Phase Angles: Consider a series circuit in which the capacitive reactance equals the inductive reactance. What is the phase angle between current and voltage for this circuit

Answer:

The total internal energy change for the entire process is  -0.94 kJ

Explanation:

Process A to B is an isothermal process, therefore, [tex]u_A[/tex] - [tex]u_B[/tex] = 0

Process B to C

P = -mV + C

When P = 12, V = 0.12

When P = 4, V = 0.135

Therefore, we have;

12 = -m·0.12 + C

4 = -m·0.135 + C

Solving gives

m = 533.33

C = 76

[tex]T = \dfrac{1}{nR} \times (-533.33 \times V^2 + 76 \times V)[/tex]

p₂ = p₁V₁/V₂ = 12*0.1/0.12 = 10 kPa

The work done = 0.5*(0.135 - 0.12)*(4 - 10.0) = -0.045 kJ = -45 J

For heat supplied

Assuming an approximate polytropic process, we have;

Work done = (p₃×v₃ - p₂×v₂)/(n - 1)

Which gives;

-45 = (4*0.135 - 10*0.12)/(n -1)

∴ n -1 = (4*0.135 - 10*0.12)/-45 =   14.67

n = 15.67

Q = W×(n - γ)/(γ - 1)

Q = -45*(15.67 - 1.4)/(1.4 - 1) = -1,605.375 J

u₃ - u₂ = Q + W = -1,605.375 J - 45 J = -1650 J = -1.65 kJ

For the constant pressure process D to C, we have;

[tex]Q = c_p \times \dfrac{p}{R} \times (V_4 -V_3) = \dfrac{5}{2} \times p \times (V_4 -V_3)[/tex]

Q₄₋₃ = (0.1 - 0.135) * 4*5/2 = -0.35 kJ

W₄₋₃ = 4*(0.1 - 0.135) = -0.14 kJ

u₄ - u₃ = Q₄₋₃ + W₄₋₃ = -0.14 kJ + -0.35 kJ = -0.49 kJ

For the process D to A, we have a constant volume process

[tex]Q_{1-4} = \dfrac{c_v}{R} \times V \times (p_1 - p_4) = \dfrac{3}{2} \times 0.1 \times (12 - 4) = 1.2 \ kJ[/tex]

W₁₋₄ = 0 for constant volume process, therefore, u₁ - u₄ = 1.2 kJ

The total internal energy change Δ[tex]u_{process}[/tex] for the entire process is therefore;

Δ[tex]u_{process}[/tex] = u₂ - u₁ + u₃ - u₂ + u₄ - u₃ + u₁ - u₄ = 0  - 1.65 - 0.49 + 1.2 = -0.94 kJ.

Answer: According to Newtons second Law of motion ;

F = ma (Force  equals  mass multiplied by acceleration.)

The acceleration is directly proportional to the net force; the net force equals mass times acceleration; the acceleration in the same direction as the net force; an acceleration is produced by a net force

Explanation:

Answer:

The amount of liters of fuel should have been added is  20093 L

Explanation:

Given that;

a mechanic used a dipstick to determine that 7682 L of fuel were left on the plane

i.e the volume of fuel left in the plane = 7682 L

Required fuel to make  a trip = 22,300 kg of fuel

Also from the question; we are being told that in order for the pilot to determine the volume ; he asked for the density of the fuel and the mechanic said 1.77.

This volume of fuel was added and the plane subsequently ran out of fuel, but landed safely by gliding into Gimli Airport near Winnipeg. The error arose because the factor 1.77 was in units of pounds per liter (lbs/L).

Now; we can understand that the density of the fuel was 1.77 pound /litre.

SO , let convert 1.77 pound /litre to kg/Litre;

we all know that

1 pound = 0.4536 kg

1.77 pound/litre  = x kg

If we cross multiply ; we will have:

1.77 pound/litre  × 0.4536 kg = 1 pound × x kg

x kg = (1.77 pound/litre  × 0.4536 kg) /1 pound

x = 0.802872 kg/litre

[tex]\mathbf{Density = \dfrac{mass}{volume}}[/tex]

where ;

mass =  22,300 kg of fuel

volume = unknown ???

density = 0.802872 kg/litre

making volume the subject of the formula from above; we have:

[tex]\mathbf{volume = \dfrac{mass}{Density}}[/tex]

[tex]\mathbf{volume = 22300 \ kg \ of \ fuel *\dfrac{1 \ litre }{0.802872 \ kg \ of \ fuel}}[/tex]

volume = 27775.28672 litre

volume [tex]\approx[/tex] 27775 L

Let not forget that we are being told as well that the volume of fuel left in the plane = 7682 L

Now;

The amount of liters of fuel should have been added is: =  27775 L - 7682 L

The amount of liters of fuel should have been added is  20093 L

Answer:

mass of the second ball is 0.379m

Explanation:

Given;

mass of first ball = m

let initial velocity of first ball = u₁

let final velocity of first ball  = v₁ = 0.45u₁

let the mass of the second ball = m₂

initial velocity of the second ball, u₂ = 0

let the final velocity of the second ball = v₂

Apply the principle of conservation of linear momentum;

mu₁ + m₂u₂ = mv₁ + m₂v₂

mu₁  +  0  = 0.45u₁m + m₂v₂

mu₁  = 0.45u₁m + m₂v₂ -------- equation (i)

Velocity for elastic collision in one dimension;

u₁ + v₁ = u₂ + v₂

u₁ + 0.45u₁ = 0 + v₂

1.45u₁ = v₂ (final velocity of the second ball)

Substitute in v₂ into equation (i)

mu₁  = 0.45u₁m + m₂(1.45u₁)

mu₁ = 0.45u₁m + 1.45m₂u₁

mu₁ - 0.45u₁m = 1.45m₂u₁

0.55mu₁ = 1.45m₂u₁

divide both sides by u₁

0.55m = 1.45m₂

m₂ = 0.55m / 1.45

m₂ = 0.379m

Therefore, mass of the second ball is 0.379m (where m is mass of the first ball)

The mass of the second ball is 0.379m and this can be determined by conserving the momentum.

Given :

A ball of mass 'm' makes a head-on elastic collision with a second ball (at rest) and rebounds with a speed equal to 0.450 its original speed.

In order to determine the mass of the second ball, apply conservation of linear momentum.

[tex]\rm m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]

where [tex]m_1[/tex] is the mass of the first ball, [tex]m_2[/tex] is the mass of the second ball, [tex]u_1[/tex] is the initial velocity of the first ball, [tex]u_2[/tex] is the initial velocity of the second ball, [tex]\rm v_1[/tex] is the final velocity of the first ball, and [tex]\rm v_2[/tex] is the final velocity of the second ball.

Now, substitute the known terms in the above formula.

[tex]\rm mu_1+0=0.45u_1m+m_2v_2[/tex]

[tex]\rm mu_1=0.45u_1m+m_2v_2[/tex]   ----  (1)

For elastic collision, the velocity is given by:

[tex]\rm v_1+u_1=v_2+u_2[/tex]

[tex]\rm 0.45u_1+u_1=0+v_2[/tex]

[tex]\rm v_2 = 1.45u_1[/tex]

Now, substitute the value of [tex]\rm v_2[/tex] in the equation (1).

[tex]\rm mu_1=0.45u_1m+1.45u_1m_2[/tex]

[tex]\rm 0.55mu_1=1.45m_2u_1[/tex]

[tex]\rm m_2=0.379m[/tex]

So, the mass of the second ball is 0.379m.

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Answer:

The height of the building is 158.140 meters.

Explanation:

A barometer is system that helps measuring atmospheric pressure. Manometric pressure is the difference between total and atmospheric pressures. Manometric pressure difference is directly proportional to fluid density and height difference. That is:

[tex]\Delta P \propto \rho \cdot \Delta h[/tex]

[tex]\Delta P = k \cdot \rho \cdot \Delta h[/tex]

Where:

[tex]\Delta P[/tex] - Manometric pressure difference, measured in kilopascals.

[tex]\rho[/tex] - Fluid density, measured in kilograms per cubic meter.

[tex]\Delta h[/tex] - Height difference, measured in meters.

Now, an equivalent height difference with a different fluid can be found by eliminating manometric pressure and proportionality constant:

[tex]\rho_{air} \cdot \Delta h_{air} = \rho_{Hg} \cdot \Delta h_{Hg}[/tex]

[tex]\Delta h_{air} = \frac{\rho_{Hg}}{\rho_{air}} \cdot \Delta h_{Hg}[/tex]

Where:

[tex]\Delta h_{air}[/tex] - Height difference of the air column, measured in meters.

[tex]\Delta h_{Hg}[/tex] - Height difference of the mercury column, measured in meters.

[tex]\rho_{air}[/tex] - Density of air, measured in kilograms per cubic meter.

[tex]\rho_{Hg}[/tex] - Density of mercury, measured in kilograms per cubic meter.

If [tex]\Delta h_{Hg} = 0.015\,m[/tex], [tex]\rho_{air} = 1.29\,\frac{kg}{m^{3}}[/tex] and [tex]\rho_{Hg} = 13600\,\frac{kg}{m^{3}}[/tex], the height difference of the air column is:

[tex]\Delta h_{air} = \frac{13600\,\frac{kg}{m^{3}} }{1.29\,\frac{kg}{m^{3}} }\times (0.015\,m)[/tex]

[tex]\Delta h_{air} = 158.140\,m[/tex]

The height of the building is 158.140 meters.

158.13m

Force exerted over a unit area is called Pressure. Also, in a given column of air, the pressure(P) is given as the product of the density(ρ) of the air, the height(h) of the column of air and the acceleration due to gravity(g). i.e

P = ρhg

Let;

Pressure measured at the roof top =  ([tex]P_{R}[/tex])

Pressure measured at the ground level =  ([tex]P_{G}[/tex])

Pressure at the ground level = Pressure at the roof + Pressure at the column height of air.

[tex]P_{G}[/tex] = [tex]P_{R}[/tex] + P               ---------------(i)

(a) P = ρhg             -----------(***)

But;

ρ = density of air = 1.29kg/m³  

h = height of column of air = height of building

g = acceleration due to gravity = 10m/s²

Substitute these values into equation (***)

P = 1.29 x h x 10

P = 12.9h Pa

(b) [tex]P_{G}[/tex] =  ρ[tex]_{mercury}[/tex] x h[tex]_{(mercury)_{ground} }[/tex] x g ------------(*)

But;

ρ[tex]_{mercury}[/tex] = density of mercury = 13600kg/m³  

h[tex]_{(mercury)_{ground} }[/tex] = height of mercury on the ground = 760.0mm = 0.76m

g = acceleration due to gravity = 10m/s²

Substitute these values into equation (*)

[tex]P_{G}[/tex] =  13600 x 0.76 x 10

[tex]P_{G}[/tex] = 103360 Pa

(c) [tex]P_{R}[/tex] = ρ[tex]_{mercury}[/tex] x h[tex]_{(mercury)_{roof} }[/tex] x g       --------------(**)

But;

ρ[tex]_{mercury}[/tex] = density of mercury = 13600kg/m³  

h[tex]_{(mercury)_{roof} }[/tex] = height of mercury on the roof = 745.0mm = 0.745m

g = acceleration due to gravity = 10m/s²

Substitute these values into equation (**)

[tex]P_{R}[/tex]  =  13600 x 0.745 x 10

[tex]P_{R}[/tex]  = 101320 Pa

(d) Now that we know the values of P, [tex]P_{G}[/tex] and [tex]P_{R}[/tex] , let's substitute them into equation (i) as follows;

[tex]P_{G}[/tex] = [tex]P_{R}[/tex] + P  

103360 = 101320 + 12.9h

Solve for h;

12.9h = 103360 - 101320

12.9h = 2040

h = [tex]\frac{2040}{12.9}[/tex]

h = 158.13m

Therefore, the height of the building is 158.13m

Answer:

1.35 kJ  

Explanation:

KE = ½mv² = ½ × 0.030  kg × (300 m·s⁻¹)² = 1350 J = 1.35 kJ

Given:-

To Find: Kinetic energy of the bullet.

We know,

Eₖ = ½mv²

where,

thus,

Eₖ = ½(30 g)(300 m/s)²

= (15 g)(90000 m²/s²)

= 1350000 g m²/s²

= 1350 kg m²/s²

= 1350 J

= 1.35 kJ (Ans.)

Explanation:

F = ma, and a = Δv / Δt.

F = m Δv / Δt

Given: m = 60 kg and Δv = -30 m/s.

a) Δt = 5.0 s

F = (60 kg) (-30 m/s) / (5.0 s)

F = -360 N

b) Δt = 0.50 s

F = (60 kg) (-30 m/s) / (0.50 s)

F = -3600 N

c) Δt = 0.05 s

F = (60 kg) (-30 m/s) / (0.05 s)

F = -36000 N

360N, 3600N and 36000N forces are required to stop a 60 kg person traveling at 30 m/s during a time of a)5.0 seconds, b) 0.50 seconds, c)0.05 seconds respectively.

To find the force, we need to know about the mathematical formulation of force.

Mathematically, it is written as

F= m×a= m×(∆V/∆t)

Here, initially the velocity of the person is 30m/s. But after applying the force, he came to rest. So his final velocity is 0 m/s. ∆V= 30m/s

When ∆t=5 seconds, F= 60×(30/5)=360N

When ∆t=0.5 seconds, F= 60×(30/0.5)=3600N

When ∆t=0.05 seconds, F= 60×(30/0.05)=36000N

Thus, we can conclude that 360N, 3600N and 36000N forces are required to stop a 60 kg person traveling at 30 m/s during a time of a)5.0 seconds, b) 0.50 seconds, c)0.05 seconds respectively.

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Answer:

a

  [tex]PE = 2.3 *10^{-16} \ J[/tex]

b

 [tex]T = 1.1 *10^{7} \ K[/tex]

Explanation:

From the question we are told that

      The distance of separation is  [tex]d = 1.00 *10^{-12} \ m[/tex]

Generally the electric potential energy can be mathematically represented as

            [tex]PE = \frac{k * q_1 q_2 }{d}[/tex]

Given that in a nuclei the only charged particle is the proton who charge is

     [tex]p = 1.60 *10^{-19} \ C[/tex]

Hence

     [tex]q_1 = q_2 = 1.60 *10 ^{-19} \ C[/tex]

And k is the coulomb constant with values   [tex]k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.N/A2[/tex]

      So we have that

       [tex]PE = \frac{9*10^9 * (1.60 *10^{-19})^2}{ 1.00*10^{-12}}[/tex]

      [tex]PE = 2.3 *10^{-16} \ J[/tex]

The relationship between the electrical potential energy and the temperature is mathematically represented as

         [tex]PE = \frac{3}{2} kT[/tex]

Here  k is  the Boltzmann's constant with value  [tex]k = 1.38*10^{-23} JK^{-1}[/tex]

   making T the subject

       [tex]T = \frac{2}{3} * \frac{PE}{k}[/tex]

substituting values

      [tex]T = \frac{2}{3} * \frac{2.30 *10^{-16}}{ 1.38 *10^{-23}}[/tex]

     [tex]T = 1.1 *10^{7} \ K[/tex]

Answer:

no two electrons in the same atom can have the same set of four quantum numbers

Explanation:

Pauli 's Theory of Exclusion specifies that for all four of its quantum numbers, neither two electrons in the same atom can have similar value.

In a different way, we can say that no more than two electrons can take up the identical orbital, and two electrons must have adversely spin in the identical orbital

Therefore the second option is correct

Answer:

[tex]y_{4}-y_{3}=35.22-11.27=23.95 \mathrm{cm}[/tex]

Explanation:

Given that

Wavelength [tex]\lambda=588 \mathrm{nm}[/tex]

slit separation [tex]\mathrm{d}=2464 \mathrm{nm}[/tex]

slit screen distance [tex]\mathrm{D}=11 \mathrm{cm}[/tex]

We know that for double slit the maxima condition is that

[tex]\operatorname{dsin} \theta=m \lambda[/tex]

[tex]\sin \theta=\frac{m \lambda}{d}[/tex]

[tex]\theta=\sin ^{-1}\left(\frac{\mathrm{m} \lambda}{\mathrm{d}}\right)[/tex]

For small angle approximation, [tex]\sin \theta \approx \tan \theta \approx \theta[/tex]

[tex]\tan \theta=\frac{y_{m}}{D}[/tex]

[tex]y_{m}=D \times \tan \left[\sin ^{-1}\left(\frac{m \lambda}{d}\right)\right][/tex]

Now [tex]y_{4}[/tex] [tex]y_{4}=D \times \tan \left[\sin ^{-1}\left(\frac{4 \lambda}{d}\right)\right]=11 \times \tan \left[\sin ^{-1}\left(\frac{4 \times 588 \mathrm{nm}}{2464 \mathrm{nm}}\right)\right]=35.22 \mathrm{cm}[/tex]

Again [tex]y_{3}=D \times \tan \left[\sin ^{-1}\left(\frac{3 \lambda}{d}\right)\right]=11 \times \tan \left[\sin ^{-1}\left(\frac{3 \times 588 \mathrm{nm}}{2464 \mathrm{nm}}\right)\right]=11.27 \mathrm{cm}[/tex]

Hence [tex]y_{4}-y_{3}=35.22-11.27=23.95 \mathrm{cm}[/tex]

Answer:

4.9 m/s

Hope this helps! ;)

Explanation:

Answer:

 I = 89 N s

Explanation:

Momentum is a concept that tells us how much the amount of movement of a system changes, it is described by the expression

      I = ∫ F dt = F_average   t

where F is the force and t is the time

let's calculate

      I = 445  0.2

      I = 89 N s

Answer:

The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.

Explanation:

By means of the Pascal's Principle, the hydraulic lift can be modelled by the following two equations:

Hydraulic Lift - Before change

[tex]P = \frac{F}{A}[/tex]

Hydraulic Lift - After change

[tex]P + \Delta P = \frac{F + \Delta F}{A}[/tex]

Where:

[tex]P[/tex] - Hydrostatic pressure, measured in pascals.

[tex]\Delta P[/tex] - Change in hydrostatic pressure, measured in pascals.

[tex]A[/tex] - Cross sectional area of the hydraulic lift, measured in square meters.

[tex]F[/tex] - Hydrostatic force, measured in newtons.

[tex]\Delta F[/tex] - Change in hydrostatic force, measured in newtons.

The additional weight is obtained after some algebraic handling and the replacing of all inputs:

[tex]\frac{F}{A} + \Delta P = \frac{F}{A} + \frac{\Delta F}{A}[/tex]

[tex]\Delta P = \frac{\Delta F}{A}[/tex]

[tex]\Delta F = A\cdot \Delta P[/tex]

Given that [tex]\Delta P = 100\,Pa[/tex] and [tex]A = 25\,m^{2}[/tex], the additional weight is:

[tex]\Delta F = (25\,m^{2})\cdot (100\,Pa)[/tex]

[tex]\Delta F = 2500\,N[/tex]

The additional mass needed for the additional weight is:

[tex]\Delta m = \frac{\Delta F}{g}[/tex]

Where:

[tex]\Delta F[/tex] - Additional weight, measured in newtons.

[tex]\Delta m[/tex] - Additional mass, measured in kilograms.

[tex]g[/tex] - Gravitational constant, measured in meters per square second.

If [tex]\Delta F = 2500\,N[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then:

[tex]\Delta m = \frac{2500\,N}{9.807\,\frac{m}{s^{2}} }[/tex]

[tex]\Delta m = 254.92\,kg[/tex]

The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.

Answer:

The radius of the sphere is 4.05 m

Explanation:

Given;

potential at surface, [tex]V_s[/tex] = 450 V

potential at radial distance, [tex]V_r[/tex] = 150

radial distance, l = 8.1 m

Apply Coulomb's law of electrostatic force;

[tex]V = \frac{KQ}{r} \\\\V_s = \frac{KQ}{r} \\\\V_r = \frac{KQ}{r+ l}[/tex]

[tex]450 = \frac{KQ}{r} ------equation (i)\\\\150 = \frac{KQ}{r+8.1} ------equation (ii)\\\\divide \ equation (i)\ by \ equation \ (ii)\\\\\frac{450}{150} = (\frac{KQ}{r} )*(\frac{r+8.1}{KQ} )\\\\3 = \frac{r+8.1}{r} \\\\3r = r + 8.1\\\\2r = 8.1\\\\r = \frac{8.1}{2} \\\\r = 4.05 \ m[/tex]

Therefore, the radius of the sphere is 4.05 m

Answer:

E = (1 / 4π ε₀ )  q r / R³

Explanation:

Thomson's stable model that the negative charge is mobile within the atom and the positive charge is uniformly distributed, to calculate the force we can use Coulomb's law

       F = K q₁ q₂ / r²

we used law Gauss

Ф = ∫ E .dA = q_{int} /ε₀

E 4π r² = q_{int} /ε₀  

E = q_{int} / 4π ε₀ r²

we replace the charge inside  

E = (1 / 4π ε₀ r²) ρ 4/3 π r³  

E = ρ r / 3 ε₀

the density for the entire atom is  

ρ = Q / V  

V = 4/3 π R³  

we substitute  

E = (r / 3ε₀ ) Q 3/4π R³  

E = (1 / 4π ε₀ ) q r / R³

Answer:

(e) 3.2

Explanation:

We are given that vector C and D.

Let R be the magnitude of C+D.

According to question

R=3D

We have to find the ratio of the magnitude of C to that of D.

By using right triangle property

[tex]C^2=R^2+D^2[/tex]

[tex]C^2=(3D)^2+D^2[/tex]

[tex]C^2=9D^2+D^2[/tex]

[tex]C^2=10D^2[/tex]

[tex]C=\sqrt{10D^2}=3.2D[/tex]

[tex]\frac{C}{D}=3.2[/tex]

Hence, the ratio of the magnitude of C to that of D=3.2

(e) 3.2

Answer:

20 J

Explanation:

From the question, since there is a lost in kinetic energy, Then the collision is an inelastic collision.

m'u'+mu = V(m+m')........... Equation 1

Where m' = mass of the moving object, m = mass of the stick, u' = initial velocity of the moving object, initial velocity of the stick, V = common velocity after collision.

make V the subject of the equation above

V = (m'u'+mu)/(m+m')............. Equation 2

Given: m' = 2 kg, m = 8 kg, u' = 5 m/s, u = 0 m/s (at rest).

Substitute into equation 2

V = [(2×5)+(8×0)]/(2+8)

V = 10/10

V = 1 m/s.

Lost in kinetic energy = Total kinetic energy before collision- total kinetic energy after collision

Total kinetic energy before collision = 1/2(2)(5²) = 25 J

Total kinetic energy after collision = 1/2(2)(1²) +1/2(8)(1²) = 1+4 = 5 J

Lost in kinetic energy = 25-5 = 20 J

The collision is inelastic collision. As a result of collision the kinetic energy lost by the given system is 20 J.

Since there is a lost in kinetic energy, the collision is inelastic collision.  

m'u'+mu = V(m+m')

[tex]\bold {V =\dfrac { (m'u'+mu)}{(m+m')} }[/tex]  

Where

m' = mass of the moving object = 2 kg

m = mass of the stick = 8 kg,

u' = initial velocity of the moving object = 5 m/s

V = common velocity after collision= ?    

u = 0 m/s (at rest).

put the values in the formula,  

[tex]\bold {V = \dfrac {(2\times 5)+(8\times 0)}{(2+8)}}\\\\\bold {V = \dfrac {10}{10}}\\\\\bold {V = 1\\ m/s.}[/tex]

  kinetic energy before collision

[tex]\bold { = \dfrac 1{2} (2)(5^2) = 25 J}[/tex]  

kinetic energy after collision

[tex]\bold { = \dfrac 12(2)(1^2) + \dfrac 12(8)(1^2) = 5\ J}[/tex]  

Lost in kinetic energy = 25-5 = 20 J

Therefore, As a result of collision the kinetic energy lost by the given system is 20 J.

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Answer:

Differences between regular and irregular objects are:

Hope this helps...

Good luck on your assignment..

Answer:

Φ = 361872 N.m^2 / C

Explanation:

Given:-

- The area of the two plates, [tex]A_p = 180 cm^2[/tex]

- The charge on each plate, [tex]q = 17 * 10^-^6 C[/tex]

- Permittivity of free space, [tex]e_o = 8.85 * 10^-^1^2 \frac{C^2}{N.m^2}[/tex]

- The radius for the flux region, [tex]r = 3.3 cm[/tex]

- The angle between normal to region and perpendicular to plates, θ = 4°

Find:-

Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates.

Solution:-

- First we will determine the area of the region ( Ar ) by using the formula for the area of a circle as follows. The region has a radius of r = 3.3 cm:

                             [tex]A_r = \pi *r^2\\\\A_r = \pi *(0.033)^2\\\\A_r = 0.00342 m^2[/tex]

- The charge density ( σ ) would be considered to be uniform for both plates. It is expressed as the ratio of the charge ( q ) on each plate and its area ( A_p ):

                           σ = [tex]\frac{q}{A_p} = \frac{17*10^-^6}{0.018} \\[/tex]

                           σ = 0.00094 C / m^2

- We will assume the electric field due to the positive charged plate ( E+ ) / negative charged plate ( E- ) to be equivalent to the electric field ( E ) of an infinitely large charged plate with uniform charge density.

                         [tex]E+ = E- = \frac{sigma}{2*e_o} \\\\[/tex]

- The electric field experienced by a region between two infinitely long charged plates with uniform charge density is the resultant effect of both plates. So from the principle of super-position we have the following net uniform electric field ( E_net ) between the two plates:

                        [tex]E_n_e_t = (E+) + ( E-)\\\\E_n_e_t = \frac{0.00094}{8.85*10^-^1^2} \\\\E_n_e_t = 106214689.26553 \frac{N}{C} \\[/tex]

- From the Gauss-Law the flux ( Φ ) through a region under uniform electric field ( E_net ) at an angle of ( θ ) is:

                        Φ = E_net * Ar * cos ( θ )

                        Φ = (106214689.26553) * (0.00342) * cos ( 5 )

                        Φ = 361872 N.m^2 / C

Answer:

F₂ = 925.92 N

Explanation:

In a hydraulic lift the normal stress applied to one arm must be equally transmitted to the other arm. Therefore,

σ₁ = σ₂

F₁/A₁ = F₂/A₂

F₂ = F₁ A₂/A₁

where,

F₂ = Initial force that must be applied to narrow arm = ?

F₁ = Load on Wider Arm to be raised = 12000 N

A₁ = Area of wider arm = πr₁² = π(18 cm)² = 324π cm²

A₂ = Area of narrow arm = πr₂² = π(5 cm)² = 25π cm²

Therefore,

F₂ = (12000 N)(25π cm²)/(324π cm²)

F₂ = 925.92 N

Answer: Acceleration of the car at time = 10 sec is 108 [tex]m/s^{2}[/tex] and velocity of the car at time t = 10 sec is 918.34 m/s.

Explanation:

The expression used will be as follows.

[tex]M\frac{dv}{dt} = u\frac{dM}{dt}[/tex]

[tex]\int_{t_{o}}^{t_{f}} \frac{dv}{dt} dt = u\int_{t_{o}}^{t_{f}} \frac{1}{M} \frac{dM}{dt} dt[/tex]

       = [tex]u\int_{M_{o}}^{M_{f}} \frac{dM}{M}[/tex]

[tex]v_{f} - v_{o} = u ln \frac{M_{f}}{M_{o}}[/tex]

[tex]v_{o} = 0[/tex]

As, [tex]v_{f} = u ln (\frac{M_{f}}{M_{o}})[/tex]

u = -2900 m/s

[tex]M_{f} = M_{o} - m \times t_{f}[/tex]

           = [tex]2500 kg + 1000 kg - 95 kg \times t_{f}s[/tex]

           = [tex](3500 - 95t_{f})s[/tex]

[tex]v_{f} = -2900 ln(\frac{3500 - 95 t_{f}}{3500}) m/s[/tex]

Also, we know that

     a = [tex]\frac{dv_{f}}{dt_{f}} = \frac{u}{M} \frac{dM}{dt}[/tex]

        = [tex]\frac{u}{3500 - 95 t} \times (-95) m/s^{2}[/tex]

        = [tex]\frac{95 \times 2900}{3500 - 95t} m/s^{2}[/tex]

At t = 10 sec,

[tex]v_{f}[/tex] = 918.34 m/s

and,   a = 108 [tex]m/s^{2}[/tex]

Answer:

0.218

Explanation:

Given that

Total vibrations completed by the wave is 43 vibrations

Time taken to complete the vibrations is 33 seconds

Length of the wave is 424 cm = 4.24 m

to solve this problem, we first find the frequency.

Frequency, F = 43 / 33 hz

Frequency, F = 1.3 hz

Also, we find the wave velocity. Which is gotten using the relation,

Wave velocity = 4.24 / 15

Wave velocity = 0.283 m/s

Now, to get our answer, we use the formula.

Frequency * Wavelength = Wave Velocity

Wavelength = Wave Velocity / Frequency

Wavelength = 0.283 / 1.3

Wavelength = 0.218

Answer:

t = 0.414s

Explanation:

In order to calculate the time that the ball takes to reach home plate, you assume that the speed of the ball is constant, and you use the following formula:

[tex]t=\frac{d}{v}[/tex]         (1)

d: distance to the plate = 18.4m

v: speed of the ball = 160.0km/h

You first convert the units of the sped of the ball to appropriate units (m/s)

[tex]160.0\frac{km}{h}*\frac{1h}{3600s}*\frac{1000m}{1km}=44.44\frac{m}{s}[/tex]

Then, you replace the values of the speed v and distance s in the equation (1):

[tex]t=\frac{18.4m}{44.44m/s}=0.414s[/tex]

THe ball takes 0.414s to reach the home plate

Answer:

The distance is  [tex]d = 1.5 *10^{15} \ km[/tex]

Explanation:

From the question we are told that

        The smallest shift is [tex]d = 0.2 \ grid \ units[/tex]

Generally a grid unit is  [tex]\frac{1}{10}[/tex] of  an arcsec

  This implies that  0.2 grid unit is  [tex]k = \frac{0.2}{10} = 0.02 \ arc sec[/tex]

The maximum distance at which a star can be located and still have a measurable parallax is mathematically represented as

           [tex]d = \frac{1}{k}[/tex]

substituting values

           [tex]d = \frac{1}{0.02}[/tex]

           [tex]d = 50 \ parsec[/tex]

Note  [tex]1 \ parsec \ \to 3.26 \ light \ year \ \to 3.086*10^{13} \ km[/tex]

So  [tex]d = 50 * 3.08 *10^{13}[/tex]

     [tex]d = 1.5 *10^{15} \ km[/tex]

Answer:

l these errors believe that the speed of the system is less than that calculated

Explanation:

When we carry out any measurement in addition to the magnitude, the sources of uncertainty must also be analyzed.

We can have random uncertainties, correspondin

g to momentary errors, for example early warps during medicine, parallax errors, errors in the starting and ending points of the movement; I mean every possible random error. This error is the one that is analyzed and calculated in the statistical equations

There is another source of error, the systematic ones, these are much more complicated, they can be an error in the pendulum length, friction in the pendulum movement mechanism, deformities in the support systems, this errors are not analyzed by the statistic, in general They discover by looking at the results and comparing with the tabulated or real ones.

tith the explanation we see that the errors described are systematic.

In general these errors believe that the speed of the system is less than that calculated

Given that,

Sun exposure = 30%, 45%, 60%, 75%, 90%

Stem mass (g) = 275, 415, 563, 815, 954

Stem volume (ml) = 1100, 1215, 1425, 1610, 1742

(a). We need to convert the mass measurements to kilograms (kg) and the volume measurements to cubic meters

Using conversion of mass

[tex]1\ g=0.001\ kg[/tex]

Conservation of volume

[tex]1\ Lt=0.001\ m^3[/tex]

[tex]1\ mL=1\times10^{-6}\ m^3[/tex]

So, mass in kg

Stem mass (kg) = 0.275, 0.415, 0.563, 0.815, 0.954

Volume in m³,

Stem volume (m³) = 0.0011, 0.001215, 0.001425, 0.001610, 0.001742

(b). We need to calculate the density of the samples

Using formula of density

[tex]\rho=\dfrac{m}{V}[/tex]

Where, m = mass

V = volume

If the m = 0.275 kg and V = 0.0011 m³

Put the value into the formula

[tex]\rho=\dfrac{0.275}{0.0011}[/tex]

[tex]\rho=250\ kg/m^3[/tex]

If the m = 0.415 kg and V = 0.001215 m³

Put the value into the formula

[tex]\rho=\dfrac{0.415}{0.001215}[/tex]

[tex]\rho=341.56\ kg/m^3[/tex]

[tex]\rho=342\ kg/m^3[/tex]

If the m = 0.563 kg and V = 0.001425 m³

Put the value into the formula

[tex]\rho=\dfrac{0.563}{0.001425}[/tex]

[tex]\rho=395.08\ kg/m^3[/tex]

If the m = 0.815 kg and V = 0.001610 m³

Put the value into the formula

[tex]\rho=\dfrac{0.815}{0.001610}[/tex]

[tex]\rho=506.21\ kg/m^3[/tex]

If the m = 0.954 kg and V = 0.001742 m³

Put the value into the formula

[tex]\rho=\dfrac{0.954}{0.001742}[/tex]

[tex]\rho=547.6\ kg/m^3[/tex]

[tex]\rho=548\ kg/m^3[/tex]

(c). We need to convert the density values to scientific notation

In scientific notation

The densities are

[tex]\rho\ (kg/m^3)= 2.50\times10^{2}, 3.42\times10^{2}, 3.95\times10^{2}, 5.06\times10^{2}, 5.48\times10^{2}[/tex]

Hence, This is required solution.

Complete Question:

The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 43 mm , while nonathletes' stretch only 32 mm . The spring constant for the tendon is the same for both groups, [tex]31 {\rm {N}/{mm}}[/tex]. What is the difference in maximum stored energy between the sprinters and the nonathlethes?

Answer:

[tex]\triangle E = 12.79 J[/tex]

Explanation:

Sprinters' tendons stretch, [tex]x_s = 43 mm = 0.043 m[/tex]

Non athletes' stretch, [tex]x_n = 32 mm = 0.032 m[/tex]

Spring constant for the two groups, k = 31 N/mm = 3100 N/m

Maximum Energy stored in the sprinter, [tex]E_s = 0.5kx_s^2[/tex]

Maximum energy stored in the non athletes, [tex]E_m = 0.5kx_n^2[/tex]

Difference in maximum stored energy between the sprinters and the non-athlethes:

[tex]\triangle E = E_s - E_n = 0.5k(x_s^2 - x_n^2)\\\triangle E = 0.5*3100* (0.043^2 - 0.032^2)\\\triangle E = 0.5*31000*0.000825\\\triangle E = 12.79 J[/tex]

Answer:

Final speed = 2.067 m/s

Explanation:

We are told that the child weighs 26 kg.

Also, that the wagon weighs 5kg.

Thus,initial mass of the child and wagon with ball is;

m_i = 26 + 5 = 31 kg.

Also, we are told that the child now dropped 1.5 kg ball from the wagon. So,

Final mass is;

m_f = 26 + 5 - 1 = 30 kg

Now, from conservation of linear momentum, we know that;

Initial momentum = final momentum

Thus;

m_i * v_i = m_f * v_f

Where v_i is initial velocity and v_f is final velocity.

Making v_f the subject, we have;

v_f = (m_i * v_i)/m_f

We are given that initial velocity v_i = 2 m/s

Plugging in the relevant values, we have;

v_f = (31 * 2)/30

v_f = 2.067 m/s

Answer:

The found acceleration in terms of h and t is:

[tex]a=\frac{h}{5(t_1)^2}[/tex]

Explanation:

(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)

We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.

Constant acceleration, starts from rest.

Distance = [tex]y = \frac{1}{2}a(t_1)^2[/tex]

Velocity = [tex]v_1=at_1[/tex]

Constant velocity where

Velocity = [tex]v_o=v_1=at_1[/tex]

Distance =

Constant deceleration where

Velocity = [tex]v_0=v_1=at_1[/tex]

Distance =

[tex]y_3=v_1t_3-\frac{1}{2}a(t_3)^2\\\text{Where}~t_3=t_1\\y_3=v_1t_1-\frac{1}{2}a(t_1)^2\\\text{Where}~ v_1t_1=a(t_1)^2\\y_3=a(t_1)^2-\frac{1}{2}a(t_1)^2\\\text{Subtracting both terms:}\\y_3=\frac{1}{2}a(t_1)^2[/tex]

Total height = y₁ + y₂ + y₃

Total height = [tex]\frac{1}{2}a(t_1)^2+4a(t_1)^2+\frac{1}{2}a(t_1)^2 = 5a(t_1)^2[/tex]

Find acceleration by rearranging the found equation of total height.

Total Height = h

h = 5a(t₁)²

[tex]a=\frac{h}{5(t_1)^2}[/tex]