839.07 + 300)

= 1.005 × 360.93

= 362.74 kJ/kg

Net heat, Qnet = 362.74 kJ/kg

Using the ideal gas equation,

P V = n R T

But n = mass/molar mass,

P = ρ R T

By putting the values

P = ρ R T

Inputting values,

100 = ρ x 0.287 x 300

ρ = 1.16 kg/m³

mass flow rate, m = ρ × Q

= 1.16 × 5

= 5.80 kg/s

Net power, Pnet = ms × Net heat, Qnet

= 5.8 × 362.74

= 2103.9 kW.

At r = 8,

For Brayton cycle,

T2/T1 = r ^ (γ - 1)/γ

T3/T4 = r ^ (γ - 1)/γ

Now by putting the values

T2/300 = 8 ^ (1.4 - 1)/1.4

1400/T4 = 8 ^ (1.4 - 1)/1.4

T₂ = 1.81 × 300

= 543.4 K

T₄ = 1400/1.81

= 772.9 K

Efficiency, η = 1 - ((T4 - T1)/(T3 - T2)

Inputting values,

= 1 - ((772.9 - 300)/(1400 - 543.4))

= 0.448

= 45%

Bwr = Wcomp/Wturb

Where,

Wcomp = workdone by compressor

Wturb = workdone by turbine

= ((T2 - T1)/(T3 - T4))

= ((543.4 - 300)/(1400 - 772.9))

= 0.39

Net work = Net heat

Net heat = Qa - Qr

Qr = Cp ( T₄-T₁)

Qa = Cp ( T₃-T₂)

Imputting values,

Net heat, Qnet = 1.005 (1400 - 543.4 - 772.9 + 300)

= 1.005 × 383.7

= 385.62 kJ/kg

Net heat, Qnet = 385.62 kJ/kg

Using the ideal gas equation,

P V = n R T

But n = mass/molar mass,

P = ρ R T

By putting the values

P = ρ R T

Inputting values,

100 = ρ x 0.287 x 300

ρ = 1.16 kg/m³

mass flow rate, m = ρ × Q

= 1.16 × 5

= 5.80 kg/s

Net power, Pnet = ms × Net heat, Qnet

= 5.8 × 385.62

= 2236.59 kW.

At r = 12,

For Brayton cycle,

T2/T1 = r ^ (γ - 1)/γ

T3/T4 = r ^ (γ - 1)/γ

Now by putting the values

T2/300 = 12 ^ (1.4 - 1)/1.4

1400/T4 = 12 ^ (1.4 - 1)/1.4

T₂ = 2.03 × 300

= 610 K

T₄ = 1400/2.03

= 688.32 K

Efficiency, η = 1 - ((T4 - T1)/(T3 - T2)

Inputting values,

= 1 - ((688.32 - 300)/(1400 - 610))

= 0.509

= 51%

Bwr = Wcomp/Wturb

Where,

Wcomp = workdone by compressor

Wturb = workdone by turbine

= ((T2 - T1)/(T3 - T4))

= ((610 - 300)/(1400 - 688.32))

= 0.44

Net work = Net heat

Net heat = Qa - Qr

Qr = Cp ( T₄-T₁)

Qa = Cp ( T₃-T₂)

Imputting values,

Net heat, Qnet = 1.005 (1400 - 610 - 688.32 + 300)

= 1.005 × 401.68

= 403.7 kJ/kg

Net heat, Qnet = 403.7 kJ/kg

Using the ideal gas equation,

P V = n R T

But n = mass/molar mass,

P = ρ R T

By putting the values

P = ρ R T

Inputting values,

100 = ρ x 0.287 x 300

ρ = 1.16 kg/m³

mass flow rate, m = ρ × Q

= 1.16 × 5

= 5.80 kg/s

Net power, Pnet = ms × Net heat, Qnet

= 5.8 × 403.7

= 2341.39 kW.

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