A parallel-plate, air-filled capacitor is being charged as in Fig. 29.23. The circular plates have radius 4.10 cm, and at a .. 29.47 particular instant the conduction current in the wires is 0.276 A. (a) What is the displacement current density jD in the air space between the plates? (b) What is the rate at which the electric field between the plates is changing? (c) W

Answer: Their final relative velocity is -0.412 m/s.

Explanation:

According to the law of conservation,

      [tex]m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v[/tex]

Putting the given values into the above formula as follows.

      [tex]m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v[/tex]

     [tex]2.50 \times 10^{3} kg \times 0 m/s + 7.50 \times 10^{3} kg \times -0.550 m/s = (2.50 \times 10^{3} kg + 7.50 \times 10^{3} kg)v[/tex]

           [tex]-4.12 \times 10^{3} kg m/s = (10^{4} kg) v[/tex]

                   v = [tex]\frac{-4.12 \times 10^{3} kg m/s}{10^{4} kg}[/tex]

                      = -0.412 m/s

Thus, we can conclude that their final relative velocity is -0.412 m/s.

Answer:

18.8m

Explanation:

Let the distance traveled before stopping be 'd' m.

Given:

Mass of the skater (m) = 90 kg

Initial velocity of the skater (u) = 12.0 m/s

Final velocity of the skater (v) = 0 m/s (Stops finally)

Coefficient of kinetic friction (μ) = 0.390

Acceleration due to gravity (g) = 9.8 m/s²

Now, we know  from work-energy theorem, that the work done by the net force on a body is equal to the change in its kinetic energy.

Hence, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.

Frictional force is given as:

[tex]f=\mu N[/tex]

Where, 'N' is the normal force acting on the skater. As there is no vertical motion, N=mg

[tex]f=\mu mg\\=0.390\times 90\times 9.8\\=343.98\ N[/tex]

work done by friction is a negative work as friction and displacement are in opposite direction and is given as

[tex]W=-fd=-343.98d[/tex]

Now, change in kinetic energy is given as:

[tex]\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J[/tex]

Therefore, from work-energy theorem,

[tex]W=\Delta K\\\\-343.98d=6480\\\\d=\frac{6480}{343.98}\\\\d=18.8m[/tex]

Hence, the skater covers a distance of 18.8 m before stopping.

Answer:

18.84m

Explanation:

We are given;

Mass; m = 90 Kg

Initial velocity; u = 12 m/s

Coefficient of friction; μ = 0.390

Let,the combined mass of the ice skater and the skate be M

Thus, So,if the coefficient of frictional force is μ,then frictional force acting is; μN. N= Mg. Thus F_friction = μMg

Now, the deceleration due to friction will be, F_friction/M

Thus, deceleration = μMg/M

M will cancel out and we have; μg

Now, deceleration means negative acceleration. Thus acceleration (a) =

-μg

Now, to find the distance, let's use equation of motion which is;

v² = u² + 2as

Putting -μg for a, we have;

v² = u² + 2(-μg)s

v² = u² — 2μgs

We want to know the distance covered before coming to rest. Thus, v = 0m/s

So,

0² = 12² - (2 x 0.39 x 9.8 x s)

0 = 144 - 7.644s

Thus,

7.644s = 144

Thus, s = 144/7.644 = 18.84m

if two objects each have a mass of 10 kg, then the force of gravity between them C) is greater when they are closer together.

We have two objects, each with a mass of 10 kg. We can calculate the force of gravity (F) between them using Newton's law of universal gravitation.

[tex]F = G \frac{m_1m_2}{r^{2} }[/tex]

where,

As we can see from this expression. the gravitational force between the objects depends on their masses and the distance between them. The closer they are, the stronger the gravity force.

if two objects each have a mass of 10 kg, then the force of gravity between them...

A) is constant. No. It varies with the distance.

B) is 100 kg. No. kg is not a unit of force.

C) is greater when they are closer together. Yes.

D) depends only on their masses. No. It also depends on the distance between them.

if two objects each have a mass of 10 kg, then the force of gravity between them C) is greater when they are closer together.

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Answer:

The frequency of the sona-pulse reflected back is  [tex]f_b = 48109.22Hz[/tex]

Explanation:

From the question we are told that

     The speed of the bat is  [tex]v = 5.1 m/s[/tex]

      The speed of the insect is [tex]v_i = 1.1 m/s[/tex]

       The frequency emitted by the bat is [tex]f = 47000 \ Hz[/tex]

        The speed of sound is  [tex]v_s = 343 m/s[/tex]

Let look at this question in this manner

At the first instant the that the bat  emits the sonar pulse

     Let the bat be the source of sound

      Let the insect be the observer

This implies that the frequency of sound the the insect would receive is mathematically represented as

                    [tex]f_a = [\frac{v_s - v_i}{v_s - v}] f[/tex]

Substituting values

                 [tex]f_a = [\frac{343 - 1.1}{345 -5.1} ] * 47000[/tex]

                     [tex]f_a = 47556.4 Hz[/tex]

Now at the instant the sonar pules reaches the insect

            Let the bat be the observer

            Let the insect be the source of the sound

Here the sound wave is reflected back to the bat

This implies that the frequency of sound the the bat  would receive is mathematically represented as

                   [tex]f_b =[ \frac{x}{y} ] * f_a[/tex]

                   [tex]f_b =[ \frac{v_s + v}{v_s + v_i} ] * f_a[/tex]

                  [tex]f_b =[ \frac{343 + 5.1}{343 + 1.1} ] * 47556.4[/tex]

                  [tex]f_b = 48109.22Hz[/tex]

Final answer:

To calculate the frequency a bat hears from a sonar pulse reflected back by an insect, the Doppler effect formula is used. By substituting the given values into the formula, the result is approximately 47,544 Hz, which is the frequency at which the bat detects the echo from the insect.

Explanation:

To calculate the frequency at which a bat hears the sonar pulse reflected back from an insect, we can use the Doppler effect formula for sound in the same direction as the source of the sound is moving:

[tex]f' =\frac{f (v + v_d)}{(v + v_s)}[/tex]

Where:

f' is the frequency heard by the bat.

f is the emitted frequency of the bat's sonar pulse (47,000 Hz).

v is the speed of sound (343 m/s).

[tex]v_d[/tex] is the speed of the detector (bat), which is 5.1 m/s.

[tex]v_s[/tex] is the speed of the source (insect), which is 1.1 m/s.

Substituting the values into the equation:

[tex]f' = \frac{47000 Hz \times (343 m/s + 5.1 m/s)}{(343 m/s + 1.1 m/s)}[/tex]

[tex]f' = \frac{47000 Hz * 348.1 m/s}{344.1 m/s}[/tex]

[tex]f' = 47000 Hz \times 1.0116[/tex]

f' = 47544.2 Hz

Therefore, the bat hears the pulse reflected back from the insect at approximately 47,544 Hz.

Answer:

True.

The day a huge asteroid strikes the earth will likely be a weekday.

Explanation:

Mathematically, there are 5 weekdays in the week and there are only 2 weekend days.

Total number of days in a week = 7

Since, the asteroids can hit on any day,

The probability of the asteroid hitting on a weekend day = (2/7) = 0.2857

The probability that an asteroid hitting on a weekday = (5/7) = 0.7143.

So, by the virtue of the probability of the asteroid hitting on a weekday or on a weekend day, the asteroid is more likely to hit on a weekday as

P(weekday) > P(weekend)

0.7143 > 0.2857.

Hope this Helps!!!

The work done by the gas when the volume increases from 12.0 m³ to 16.2 m³ at atmospheric pressure is 425.565 kJ. The change in internal energy of the gas when 254 kcal of heat is added is 636.931 kJ.

To calculate the work done by the gas during a quasi-static expansion, we can use the formula W = P ΔV, where W is work, P is pressure, and ΔV is the change in volume. Given the atmospheric pressure is 1 atm or 101,325 Pa, and the volume change is from 12.0 m³ to 16.2 m³, the work done by the gas can be calculated as:

W = P ΔV = 101,325 Pa × (16.2 m³ - 12.0 m³)

W = 101,325 Pa × 4.2 m³

W = 425,565 J or 425.565 kJ

To calculate the change in internal energy (ΔU) of the gas, we can use the first law of thermodynamics, which states that ΔU = Q - W, where Q is the heat added to the system. The heat is given as 254 kcal, which needs to be converted to joules (1 kcal = 4.184 kJ).

ΔU = Q - W = (254 kcal × 4.184 kJ/kcal) - 425.565 kJ

ΔU = 1062.496 kJ - 425.565 kJ

ΔU = 636.931 kJ

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Answer:

The strength of the magnetic field is 0.0842 mT

Explanation:

Given:

Velocity of rod [tex]v = 3.07[/tex] [tex]\frac{m}{s}[/tex]

Length of rod [tex]l = 1.11[/tex] m

Induced emf across the rod [tex]\epsilon = 0.287 \times 10^{-3}[/tex] V

According to the faraday's law

We have a special case for moving rod in magnetic field.

Induced emf in moving rod is given by,

   [tex]\epsilon = Blv[/tex]

Where [tex]B =[/tex] strength of magnetic field

  [tex]B = \frac{\epsilon}{lv}[/tex]

  [tex]B = \frac{0.287 \times 10^{-3} }{3.07 \times 1.11}[/tex]

  [tex]B = 0.0842 \times 10^{-3}[/tex] T

  [tex]B = 0.0842[/tex] mT

Therefore, the strength of the magnetic field is 0.0842 mT

Answer:

About eq (1)

[tex]mv^2/r = eVB[/tex]

when a charged particle (electron) enters into the magnetic field which is perpendicular to direction of motion than there will be magnetic force on particle and particle will travel in circular path in with constant speed.

So using force balance on charged particle:

[tex]F_{net} = Fc - Fm[/tex]

Since particle is traveling at constant speed, So acceleration is zero, and

[tex]F_{net} = 0[/tex]

[tex]Fc - Fm=0[/tex]

[tex]Fc = Fm[/tex]

Fc = centripetal force on particle [tex]= m*v^2/r[/tex]

Fm = magnetic force on electron = [tex]q*VxB = q*V*B*sin \theta[/tex]

q = charge on electron = e

since magnetic field is perpendicular to the velocity of particle, So theta = 90 deg

sin 90 deg = 1

So,

[tex]m*v^2/r = e*v*B[/tex]

About equation 2:

When this charged particle is released from rest in a potential difference V, and then it enters into above magnetic field, then using energy conservation on charge particle

[tex]KEi + PEi = KEf + PEf[/tex]

KEi = 0, since charged particle started from rest

[tex]PEi - PEf = q*dV[/tex]

[tex]PEi - PEf = eV[/tex]

KEf = final kinetic energy of particle when it leaves [tex]= (1/2)*m*v^2[/tex]

So,

[tex]0 + eV = (1/2)*m*v^2[/tex]

[tex]eV = (1/2)*m*v^2[/tex]

From above equation (1) and (2)

[tex]m*v^2/r = evB[/tex]

[tex]e/m = v/(r*B)[/tex]

Now

[tex]eV = (1/2)*m*v^2[/tex]

[tex]v = \sqrt{(2*e*V/m)}[/tex]

[tex]e/m = \sqrt{ (2*e*V/m)/(r*B)}[/tex]

[tex]\frac{e^2}{m^2} = \frac{2*e*V}{(m*r^2*B^2)}[/tex]

[tex]e/m = 2*V/(r^2*B^2)[/tex]

[tex]e/m = 2V/(Br)^2[/tex]

Answer:

0.2923 V

Explanation:

Given that

Angle between the rails, θ = 19°

Speed of the rod, v = 0.6 m/s

Magnetic field present, B = 0.63 T

Time used, t = 7.5 s

E = -ΔΦ/Δt

where, Φ = BA, so

E = -BΔA / Δt

To get the area, if we assume the rails are joined in a triangular fashion(see attachment)

E = -B(1/2 * AC * BC) / Δt

E = -B(vΔt * vΔt tanθ) / 2Δt

E = -(B * v² * Δt² * tanθ) / 2Δt

E = -Bv²Δt.tanθ/2

E = -(0.63 * 0.6² * 7.5 * tan 19) / 2

E = -0.5857 / 2

E = -0.2923

Thus, the magnitude of average emf induced if 0.2923 V

Answer:

 m_{p} = 0.3506 kg

Explanation:

For this exercise we use Newton's equilibrium equation

      B - Wc-Wp = 0

where B is the thrust of the water, Wc is the weight of the coins and Wp is the weight of the plastic block

       B = Wc + Wp

the state push for the Archimeas equation

      B = ρ_water g V

the volume of the water is the area of ​​the block times the submerged height h, which is

        h´ = 8 - h

where h is the height out of the water

      ρ_water g A h´ = [tex]m_{c}[/tex] g + [tex]m_{p}[/tex] g

      ρ_water A h´ = m_{c} + m_{p}

write this equation to make the graph

        h´= 1 /ρ_water A    (m_{c} +m_{p})

        h´ = 1 /ρ_waterA    (m_{c} + m_{p})

if we graph this expression, we get an equation of the line

        y = m x + b

where

        y = h´

        m = 1 /ρ_water A

        b = mp /ρ_water A

whereby

       m_{p} = b ρ_water A

       ρ_water = 1000 kg / m³

       b = 0.0312 m

       m = 0.0890 m / kg

we substitute the slope equation

       b = m_{p} / m

calculate

       m_{p}= 0.0312 / 0.0890

       m_{p} = 0.3506 kg

Final answer:

To find the voltage across the secondary winding of a step-up transformer with a primary voltage of 141 V and a turns ratio of 1110 to 5, the secondary voltage is calculated to be 31302 V.

Explanation:

Calculating the Voltage Across the Secondary in a Step-Up Transformer

To calculate the voltage across the secondary winding of a step-up transformer, we use the transformer equation that relates the primary and secondary voltages with the number of turns on the primary (Np) and secondary (Ns) windings:

Vs / Vp = Ns / Np

Given that the primary voltage (Vp) is 141 V, and the ratio of the number of turns is 1110 to 5 (Ns / Np), we can rearrange the equation to solve for the secondary voltage (Vs) as follows:

Vs = Vp × (Ns / Np)

Vs = 141 V × (1110 / 5)

Vs = 141 V × 222

Vs = 31302 V

Thus, the voltage across the secondary winding is 31302 V.

Answer:

Explanation:

A police car's radar gun emits microwaves with frequency f. 1. =36 GHz. The beam reflects from a car that speeds away from the cruiser with 43 m/s. The receiver ...

Answer: The length L of the wire used to construct the coil is 191.4m

Explanation: Please see the attachments below

Answer:

0.002638 C or 2.6388*10^-3 C

Explanation:

Given that

Quantity of the charge, q = -0.0005 C

Force on the charge of magnitude, F1 = 19 N

Distance from the second charge, r = 25 m

Magnitude of force of the second charge, q2 = ? N

F = (kq1q2) / r², where

k = 9*10^9

19 = (9*10^9 * 0.0005 * q2) / 25²

19 * 625 = 4.5*10^6 * q2

q2 = 11875 / 4.5*10^6

q2 = 0.002638 C or 2.6388*10^-3 C

Thus, the magnitude of the second charge is 0.002638 C or 2.6388*10^-3 C

Answer:

False

Explanation:

Friction is the resisting force of motion that converts kinetic energy into heat energy. Kinetic energy is the energy of movement. Removing kinetic energy makes the object slower.

Final answer:

The calculated density ratio is 0.5.

The ratio of the densities of two blocks, A and B, is found by comparing the apparent weight reductions when they are submerged in a fluid. According to Archimedes' principle, this reduction is due to the buoyant force, which is equal to the weight of the fluid displaced.

Explanation:

The question revolves around calculating the ratio of the densities of two blocks, A and B, using the concept of buoyant force according to Archimedes' principle. The apparent weight of each block when submerged in a fluid gives us the buoyant force acting upon it, which is equal to the weight of the fluid displaced. Since both blocks are submerged in the same fluid, the difference in weights in air versus submerged gives us a measure of the volume of fluid displaced due to the respective densities of the blocks.

To determine the ratio of densities (A/B), we use the formula:

Ratio of densities = (Weight of A in air - Apparent weight of A) / (Weight of B in air - Apparent weight of B)

Substituting the given weights:

Ratio of densities = (12 N - 8 N) / (20 N - 12 N) = 4 N / 8 N = 0.5

Thus, the ratio of the densities of block A to block B is 0.5.

Answer:

The answer is C.

Explanation:

Current can be represented in the symbol of I.

The symbol for voltage is V.

The symbol for power is W.

The symbol for resistance is R.

Answer:

The cart would speed up.

Explanation:

According to Newton's 1st law, object subjected to no force, or net force 0, would have a constant speed. In our case the cart is initially at constant speed, meaning the man exerts a force that is equal to friction force. If he increases the force on the cart, the net force would no longer be 0. The cart would gain an acceleration and increases its speed.

Answer:

A. The cart accelerates

Explanation:

Answer:

Explanation:

According to  energy conservation principle;

[tex]\delta \ K.E = \delta \ P.E[/tex]

[tex]8.00*10^{-19} \ J = (1.6*10^{-19} \ C ) |V_b-V_a|[/tex]

[tex]|V_b-V_a|= \frac {8.00*10^{-19} }{(1.6*10^{-19} ) }[/tex]

[tex]|V_b-V_a|= 5 \ V[/tex]

The potential is positive hence the electric fielf=d is negative along the x-axis.

We can then say that the movement of the particle goes to the left through a potential difference of  [tex]|V_b-V_a|= 5 \ V[/tex].

There will be no significant change in the y-direction of the potential energy when the particle moves from point b to point c in the y-direction.

Answer:

Explanation:

velocity of proton  v = 1.5 x 10³ i  m /s

charge on proton e = 1.6 x 10⁻¹⁹ C

Let the magnetic field be B = Bx i + Bz k

force on charged particle ( proton )

F = e ( v x B )

2.06  x10⁻¹⁶ j = 1.6 x 10⁻¹⁹ [ 1.5 x 10³ i x ( Bx i + Bz k) ]

2.06  x10⁻¹⁶ j = - 1.6 x 10⁻¹⁹ x 1.5 x 10³  Bz j) ]

2.06  x10⁻¹⁶ = - 1.6 x 10⁻¹⁹ x 1.5 x 10³  Bz

Bz = -  .8583  

force on charged particle ( electron )

F = e ( v x B )

8.40  x10⁻¹⁶ j = -1.6 x 10⁻¹⁹ [ - 4.4 x 10³ k  x ( Bx i + Bz k) ]

8.4  x10⁻¹⁶ j =  1.6 x 10⁻¹⁹ x 4.4 x 10³  Bx j ]

- 8.4  x10⁻¹⁶ =  1.6 x 10⁻¹⁹ x 4.4 x 10³  Bx

Bx =  - 1.19

Magnetic field = - 1.19 i - .8583 k

magnitude = √ (1.19² + .8583²)

= 1.467 T

If it is making angle θ with x - axis in x -z plane

Tanθ = (.8583 / 1.19 )

36⁰ .

C )

v = - 3.7 x 10³j m /s

e = - 1.6 x 10⁻¹⁶ C

Force = F = e ( v x B )

= -1.6 x 10⁻¹⁹ [ -3.7 x 10³ j  x ( Bx i + Bz k) ]

=  - 1.6 x 10⁻¹⁹ x 3.7 x 10³  Bx  k -1.6 x 10⁻¹⁹ x 3.7 x 10³Bzi ]

=    5.08 i - 7.04 k

Tanθ = 54 ° .

Answer:

A)P = 1.2 × 10⁵Pa

B)F = 3.2 × 10⁷N

C) P = 1.2 × 10⁵Pa

Explanation:

Part A)

The relative pressure at the bottom of a column of fluid is given by

[tex]p_r = \rho g h[/tex]

where

[tex]\rho[/tex] is the fluid density

g is the gravitational acceleration  

h is the height of the column of fluid

At the bottom of the swimming pool, h=1.9 m, and the water density is  

[tex]\rho[/tex] = 1000 kg/m^3, therefore the relative pressure is

[tex]p_r = (1000 kg/m^3)(9.81 m/s^2)(1.9 m)=1.86 \cdot 10^4 Pa[/tex]

To find the absolute pressure, we must add to this the atmospheric pressure, [tex]p_a[/tex] :

[tex]p= p_r + p_a\\= 1.86 \cdot 10^4 Pa + 1.01 \cdot 10^5 Pa \\=1.2 \times 10^5 Pa[/tex]

part B

Total force acting on the bottom

force = pressure * area

area of pool = 30.0 m × 8.9 m

= 267m²

Force F =

1.2 × 10⁵ * 267m² N

= 32040000 N

F = 3.2 × 10⁷N

Part C

The pressure acting on the side wall will be

now the pressure at the side of the pool at the bottom is simply equal to absolute pressure as they are at same level

P = 1.2 × 10⁵

Answer:

0.2cm towards the retina.

Explanation:

the focal length of the frog eye is

(1/f) = (1/10) + (1/0.8)

f = 0.74cm

Since the distance of the object is 15cm Hence

(1/0.74) = (1/15) + (1/V)

V = 0.78cm

Therefore the distance the retina is to move is

0.78cm - 0.8cm = 0.02cm towards the retina.

Answer:

a) [tex]t = 4.6\tau[/tex]

b) [tex]L = 0.0187 \: H[/tex]

Explanation:

The current flowing in a R-L series circuit is given by

[tex]I = I_{0} (1 - e^{\frac{-t}{\tau} })[/tex]

Where τ is the time constant and is given by

[tex]\tau = \frac{L}{R}[/tex]

Where L is the inductance and R is the resistance

Assuming the current has reached steady state when it is at 99% of its maximum value,

[tex]0.99I_{0} = I_{0} (1 - e^{\frac{-t}{\tau} })\\0.99 = (1 - e^{\frac{-t}{\tau} })\\1 - 0.99 = e^{\frac{-t}{\tau}}\\ln(0.01) = ln(e^{\frac{-t}{\tau}})\\-4.6 = \frac{-t}{\tau}\\t = 4.6\tau\\[/tex]

Therefore, it would take t = 4.6τ to reach the steady state.

(b) If an emergency power circuit needs to reach steady state within 1.2 ms of turning on and the circuit has a total resistance of 72 Ω, what values of the total inductance of the circuit are needed to satisfy the requirement?

[tex]t = 4.6\frac{L}{R}\\t = 4.6\frac{L}{R}\\0.0012 = 4.6\frac{L}{72}\\0.0864 = 4.6 L\\L = 0.0864/4.6\\L = 0.0187 \: H[/tex]

Therefore, an inductance of 0.0187 H is needed to satisfy the requirement.

The ranking from largest to smallest the values of the magnetic field should be like

Rb = 4.95 cm

r= 5.40 cm

Ra =7.75 cm

r> Ra

It is a vector field that explained the magnetic impact on moving electric charges, electric currents, and magnetic materials.

In this, the force should be perpendicular to the velocity and the magnetic field. Also, it should be inversely proportional with respect to the distance with the axis of the cylinder.

So,

The largest magnetic field =Rb =4.95 cm

And,

Smallest magnetic field = r>Ra

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The magnetic field inside a conducting cylinder is maximized at certain points before dropping off outside. The ranked values from largest to smallest  are: r = 5.40 cm, Rb = 4.95 cm, Ra = 7.75 cm, and r > Ra.

To determine the magnetic field  at different distances from the axis of a conducting cylinder, we need to understand how the magnetic field varies inside and outside a cylinder carrying a current.

Given the distances Ra = 7.75 cm, Rb = 4.95 cm, r = 5.40 cm, and r > Ra, we will rank the magnetic field intensities:

1.) r = 5.40 cm:

Since r is within the cylinder where current is distributed uniformly, the magnetic field at this point can be calculated using Ampère's Law.

2.) Rb = 4.95 cm:

Still within the cylinder but closer to the center, thus the field here will not yet be maximized.

3.) Ra = 7.75 cm:

Outside the cylinder where the current configuration influences how the field drops off with increasing distance.

4.) r > Ra:

This point is furthest from the axis, hence the magnetic field will be the smallest.

The magnetic field reaches a maximum inside the conducting cylinder before decreasing outside of it.

Answer:

The work done on the jet by the catapult is 1.53 x  10⁷ J

Explanation:

Given;

force of thrust engines, F =  2.3 x 10⁵ N

distance moved by the thrust engines, d =  90 m

kinetic energy of the plane, K.E = 3.60 x 10⁷ J

Based on work-energy theorem;

the net work done on the plane by catapult and the thrust engine is given as;

[tex]W_n = K.E_f = 3.6 *10^7 \ J[/tex]

work done by the thrust engine on the jet;

[tex]W_T_._e_n = fd\\\\W_T_._e_n = 2.3*10^5 * 90\\\\W_T_._e_n = 2.07*10^7 \ J[/tex]

Work done on the jet by the catapult;

[tex]W_c_p= W_n - W_T_._e_g\\\\W_c_p= (3.6*10^7 - 2.07*10^7) J\\\\W_c_p= 1.53*10^7 \ J[/tex]

Therefore, the work done on the jet by the catapult is 1.53 x  10⁷ J

Answer:

15.4 kg.

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

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

Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, V = common velocity of both sphere.

Given: m = 7.7 kg, u' = 0 m/s (at rest)

Let: u = x m/s, and V = 1/3x m/s

Substitute into equation 1

7.7(x)+m'(0) = 1/3x(7.7+m')

7.7x = 1/3x(7.7+m')

7.7 = 1/3(7.7+m')

23.1 = 7.7+m'

m' = 23.1-7.7

m' = 15.4 kg.

Hence the mass of the second sphere = 15.4 kg

Answer:

The mass of the second sphere is 15.4 kg

Explanation:

Given;

mass of the first sphere, m₁ = 7.7 kg

initial velocity of the second sphere, u₂ = 0

let mass of the second sphere =  m₂

let the initial velocity of the first sphere = u₁

final velocity of the composite system, v = ¹/₃ x u₁ = [tex]\frac{u_1}{3}[/tex]

From the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = v(m₁ + m₂)

Substitute the given values;

[tex]7.7u_1 + 0=\frac{u_1}{3} (7.7+m_2)[/tex]

Divide through by u₁

7.7 = ¹/₃(7.7 + m₂)

multiply both sides by 3

23.1 = 7.7 + m₂

m₂ = 23.1 - 7.7

m₂ = 15.4 kg

Therefore, the mass of the second sphere is 15.4 kg

Answer:

0.372 kg

Explanation:

The collision between the bullet and the block is inelastic, so only the total momentum of the system is conserved. So we can write:

[tex]mu=(M+m)v[/tex] (1)

where

[tex]m=11.9 g = 11.9\cdot 10^{-3}kg[/tex] is the mass of the bullet

[tex]u=261 m/s[/tex] is the initial velocity of the bullet

[tex]M[/tex] is the mass of the block

[tex]v[/tex] is the velocity at which the bullet and the block travels after the collision

We also know that the block is attached to a spring, and that the surface over which the block slides after the collision is frictionless. This means that the energy is conserved: so, the total kinetic energy of the block+bullet system just after the collision will entirely convert into elastic potential energy of the spring when the system comes to rest. So we can write

[tex]\frac{1}{2}(M+m)v^2 = \frac{1}{2}kx^2[/tex] (2)

where

k = 205 N/m is the spring constant

x = 35.0 cm = 0.35 m is the compression of the spring

From eq(1) we get

[tex]v=\frac{mu}{M+m}[/tex]

And substituting into eq(2), we can solve to find the mass of the block:

[tex](M+m) \frac{(mu)^2}{(M+m)^2}=kx^2\\\frac{(mu)^2}{M+m}=kx^2\\M+m=\frac{(mu)^2}{kx^2}\\M=\frac{(mu)^2}{kx^2}-m=\frac{(11.9\cdot 10^{-3}\cdot 261)^2}{(205)(0.35)^2}-11.9\cdot 10^{-3}=0.372 kg[/tex]

The mass of the wooden block is approximately 0.404 kg.

The initial kinetic energy of the bullet is given by:

KE_initial = 0.5 * mb * vb^2

where:

`m_b` is the mass of the bullet (11.9 g = 0.0119 kg)

`v_b` is the velocity of the bullet (261 m/s)

The potential energy stored in the compressed spring is given by:

PE_spring = 0.5 * k * x^2

where:

`k` is the spring constant (205 N/m)

`x` is the compression of the spring (0.35 m)

Since the system comes to a stop, the initial kinetic energy is converted to potential energy stored in the spring:

KE_initial = PE_spring

Solving for the mass of the wooden block, we get:

m_w = (2 * KE_initial) / (v_b^2 - (k * x) / m_b)

Plugging in the values, we get,

m_w = (2 * 0.5 * 0.0119 kg * (261 m/s)^2) / ((261 m/s)^2 - (205 N/m * 0.35 m) / 0.0119 kg)

m_w ≈ 0.404 kg

Therefore, the mass of the wooden block is approximately 0.404 kg.

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Answer:

1. Our ears can sort out the individual sine waves from a mixture of two or more sine waves, so we hear the pure tones that make up a complex tone.

Explanation:

A complex tone is a sound wave that consist of two or more forms of audible sound frequencies. Sound wave is a mechanical wave that is longitudinal, and could be represented by a sine wave because of it sinusoidal manner of propagation.

A Fourier analyzer can be used to differentiate individual sine waves from a combination of two or more of it; which is as the same function performed by human ear. To the human ear, a sound wave that consist of more than one sine wave will have perceptible harmonics which would be distorted and turn to a noise.

Thus, the human ear makes it possible to hear the pure tones that make up a complex tone.

Answer:

1. A Fourier analyzer sorts out the individual sine waves from a mixture of two or more sine waves, so we hear the pure tones that make up a complex tone.

Explanation:

Fourier analysis is a technique that is used to determine which sine waves constitute a given signal, i.e. to deconstruct the signal into its individual sine waves.  It is the process of decomposing a periodic function into its constituent sine or cosine waves.

What goes on inside our ears is a mathematical process called a Fourier transform. In the ear, there's a combination of different waves, Fourier analysis identifies contributions at different frequencies, allowing us to reconstruct the individual signals that go into it.

A complex tone perceived by the air is is an individual sine wave that the ear, by acting as a Fourier analyzer, decomposes to serious of sine waves that we hear as pure tones.

Answer:

a) 22.06m/s

b)45°

Explanation:

Let 'm' be  mass of 1st and 2nd pieces

mass of 3rd piece=3m

[tex]v_{1}[/tex]=velocity of 1st piece = -47iˆ  m/s

[tex]v_{2}[/tex]=velocity of 2nd piece = -47jˆ  m/s

[tex]v_{3}[/tex]=velocity of 3rd piece=?

By considering conservation of linear momentum, we have

m[tex]v_{1}[/tex] + m[tex]v_{2}[/tex] + 3m[tex]v_{3}[/tex]=0

[tex]v_{1}[/tex] + [tex]v_{2}[/tex] + 3[tex]v_{3}[/tex]=0

3[tex]v_{3}[/tex]= - ([tex]v_{1}[/tex] + [tex]v_{2}[/tex] )

[tex]v_{3}[/tex] = -[tex]\frac{1}{3}[/tex] ([tex]v_{1}[/tex] + [tex]v_{2}[/tex] )

Substituting the values of [tex]v_{1}[/tex] and [tex]v_{2}[/tex] in above equation

[tex]v_{3}[/tex] = -[tex]\frac{1}{3}[/tex] (-47iˆ -47jˆ ) => 15.6iˆ + 15.6jˆ

(a)Magnitude of the velocity of  the third piece is given by

|[tex]v_{3}[/tex]| =√15.6²+15.6² => 22.06m/s

(b) its direction (as an angle relative to the +x axis)  'θ'

θ= [tex]tan^{-1} (\frac{15.6}{15.6} )[/tex] => [tex]tan^{-1} (1 )[/tex] =>45°

Answer:

a. Air fuel Ratio = 19.76 kg air/kg fuel

b. % Theoretical air used = 131%

c. Amount of H2O that condenses as the products are cooled to 25°C at 100kPa = 6.59 kmol

Explanation: