A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m

Solution :

Given : Mass of ladder = 10 kg

Length of ladder = 5 m

Weight of window cleaner = 75 kg

a). Now equate the torque about the lowermost point of the ladder is given by :

[tex]$=10 \times 9.8 \times \frac{2.5}{2} + 75 \times 9.8 \times \frac{3}{5} \times 2.5 = N \times \sqrt{5^2 - 2.5^2}$[/tex]

Here, N = normal force that the glass exerts on the ladder

Therefore, [tex]$N = 282.9 \ N$[/tex]

                      = 280 N (in 2  significant figures)

b). Equate the forces along horizontal direction,

The horizontal component of the friction, [tex]$F_x = N = 282.9 \ N $[/tex]

The vertical component of the friction, [tex]$F_y = (10+75) \times 9.8$[/tex]

                                                                     = 833 N

Therefore, the net frictional force, [tex]$F = \sqrt{F_x^2+F_y^2}$[/tex]

[tex]$F = \sqrt{(282.9)^2+(833)^2}$[/tex]

    = 879.7 N

    = 880 N (in 2 significant figures)  

c). The angle the forces makes [tex]$= \tan \frac{833}{282.9} $[/tex]

                                                    [tex]$= 71.2 ^\circ $[/tex]

  Therefore in 2 significant figures = [tex]$71 ^\circ$[/tex]

Answer:

cos 0 = 1.

Fs = 7×8 = 56 J

Explanation:

Answer:

Explanation:

The lift is going down with acceleration

Initial speed u = 0

Final speed v = 6 m/s

distance s = 15.25 m

acceleration a = ?

v² = u² + 2 a s

6² = 0 + 2 x a x 15.25

a = 1.18 m /s²

Elevator is going down with acceleration  .

mg - T = ma where T is tension in the cable .

722 x 9.8 - T = 722 x 1.18

7075.6 - T = 851.96

T = 6223.64 N .

Answer:

A. Retina

Explanation:

Answer:

it will option option A hope it helps

Answer:planning

Explanation:

The person is in the stage of planning due to its action of planning to be active.

The person is in the planning stage among the five stages of change for physical activity because the person just started planning to be active not yet started the activity. If a person is in the state of looking thoughtfully at something for a very long time then it is said to be Contemplation.

While on the other hand, if a person is in a stage in which there is no intention to change behavior in the foreseeable future then it is called precontemplation so we can conclude that the person is in the stage of planning due to its action of planning to be active.

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Answer:

mining

Explanation:

because they use the most heavy duty machines

Answer:

0.12 mm ; 140.50 rad/m ; 628.32 rad/sec ; +

Explanation:

Given the wave equation of the form :

y(x, t) = ym sin(kx ± ωt)

Mas per unit length (u) = 5 g/cm = (5÷1000)kg / 0.01m) = 0.005kg/0.01m = 0.5kg/m

Tension, T = 10 N

Amplitude, A = 0.12 mm

Frequency, F = 100 Hz

Comparing with the general wave equation :

y = Asin(kx ± ωt)

A = amplitude = ym = 0.12 mm

2.) k = 2π / λ

Recall :

v = fλ

v = sqrt(T/u) = sqrt(10/0.5) = sqrt(20) = 4.472

λ = v/ f = 4.472 / 100 = 0.04472

Hence,

k = (2 * π) / 0.04472

k = 140.50 rad/m

3.) Angular frequency, ω

ω = 2πf = 2 * 3.14 * 100 = 628.32 rad/sec

4.) sign is +ve

Direction of wave propagation as given is in the negative x axis

Answer:

(a) the unknown inertia is 0.388 kg

(b) the average acceleration of the heavier cart is 317 m/s²

(c) the average acceleration of the lighter cart is 539 m/s²

Explanation:

Given;

mass of the first cart, m₁ = 0.66 kg

initial speed of the first cart, u₁ = 1.85 m/s

let the mass of the cart with unknown inertia be m₂

initial velocity of the second cart, u₂ = 2.17 m/s to the left

velocity of the first cart after collision, v₁ = 1.32 m/s to the left

velocity of the second cart after collision, v₂ = 3.22 m/s

time of collision, t = 0.010 s

(a) What is the unknown inertia?

Apply the principle of conservation of linear momentum, to determine the unknown inertia.

let leftward direction be negative direction

let rightward direction be positive direction

m₁u₁ + m₂u₂ = m₁v₁  + m₂v₂

0.66(1.85) + m₂(-2.17) = 0.66(-1.32) + m₂(3.22)

1.221 - 2.17m₂ = -0.8712 + 3.22m₂

1.221 + 0.8712 = 3.22m₂ + 2.17m₂

2.0922 = 5.39m₂

m₂ = 2.0922 / 5.39

m₂ = 0.388 kg

The unknown inertia is 0.388 kg

(b) the average acceleration of the heavier cart

the heavier cart has a mass of 0.66 kg

[tex]a = \frac{v_1 - u_1}{t} \\\\a = \frac{-1.32 - 1.85}{0.01} \\\\a = -317 \ m/s^2\\\\|a| = 317 \ m/s^2[/tex]

(c) the average acceleration of the lighter cart;

the lighter cart has a mass of 0.388 kg

[tex]a = \frac{v_2 - u_2}{t} \\\\a = \frac{3.22 - (-2.17)}{0.01} \\\\a =\frac{3.22 \ +\ 2.17}{0.01} \\\\a= 539\ m/s^2[/tex]

Answer:

This idea helps students explain why more rain forms over West Ferris than East Ferris. ... Therefore, when students explain that water vapor condenses higher in the atmosphere, they are actually explaining that water vapor condenses high in the troposphere, which is relatively low in the atmosphere.

Explanation:

Plz mark me brainliest thank u> have a good day

Answer:

40m/s

Explanation:

v²=u²+2as

v²=0²+2(16)(50)

v²=160v=40m/s

Answer:

B. The rod must be an insulator

Explanation:

We have that this rod must be an insulator. After this child rubbed the balloon, the balloon acquired static charge. So holding a rid against it is going to cause it to repel, this is to say it is repelling because the rod also is carrying some static charges. If this rid was a conductor, there would be no charge in its surface. The charge would have passed through her hand as it comes in contact.

Answer:

Explanation:

equation of wave is given by the following equation

y = (2.6 cm) sin[1.8 - (5.8 s-1)t].

Comparing it with standard form of wave

y = A sin ( ωt - kx )

we get

ω = 5.8

2πn = 5.8

n = .92 per second

kx = 1.8

k x 6 = 1.8

k = 0.3

[tex]\frac{2\pi}{\lambda}[/tex] = 0.3

λ = 20.9 cm

Answer:

104.59 m

Explanation:

From the question given above, the following data were obtained:

Horizontal velocity (u) = 13 m/s

Horizontal distance (s) = 60 m

Height of cliff =?

Next, we shall determine the time taken for the car to hit the ground. This can be obtained as follow:

Horizontal velocity (u) = 13 m/s

Horizontal distance (s) = 60 m

Time (t) =?

s = ut

60 = 13 × t

Divide both side by 13

t = 60 /13

t = 4.62 s

Finally, we shall determine the height of the cliff. This can be obtained as follow:

Time (t) = 4.62 s

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

Height (h) =?

h = ½gt²

h = ½ × 9.8 × 4.62²

h = 4.9 × 21.3444

h = 104.59 m

The, the height of the cliff is 104.59 m

Answer:

75 rad/s

Explanation:

The angular acceleration is the time rate of change of angular velocity. It is given by the formula:

α(t) = d/dt[ω(t)]

Hence: ω(t) = ∫a(t) dt

Also, angular velocity is the time rate of change of displacement. It is given by:

ω(t) = d/dt[θ(t)]

θ(t) = ∫w(t) dt

θ(t) = ∫∫α(t) dtdt

Given that: α (t) = (6.0 rad/s4)t² = 6t² rad/s⁴. Hence:

θ(t) = ∫∫α(t) dtdt

θ(t) = ∫∫6t² dtdt =∫[∫6t² dt]dt

θ(t) = ∫[2t³]dt = t⁴/2 rad

θ(t) = t⁴/2 rad

At θ(t) = 10 rev = (10 *  2π) rad = 20π rad, we can find t:

20π = t⁴/2

40π = t⁴

t = ⁴√40π

t = 3.348 s

ω(t) = ∫α(t) dt = ∫6t² dt = 2t³

ω(t) = 2t³

ω(3.348) = 2(3.348)³ = 75 rad/s

The required value of angular velocity of wheel is 75 rad/s.

Given data:

The angular acceleration of the wheel is, [tex]\alpha (t) = 6.0 t^{2} \;\rm rad/s^{4}[/tex].

The turning rate of wheel is, n = 10 rev.

The angular acceleration is the time rate of change of angular velocity. It is given by the formula:

α(t) = d/dt[ω(t)

Clearly, the angular velocity is the single integral of angular acceleration.  Then,

ω(t) = ∫a(t) dt

Also, angular velocity is the time rate of change of displacement. It is given by:

ω(t) = d/dt[θ(t)]

θ(t) = ∫w(t) dt

θ(t) = ∫∫α(t) dt dt

Since,

α (t) = (6.0 rad/s4)t² = 6t² rad/s⁴.

Then solve by substituting the values as,

θ(t) = ∫∫α(t) dt dt

θ(t) = ∫∫6t² dt dt

θ(t) =∫[∫6t² dt]dt  

θ(t) = ∫[2t³]dt = t⁴/2 rad

θ(t) = t⁴/2 rad

At θ(t) = 10 rev = (10 *  2π) rad = 20π rad, we can find t:

20π = t⁴/2

40π = t⁴

t = ⁴√40π

t = 3.348 s

Also,  

ω(t) = ∫α(t) dt = ∫6t² dt = 2t³

ω(t) = 2t³ ......................................................(1)

Substitute the value of time in equation (1) as,

ω( t = 3.348) = 2(3.348)³ = 75 rad/s

Thus, we can conclude that the required value of angular velocity of wheel is 75 rad/s.

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I'm pretty sure I the third option C.

Explanation:

sorry if I'm wrong

Answer:

v = 8.1 m/s

θ = -36.4º (36.4º South of East).

Explanation:

        [tex]p_{ox} = p_{fx} (1)[/tex]

         ⇒ [tex]m_{tr} * v_{tr}= (m_{olds} + m_{tr}) * v_{fx} (2)[/tex]

       [tex]v_{fx} = \frac{m_{tr}*v_{tr} }{(m_{tr} + m_{olds)} } = \frac{4146kg*9.7m/s}{2028kg+4146 kg} = 6.5 m/s (3)[/tex]

        [tex]p_{oy} = p_{fy} (4)[/tex]

        ⇒[tex]m_{olds} * v_{olds}= (m_{olds} + m_{tr}) * v_{fy} (5)[/tex]

       [tex]v_{fy} = \frac{m_{olds}*v_{olds} }{(m_{tr} + m_{olds)} } = \frac{2028kg*(-14.5)m/s}{2028kg+4146 kg} = -4.8 m/s (6)[/tex]

       [tex]v_{f} = \sqrt{v_{fx} ^{2} +v_{fy} ^{2} }} = \sqrt{(6.5m/s)^{2} +(-4.8m/s)^{2}} = 8.1 m/s (7)[/tex]

       [tex]\frac{v_{fy} }{v_{fx} } = tg \theta (8)[/tex]

      ⇒ tg θ = vfy/vfx = (-4.8m/s) / (6.5m/s) = -0.738 (9)

      ⇒ θ = tg⁻¹ (-0.738) = -36.4º

Answer:

Explanation:

Well they told you the exact formula to use. Work is the force multiplied by  the distance through which its applied.

W = (20N)(35m)

= 700 Joules

13.) Power is the amount of work done over the time through which the work is being done.

P = W/t

= 10J/2s

= 5J/s

When a ball of mass m makes a head-on elastic collision with a second ball (at rest) and rebounds in the opposite direction with a speed equal to one-fourth its original speed, then mass of the second ball having v/3 is velocity after collision is 9m/4.

Momentum is defined as mass times velocity of body. it is denoted by p and its SI unit is Kg.m/s. It has both magnitude and direction. it is a vector quantity.  it tells about the moment of the body. it is denoted by p and expressed in kg.m/s. mathematically it is written as p = mv. A body having zero velocity or zero mass has zero momentum. its dimensions is [M¹ L¹ T⁻¹]. Momentum is conserved throughout the motion.

initial momentum = final momentum

Given,

mass of first body m₁ = m

initial velocity of first body = v₁' = v

final velocity of first body = v₁'' =v/4

mass of second body m₂ = ?

initial velocity of second body = v₂' = 0

final velocity of second body = v₂'' = v/3

According to conservation of momentum,

initial momentum = final momentum

m₁v₁' + m₂v₂' = m₁v₁'' + m₂v₂''

putting al above values

m₁v + 0 = m₁v/4 + m₂v/3

m₁v - m₁v/4 = m₂v/3

m (1 - 1/4)v = m₂v/3

3m/4 = m₂/3

m₂ = 9m/4

Hence mass of the second body is 9m/4.

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Answer:

a) Time period is 3.3 seconds

b) The frequency is 0.3030 Hz

c) amplitude is 0.25 m

d) maximum speed is 0.476 m/s

Explanation:

Given the data in the question;

a) Period

Time Period T = Time taken for one oscillation

T = 33s / 10 = 3.3 seconds

Therefore, Time period is 3.3 seconds

b) Frequency

we know that frequency is the inverse of time period

so;

Frequency f = 1/T = 1 / 3.3 s

Frequency f = 0.3030 Hz

Therefore, The frequency is 0.3030 Hz

c) amplitude

amplitude A = [tex]\frac{1}{2}[/tex]( 60 cm - 10 cm )

A = [tex]\frac{1}{2}[/tex] × 50 cm

A = 25 cm

A = 0.25 m

Therefore, amplitude is 0.25 m

d) maximum speed of the glider

maximum speed [tex]V_{max}[/tex] = ωA

and ω = 2π/T

so maximum speed [tex]V_{max}[/tex] = [tex]\frac{2\pi }{T}[/tex]A

so we substitute

so maximum speed [tex]V_{max}[/tex] = [tex]\frac{2\pi }{3.3}[/tex] × 0.25 m

so maximum speed [tex]V_{max}[/tex] = 0.476 m/s

Therefore, maximum speed is 0.476 m/s

The time period is 3.3 s, frequency is 0.303 Hz, amplitude is 0.25 m, and maximum speed is 0.47575 m/s.

Based on the given information,

• The air-track glider connected with a spring oscillates between the 10 cm mark and the 60 cm mark.  

• In 33 seconds, the glider completes 10 oscillations.  

There is a need to find, the period, frequency, amplitude, and maximum speed of the glider.

a) Time period (T) for one oscillation is,

[tex]\frac{33s}{10} = 3.3 s[/tex]

b) The frequency (f) is the reciprocal of the time period,

[tex]f = \frac{1}{T} =\frac{1}{3.3S} = 0.303 Hz[/tex]

c) The amplitude (A) is,

[tex]A = \frac{1}{2} (60 cm-10cm) = 25cm[/tex]

[tex]A = 0.25 m[/tex]

d) Maximum speed of the glider is,

[tex]Vmax = \frac{2\pi }{T} (0.25 m)[/tex]

[tex]Vmax = 0.47575 m/s[/tex]

Thus,  time period is 3.3 s, frequency is 0.303 Hz, amplitude is 0.25 m, and maximum speed is 0.47575 m/s.

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Answer:

The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.

Explanation:

Let suppose that RC car-slingshot-steelpellet is a conservative system, that is, that non-conservative forces (i.e. friction, air viscosity) can be neglected. The velocity of the steel pellet can be found by means of the Principle of Energy Conservation and under the consideration that change in gravitational potential energy is negligible and that the RC car travels at constant velocity:

[tex]\frac{1}{2}\cdot (m_{C}+m_{P})\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{C}\cdot v_{o}^{2} + \frac{1}{2}\cdot m_{P}\cdot v^{2}[/tex]

[tex]\frac{1}{2}\cdot m_{P}\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{P}\cdot v^{2}[/tex]

[tex]m_{P}\cdot v_{o}^{2} + k\cdot x^{2} = m_{P}\cdot v^{2}[/tex]

[tex]v^{2} = v_{o}^{2} + \frac{k}{m_{P}}\cdot x^{2}[/tex]

[tex]v = \sqrt{v_{o}^{2}+\frac{k}{m_{P}}\cdot x^{2} }[/tex] (1)

Where:

[tex]v_{o}[/tex] - Initial velocity of the steel pellet, measured in meters per second.

[tex]v[/tex] - Final velocity of the steel pellet, measured in meters per second.

[tex]k[/tex] - Spring constant, measured in newtons per meter.

[tex]m_{P}[/tex] - Mass of the steel pellet, measured in kilograms.

[tex]m_{C}[/tex] - Mass of the RC car, measured in kilograms.

[tex]x[/tex] - Initial deformation of the spring, measured in meters.

If we know that [tex]v_{o} = 5.6\,\frac{m}{s}[/tex], [tex]k = 85\,\frac{N}{m}[/tex], [tex]m_{P} = 0.025\,kg[/tex] and [tex]x = 0.035\,m[/tex], then the velocity of the steel pellet relative to the ground when it leaves the sling shot is:

[tex]v = \sqrt{\left(5.6\,\frac{m}{s} \right)^{2}+\frac{\left(85\,\frac{N}{m} \right)\cdot (0.035\,m)^{2}}{0.025\,kg} }[/tex]

[tex]v \approx 5.960\,\frac{m}{s}[/tex]

The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.

Answer:

It means when you look into the lens your vision magnifies by x1

Explanation:

Answer:

22.5 m

Explanation:

Using v² = u² - 2gy where u = initial velocity of BB = 21 m/s, v = final velocity of BB = 1 m/s (since this is the required speed of BB in which it will not harm the birds), g = acceleration due to gravity = 9.8 m/s² and y = minimum height of BB above the gun at which the birds may safely fly.

Substituting the values of the variables into the equation, we have

v² = u² - 2gy

(1 m/s)² = (21 m/s)² - 2(9.8 m/s²)y

collecting like terms, we have

(1 m/s)² - (21 m/s)² = - 2(9.8 m/s²)y

1 m²/s² - 441 m²/s² = -(19.6 m/s²)y

simplifying, we have

- 440 m²/s² = -(19.6 m/s²)y

dividing through by -19.6 m/s², we have

y = - 440 m²/s² ÷ -19.6 m/s²

y = 22.45 m

y ≅ 22.5 m

Answer:

-32.5 * 10^-5 J

Explanation:

The potential energy of this system of charges is;

Ue = kq1q2/r

Where;

k is the Coulumb's constant

q1 and q2 are the magnitudes of the charges

r is the distance of separation between the charges

Substituting values;

Ue = 9.0×10^9 N⋅m2/C2 * 5.5 x 10^-8 C *( -2.3 x10^-8) C/(3.5 * 10^-2)

Ue= -32.5 * 10^-5 J

The potential energy of this two particle system relative to the potential energy at infinite separation is [tex]\bold {-32.5x 10^-^5\ J}[/tex].

The potential energy of this system of charges,  

[tex]\bold {Ue = k\dfrac{q1q2}{r}}[/tex]

Where;  

k - Coulumb's constant  

q1 and q2 - magnitudes of the charges  

r - distance between the charges

Put the values in the equation,

[tex]\bold {Ue = 9.0x10^9\times \dfrac {5.5 x 10^{-8} C \times -2.3 x10^{-8} C}{3.5 \times 10^{-2}}}\\\\\bold {Ue= -32.5 x 10^-^5\ J}[/tex]

Therefore, the the potential energy of this two particle system relative to the potential energy at infinite separation is [tex]\bold {-32.5x 10^-^5\ J}[/tex].

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Answer:

Mechanical waves

Explanation:

Mechanical waves  are the waves that can travel only through a medium. Mechanical waves are disturbance of matter and require medium to transfer the energy. There are three types of mechanical waves that include  transverse wave, longitudinal wave and surface wave.

Some of the examples of mechanical waves are sound waves and seismic waves etcetera.

Hence, the correct answer is "Mechanical waves".

Answer:

6m/s

Explanation:

Using the law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the momentum after collision.

Using the expression

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are the initial velocities

v is the final velocity after collision

Substitute the given values in the formula

0.35(10)+0.35(2) = (0.35+0.35)v

3.5+0.7 = 0.7v

4.2 = 0.7v

v = 4.2/0.7

v = 6m/s

Hence the magnitude of the velocity for Ball B after the collision is 6m/s

Answer:

a)

Mass flow rate of core =  [tex]m_{e}[/tex] = 60.94 Kg/s

Mass flow rate of fan =  [tex]m_{s}[/tex]  = 73.12 kg/s

TSFC = 3.301 x [tex]10^{-5}[/tex]

b)

Exit Area of Fan = [tex]A_{e}[/tex] = 0.3624 [tex]m^{2}[/tex]

Exit Area of Core = [tex]A_{s}[/tex] = 0.2635 [tex]m^{2}[/tex]

Explanation:

Data Given:

Height = 25000 ft

Vehicle velocity = [tex]u_{a}[/tex] = 815 ft/s = 248.41 m/s

[tex]m_{s} = 1.2m_{e}[/tex]

[tex]m_{f}[/tex] = 0.0255[tex]m_{e}[/tex]  

Where,

[tex]m_{s}[/tex] = Mass flow rate of fan

[tex]m_{e}[/tex] = Mass flow rate of core

F = Thrust

Density of core = [tex]D_{e}[/tex] = 0.000578 slugs/[tex]ft^{3}[/tex] = 0.2979 kg/[tex]m^{3}[/tex]

Density of fan = [tex]D_{s}[/tex] = 0.00154 slugs/[tex]ft^{2}[/tex] = 0.7937 kg/[tex]m^{3}[/tex]

Ambient Pressure of Fan = [tex]P_{s}[/tex] = 10.07 Psi = 69430.21 Pa

Ambient Pressure of core = [tex]P_{e}[/tex] = 10.26 Psi = 70740.2 Pa

Thrust = F = 10580 lbf = 47062.2 N

Velocity of fan = [tex]u_{s}[/tex] = 1147 ft/s = 349.6 m/s

Velocity of core = [tex]u_{e}[/tex] = 1852 ft/s = 564.5 m/s

At the height of 25000 ft, P = 37600 [tex]P_{a}[/tex]

Now,

we have:

[tex]m_{e}[/tex] = [tex]u_{e}[/tex] x  [tex]D_{e}[/tex]  x [tex]A_{e}[/tex]

Plugging in the values, we get:

[tex]m_{e}[/tex]  = 168.16 [tex]A_{e}[/tex]   Equation 1

And,

[tex]m_{s}[/tex]  = [tex]D_{s}[/tex]  x [tex]A_{s}[/tex] x  [tex]u_{s}[/tex]

[tex]m_{s}[/tex]  = 277.5 [tex]A_{s}[/tex]  Equation 2

As, we know,

[tex]m_{s} = 1.2m_{e}[/tex]  

[tex]m_{s}[/tex]  = 277.5 [tex]A_{s}[/tex]

And now for Thrust, we have:

F = [tex]A_{e}[/tex] x ([tex]P_{e}[/tex]  - [tex]P_{a}[/tex] ) + [tex]A_{s}[/tex] x ([tex]P_{s}[/tex] - [tex]P_{a}[/tex] ) + [tex]m_{e}[/tex]x ([tex]u_{e}[/tex]  - [tex]u_{a}[/tex] ) + [tex]m_{s}[/tex] x ([tex]u_{s}[/tex]  -  [tex]u_{a}[/tex] ) Equation 3

Now, substitute equation 1 and 2 in equation 3, we get:

Exit Area of Fan = [tex]A_{e}[/tex] = 0.3624 [tex]m^{2}[/tex]

Exit Area of Core = [tex]A_{s}[/tex] = 0.2635 [tex]m^{2}[/tex]

Mass flow rate of core =  [tex]m_{e}[/tex] = 60.94 Kg/s

Mass flow rate of fan =  [tex]m_{s}[/tex]  = 73.12 kg/s

TSFC = Thrust Specific Fuel Consumption = fuel mass flow rate  / Thrust

TSFC = [tex]m_{f}[/tex]/F

And,

[tex]m_{f}[/tex] = 0.0255[tex]m_{e}[/tex]  

[tex]m_{e}[/tex]   =  60.94

[tex]m_{f}[/tex]  = 0.0255 x 60.94

[tex]m_{f}[/tex]  = 1.55397

TSFC = [tex]m_{f}[/tex]/F

TSFC = 1.55397/47062.2

TSFC = 3.301 x [tex]10^{-5}[/tex]

Low TSFC = High efficiency

High TSFC = Low efficiency

a)

Mass flow rate of core =  [tex]m_{e}[/tex] = 60.94 Kg/s

Mass flow rate of fan =  [tex]m_{s}[/tex]  = 73.12 kg/s

TSFC = 3.301 x [tex]10^{-5}[/tex]

b)

Exit Area of Fan = [tex]A_{e}[/tex] = 0.3624 [tex]m^{2}[/tex]

Exit Area of Core = [tex]A_{s}[/tex] = 0.2635 [tex]m^{2}[/tex]

Answer:

The work done by the force on the particle is 29.85 J.

Explanation:

The work is given by:  

[tex] W = ^{x_{2}}_{x_{1}}\int F_{x} dx [/tex]

Where:

x₁: is the lower limit = 0 m    

x₂: is the upper limit = 1.9 m

Fₓ: is the force in the horizontal direction =  (4.5 + 13.7x - 1.5x²)N

[tex]W = ^{1.9}_{0}\int (4.5 + 13.7x - 1.5x^{2}) dx[/tex]  

[tex] W = 4.5x|^{1.9}_{0} + \frac{13.7}{2}x^{2}|^{1.9}_{0} - \frac{1.5}{3}x^{3}|^{1.9}_{0} [/tex]  

[tex] W = 4.5N(1.9 m) + \frac{13.7N}{2}(1.9 m)^{2} - \frac{1.5N}{3}(1.9 m)^{3} [/tex]    

[tex] W = 4.5N(1.9 m) + \frac{13.7N}{2}(1.9 m)^{2} - \frac{1.5N}{3}(1.9 m)^{3} [/tex]      

[tex]W = 29.85 J[/tex]

Therefore, the work done by the force on the particle is 29.85 J.

I hope it helps you!