A jumbo jet has a mass of 100,000 kg. The thrust of each of its four engines is 50,000 N. What is the jet's acceleration in meters per second squared right before taking off? Neglect air resistance and friction.

Answer:

Explanation:

Bohr's atomic model was replaced by the  modern atomic model because of its limitations, which included :

(a) Only applicable for Hydrogen and like atoms ( He+1, Li+2 )

(b) Couldn't explain Zeeman Effect (splitting of spectral lines due external magnetic field ) and Stark Effect (splitting of spectral lines due to external electric field).

(c) Inconsistent with De-Broglie's Dual nature of matter and Heisenberg Uncertainty principal, etc.

Complete Question

A voltaic cell utilizes the following reaction and operates at 298 K:

3Ce4+(aq)+Cr(s)→3Ce3+(aq)+Cr3+(aq).

What is the emf of this cell under standard conditions? Express your answer using three significant figures.

Answer:

The value is [tex]E^o_{cell} = 2.35 V[/tex]

Explanation:

From the question we are told that

   The ionic equation is  

               [tex]3 Ce^{4 +} _{(aq)} + Cr _{(s)} \to 3 Ce^{3+} _{(aq)} + Cr^{3r} _{(aq)}[/tex]

Now under standard conditions the reduction  half reaction  is

      [tex]Ce^{4+} + e \to Ce^{3+} ; \ \ E^o_r = 1.61 V[/tex]

And the oxidation half reaction is

      [tex]Cr^{3+} + 3e^{-} \to Cr ; \ \ \ E^o_o = - 0.74 V[/tex]

The emf of this cell under standard conditions  is mathematically represented as

     [tex]E^o_{cell} = E^o _r - E^o _o[/tex]

substituting values

     [tex]E^o_{cell} = 1.61 - (- 0.74)[/tex]

    [tex]E^o_{cell} = 2.35 V[/tex]

Answer:

The width is [tex]w_c = 0.00422 \ m[/tex]

Explanation:

From the question we are told that

   The  wavelength is  [tex]\lambda = 6.33*10^{-7} \ m[/tex]

    The  width of the slit is  [tex]d = 0.3\ mm = 0.3 *10^{-3} \ m[/tex]

    The distance of the screen is  [tex]D = 1.0 \ m[/tex]

Generally the central maximum is mathematically represented as

      [tex]w_c = 2 * y[/tex]

Here  y is the width of the first order maxima which is mathematically represented as

      [tex]y = \frac{\lambda * D}{d}[/tex]

substituting values

      [tex]y = \frac{6.33*10^{-7} * 1.0}{ 0.30}[/tex]

       [tex]y = 0.00211 \ m[/tex]

So  

    [tex]w_c = 2 *0.00211[/tex]

     [tex]w_c = 0.00422 \ m[/tex]

Answer:

The average emf induced in the coil is 175 mV

Explanation:

Given;

number of turns of the coil, N = 1060 turns

diameter of the coil, d = 20.0 cm = 0.2 m

magnitude of the magnetic field,  B = 5.25 x 10⁻⁵ T

duration of change in field, t = 10 ms = 10 x 10⁻³ s

The average emf induced in the coil is given by;

[tex]E = N\frac{\delta \phi}{dt} \\\\E = N\frac{\delta B}{\delta t}A[/tex]

where;

A is the area of the coil

A = πr²

r is the radius of the coil = 0.2 /2 = 0.1 m

A = π(0.1)² = 0.03142 m²

[tex]E = \frac{NBA}{t} \\\\E = \frac{1060*5.25*10^{-5}*0.03142}{10*10^{-3}} \\\\E = 0.175 \ V\\\\E = 175 \ mV[/tex]

Therefore, the average emf induced in the coil is 175 mV

Answer:

See attached file

Explanation:

Answer:

a) The value of g at such location is:

[tex]g=9.8005171\,\frac{m}{s^2}[/tex]

b) the period of the pendulum with the length is 0.970 m is:

[tex]T=1.9767 sec[/tex]

Explanation:

Recall the relationship between the period (T) of a pendulum and its length (L) when it swings under  an acceleration of gravity g:

[tex]L=\frac{g}{4\,\pi^2} \,T^2[/tex]

a) Then, given that we know the period (2.0 seconds), and the pendulum's length (L=0.993 m), we can determine g at that location:

[tex]g=\frac{4\,\pi^2\,L}{T^2}\\g=\frac{4\,\pi^2\,0.993}{(2)^2}\\g=\pi^2\,(0.993)\,\frac{m}{s^2} \\g=9.8005171\,\frac{m}{s^2}[/tex]

b) for this value of g, when the pendulum is shortened to 0.970 m, the period becomes:

Answer:

tmin= lambda/2

Explanation:

See attached file pls

Complete question is;

A rope, under a tension of 153 N and fixed at both ends, oscillates in a second harmonic standing wave pattern. The displacement of the rope is given by

y = (0.15 m) sin[πx/3] sin[12π t].

where x = 0 at one end of the rope, x is in meters, and t is in seconds. What are (a) the length of the rope, (b) the speed of the waves on the rope, and (c)the mass of the rope? (d) If the rope oscillates in a third - harmonic standing wave pattern, what will be the period of oscillation?

Answer:

A) Length of rope = 4 m

B) v = 24 m/s

C) m = 1.0625 kg

D) T = 0.11 s

Explanation:

We are given;

T = 153 N

y = (0.15 m) sin[πx/3] sin[12πt]

Comparing this displacement equation with general waveform equation, we have;

k = 2π/λ = π/2 rad/m

ω = 2πf = 12π rad/s

Since, 2π/λ = π/2

Thus,wavelength; λ = 4 m

Since, 2πf = 12π

Frequency;f = 6 Hz

A) We are told the rope oscillates in a second-harmonic standing wave pattern. So, we will use the equation;

λ = 2L/n

Since second harmonic, n = 2 and λ = L = 4 m

Length of rope = 4 m

B) speed is given by the equation;

v = fλ = 6 × 4

v = 24 m/s

C) To calculate the mass, we will use;

v = √T/μ

Where μ = mass(m)/4

Thus;

v = √(T/(m/4))

Making m the subject;

m = 4T/v²

m = (4 × 153)/24²

m = 1.0625 kg

D) Now, the rope oscillates in a third harmonic.

So n = 3.

Using the formula f = 1/T = nv/2L

T = 2L/nv

T = (2 × 4)/(3 × 24)

T = 0.11 s

Answer:

v_{f} = 74 m/s, F = 230 N

Explanation:

We can work on this exercise using the relationship between momentum and moment

        I = ∫ F dt = Δp

bold indicates vectors

we can write this equations in its components

X axis

       Fₓ t = m ( -v_{xo})

Y axis  

        t = m (v_{yf} - v_{yo})

in this case with the ball it travels horizontally v_{yo} = 0

Let's use trigonometry to write the final velocities and the force

        sin 30 = v_{yf} / vf

        cos 30 = v_{xf} / vf

        v_{yf} = vf sin 30

        v_{xf} = vf cos 30

         sin40 = F_{y} / F

         F_{y} = F sin 40

         cos 40 = Fₓ / F

         Fₓ = F cos 40

let's substitute

      F cos 40 t = m ( cos 30 - vₓ₀)

      F sin 40 t = m (v_{f} sin 30-0)

we have two equations and two unknowns, so the system can be solved

        F cos 40 0.1 = 0.4 (v_{f} cos 30 - 20)

        F sin 40 0.1 = 0.4 v_{f} sin 30

we clear fen the second equation and subtitles in the first

         F = 4 sin30 /sin40     v_{f}

         F = 3.111 v_{f}

        (3,111 v_{f}) cos 40 = 4 v_{f} cos 30 - 80

        v_{f} (3,111 cos 40 -4 cos30) = - 80

        v_{f} (- 1.0812) = - 80

        v_{f} = 73.99

        v_{f} = 74 m/s

now we can calculate the force

          F = 3.111 73.99

          F = 230 N

Answer:

B. If the container is cooled, the gas particles will lose kinetic energy and temperature will decrease.

C. If the gas particles move more quickly, they will collide more frequently with the walls of the container and pressure will increase.

E. If the gas particles move more quickly, they will collide with the walls of the container more often and with more force, and pressure will increase.

#FreeMelvin

Answer:

The angular accelerations of the hoops are related by the following equation [tex]\alpha _1 = 2\alpha_2[/tex].

Explanation:

Net force on the hoop is given by;

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

where;

a is linear acceleration

m is the mass

Net torque on the hoop is given by;

[tex]\tau_{net} =I\alpha[/tex]

where;

I is moment of inertia

α is the angular acceleration

But, τ = Fr

[tex]Fr = I \alpha\\\\\alpha = \frac{Fr}{I} \\\\\alpha = \frac{Fr}{mr^2} \\\\\alpha = \frac{F}{mr} \\\\\alpha = \frac{1}{r} (\frac{F}{m} )\\\\(since\ the \ force\ and \ mass \ are \ the \ same, \frac{F}{m} = constant=k)\\\\ \alpha = \frac{k}{r}\\\\k = \alpha r[/tex]

[tex]\alpha _1 r_1= \alpha_2 r_2[/tex]

let the angular acceleration of the smaller hoop = α₁

let the radius of the smaller hoop = r₁

then, the radius of the larger loop, r₂ = 2r₁

let the angular acceleration of the larger hoop = α₂

[tex]\alpha _1 r_1= \alpha_2 r_2\\\\\alpha_2= \frac{ \alpha _1 r_1}{r_2} \\\\\alpha_2=\frac{\alpha _1 r_1}{2r_1} \\\\\alpha_2= \frac{\alpha _1}{2} \\\\\alpha _1 = 2\alpha_2[/tex]

Therefore, the angular accelerations of the hoops are related by the following equation [tex]\alpha _1 = 2\alpha_2[/tex]

Answer:

The length of the tube is 85 cm

Explanation:

Given;

speed of sound, v = 340 m/s

first harmonic of open-closed tube is given by;

N----->A , L= λ/₄

λ₁ = 4L

v = Fλ

F = v / λ

F₁ = v/4L

Second harmonic of open-closed tube is given by;

L = N-----N + N-----A, L = (³/₄)λ

[tex]\lambda = \frac{4L}{3}\\\\ F= \frac{v}{\lambda}\\\\F_2 = \frac{3v}{4L}[/tex]

Third harmonic of open-closed tube is given by;

L = N------N + N-----N + N-----A, L = (⁵/₄)λ

[tex]\lambda = \frac{4L}{5}\\\\ F= \frac{v}{\lambda}\\\\F_3 = \frac{5v}{4L}[/tex]

The difference between second harmonic and first harmonic;

[tex]F_2 -F_1 = \frac{3v}{4L} - \frac{v}{4L}\\\\F_2 -F_1 = \frac{2v}{4L} \\\\F_2 -F_1 =\frac{v}{2L}[/tex]

The difference between third harmonic and second harmonic;

[tex]F_3 -F_2 = \frac{5v}{4L} - \frac{3v}{4L}\\\\F_3 -F_2 = \frac{2v}{4L} \\\\F_3 -F_2 =\frac{v}{2L}[/tex]

Thus, the difference between successive harmonic of open-closed tube is

v / 2L.

[tex]700H_z- 500H_z= \frac{v}{2L} \\\\200 = \frac{v}{2L}\\\\L = \frac{v}{2*200} \\\\L = \frac{340}{2*200}\\\\L = 0.85 \ m\\\\L = 85 \ cm[/tex]

Therefore, the length of the tube is 85 cm

Answer:

Explanation:

If the river moves towards the south at 3m/s and the both moves towards the east at 4.0m/s, the speed of the boat relative to me will be the resulting displacement of both velocities of the river and that of the boat. This can be gotten using pythagoras theorem.

Let Vr be the relative speed. According to the theorem;

[tex]V_r^2 = V_s^2 + V_e^2\\\\V_r^2 = 3.0^2 + 4.0^2\\\\V_r^2 = 9+16\\\\V_r^2 = \sqrt{25}\\ \\V_r = 5.0m/s[/tex]

Hence this relative speed is 5.0 m/s

Answer:

198 ms-1

Explanation:

According to the Heisenberg uncertainty principle; it is not possible to simultaneously measure the momentum and position of a particle with precision.

The uncertainty associated with each measurement is given by;

∆x∆p≥h/4π

Where;

∆x = uncertainty in the measurement of position

∆p = uncertainty in the measurement of momentum

h= Plank's constant

But ∆p= mΔv

And;

m= 1.770×10^-27 kg

∆x = 0.15 nm

Making ∆v the subject of the formula;

∆v≥h/m∆x4π

∆v≥ 6.6 ×10^-34/1.770×10^-27 × 1.5×10^-10 ×4×3.142

∆v≥198 ms-1

Answer:

The current is  [tex]I = 8.9 *10^{-5} \ A[/tex]

Explanation:

From the question we are told that

     The  radius is [tex]r = 3.17 \ mm = 3.17 *10^{-3} \ m[/tex]

      The current density is  [tex]J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2[/tex]

      The distance we are considering is  [tex]r = 0.5 R = 0.001585[/tex]

Generally current density is mathematically represented as

          [tex]J = \frac{I}{A }[/tex]

Where A is the cross-sectional area represented as

         [tex]A = \pi r^2[/tex]

=>      [tex]J = \frac{I}{\pi r^2 }[/tex]

=>    [tex]I = J * (\pi r^2 )[/tex]

Now the change in current per unit length is mathematically evaluated as

        [tex]dI = 2 J * \pi r dr[/tex]

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

         [tex]I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr[/tex]

         [tex]I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr[/tex]

        [tex]I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.[/tex]

        [tex]I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ][/tex]

substituting values

        [tex]I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ][/tex]

        [tex]I = 8.9 *10^{-5} \ A[/tex]

Answer:I don’t know

Explanation:

Answer:

b) the heavier one will have twice the kinetic energy of the lighter one.

Explanation:

The kinetic energy of object with mass, m

K.E₁ = ¹/₂mv²

where;

m is mass of the object

v is the velocity of the object

Since, the two objects are falling under same acceleration due to gravity, their velocity will be increasing at the same rate

The kinetic energy of object with mass, 2m

K.E₂ = ¹/₂(2m)v²

K.E₂ = 2(¹/₂mv²)

BUT K.E₁ = ¹/₂mv²

K.E₂ = 2(K.E₁)

Therefore, the heavier one will have twice the kinetic energy of the lighter one.

b) the heavier one will have twice the kinetic energy of the lighter one.

Answer:

The  apparent depth is  [tex]D' = 0.00376 \ m[/tex]

Explanation:

From the question we are told that

     The  depth of the water is  [tex]D = 5.0 \ mm = 5.0 *10^{-3} \ m[/tex]

      The  refractive index of water is  [tex]n = 1.33[/tex]

Generally the apparent depth of the coin is mathematically represented as

          [tex]D' = D * [\frac{ n_a}{n} ][/tex]

Here  [tex]n_a[/tex]  is the refractive index of  air the value is  [tex]n_a = 1[/tex]

So

        [tex]D' = 5.0 *10^{-3} * [\frac{1}{1.33} ][/tex]

        [tex]D' = 0.00376 \ m[/tex]

The apparent depth will be 0.00376 m.

The index of refraction of a substance also known as the refraction index is a dimensionless quantity that specifies how quickly light passes through it in optics.

d is the depth of the water =5.0 mm =5.0 ×10⁻³

n is the refractive index of water =1.33

[tex]\rm n_a[/tex] is the refractive index of wire=1

The apparent depth of the coin is given as;

[tex]\rm D'=D \times \frac{n_a}{n} \\\\ \rm D'=5.0 \times 10^{-3} \times \frac{1}{1.33} \\\\ \rm D'=0.00376 \ m[/tex]

Hence the apparent depth will be 0.00376 m.

To learn more about the index of refraction refer to the link;

https://brainly.com/question/23750645

Explanation:

the missing figure in the Question has been put in the attachment.

Then from the figure we can observe that

the center of the sphere is positive, therefore, negative charge will be  induced at A.

As B is grounded there will not be any charge on B

Hence the answer is A is negative and B is charge less.

Answer;

26.45cm

See attached file for explanation

Answer:

From fastest speed to slowest speed, the electromagnetic waves are ranked as(up to down):

d. X-ray

c. Green light

a. Yellow light

f. Infrared wave

b. FM radio wave

e. AM radio wave

Explanation:

Electromagnetic waves are waves produced as a result of vibrations between an electric field and a magnetic field. The waves have three properties and these properties are frequency, speed and wavelength, which are related by the relationship below

V = Fλ

where:\

V = speed (velocity)

F = frequency

λ = wavelength.

From the relationship above, it is seen that the speed of a wave is directly proportional to its frequency. The higher the frequency, the higher the speed. Therefore, from the list given, the waves with  the highest to lowest frequencies/ from left to right are:

X-ray (3×10¹⁹ Hz to 3×10¹⁶Hz), Green light (5.66×10¹⁴Hz), Yellow light (5.17×10¹⁴Hz), Infrared wave (3×10¹¹Hz), FM radio wave (10.8×10⁸Hz to 8.8×10⁷Hz), AM radio wave (1.72 × 10⁶Hz to 5.5×10⁵Hz).

This corresponds to the speed from highest to lowest from left to right.

Answer:

The approximate  spring constant is  [tex]k = 55533.33 \ N/m[/tex]

Explanation:

From the question we are told that

   The  mass of the person is  [tex]m = 68 \ kg[/tex]

     The  dip of the car is  [tex]x = 1.2 \ cm = 0.012 \ m[/tex]

Generally according to hooks law  

        [tex]F = k * x[/tex]

here the force F is the weight of the person which is mathematically represented as

         [tex]F = m * g[/tex]

=>    [tex]m * g = k * x[/tex]

=>     [tex]k = \frac{m * g }{x }[/tex]

=>    [tex]k = \frac{68 * 9.8}{ 0.012}[/tex]

=>   [tex]k = 55533.33 \ N/m[/tex]

Explanation:

EE = ½ kx²

EE = ½ (800 N/m) (6 m)²

EE = 14,400 J

Answer:

14,400 J

Explanation:

Its the answer

Answer:

Explanation:

We shall apply formula of Doppler's effect

Here source is fixed and observer is approaching the source

f = f₀ x [(V + v ) / V ]

f₀ is original and f is apparent frequency , V is velocity of sound and v is velocity of motorcyclist .

f = 1000 x [(331 + 27.27 ) / 331 ]

= 1082 .4 Hz

This is the frequency heard by motorcyclist .

When police car is approaching him when he is stopped

f = f₀ x [V /(V - v ) ]

v is velocity of police car .

= 1000  x 331 / (331 - 27.27)

= 1090 Hz  

Answer:

Explanation:

(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)

ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .

ΔK = 0

ΔUg  is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .

ΔUg = positive

ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .

ΔUs = 0

ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence

ΔEth = negative .

b )

work done by athlete

= 400 x 2 = 800 J

energy output = 800 J

c )

It is 25% of metabolic energy output of his body

so metalic energy output of body

= 4x 800 J .

3200 J

power = energy output / time

= 3200 / 1.6

= 2000 W .

d )

1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .

2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .

A) Applying the energy equation

B) The energy output by the athlete is ; 800 Joules

C) The metabolic power is : 2000 w

D) When he performs the task in 1.2 s

Given data :

Weight of barbell = 400 N

Height = 2.0 m

Time = 1.6 secs

efficiency of the human body = 25%

Speed = constant

A) From the energy equation the ΔK is zero because the athlete is lifting the barbell at a constant speed. ΔUg is positive because as the weight is lifted its  potential energy increases.  ΔEth ( change in energy of earth ) is negative because it exerts a force in opposite direction to displacement

B)  Determine the energy output of the athlete

weight of barbell * Height  = 400 * 2 = 800 J

C) Determine the metabolic power

Metabolic power = energy output / Time

where ; energy output = 4 * 800 = 3200

∴ Metabolic power = 3200 / 1.6

                                = 2000 w

D) When performs same task at 1.2 s

The metabolic energy he expends is  the same  and His metabolic power is  more

Hence we can conclude that the answers to your questions are as listed above

Learn more : https://brainly.com/question/17807740

Answer:

a

  [tex]Q = 5.34 *10^{19} \ J[/tex]

b

   [tex]T = 0.445 * 365 = 162. 413 \ days[/tex]

Explanation:

From the question we are told that

     The  area of  Manhattan is  [tex]a_k = 87.46 *10^{6} \ m^2[/tex]

      The area of the ice is [tex]a_i = 4* 87.46 *10^{6 } = 3.498 *10^{8}\ m^2[/tex]

        The  thickness is  [tex]t = 500 \ m \\[/tex]

Generally the volume of the ice is mathematically represented is

         [tex]V = a_i * t[/tex]

substituting value

         [tex]V = 500 * 3.498*10^{8}[/tex]

         [tex]V = 1.75 *10^{11}\ m^3[/tex]

Generally the mass of the ice is

       [tex]m_i = \rho_i * V[/tex]

Here [tex]\rho_i[/tex] is the density of ice the value is  [tex]\rho _i = 916.7 \ kg/m^3[/tex]

=>   [tex]m_i = 916.7 * 1.75*10^{11}[/tex]

=>    [tex]m_i = 1.60 *10^{14} \ kg[/tex]

Generally the energy needed for the ice to melt is mathematically represented as

        [tex]Q = m _i * H_f[/tex]

Where [tex]H_f[/tex] is the latent heat of fusion of ice and the value is  [tex]H_f = 3.33*10^{5} \ J/kg[/tex]

=>    [tex]Q = 1.60 *10^{14} * 3.33*10^{5}[/tex]

=>    [tex]Q = 5.34 *10^{19} \ J[/tex]

Considering part b

  We are told that the annual energy consumption is  [tex]G = 1.2*10^{20 } \ J / year[/tex]

So  the time taken to melt the ice is

      [tex]T = \frac{ 5.34 *10^{19}}{ 1.2 *10^{20}}[/tex]

        [tex]T = 0.445 \ years[/tex]

converting to days

      [tex]T = 0.445 * 365 = 162. 413 \ days[/tex]

Answer:A  200

Explanation:

Vp=1.41*Vrms

Vp=169.7 v

Z=Vp/Ip

Z=169.7/.8484

Z=200.03 ohm

Answer:

The frequency of the coil is 7.07 Hz

Explanation:

Given;

number of turns of the coil, 200 turn

cross sectional area of the coil, A = 300 cm² = 0.03 m²

magnitude of the magnetic field, B = 30 mT = 0.03 T

Maximum value of the induced emf, E = 8 V

The maximum induced emf in the coil is given by;

E = NBAω

Where;

ω is angular frequency = 2πf

E = NBA(2πf)

f = E / 2πNBA

f = (8) / (2π x 200 x 0.03 x 0.03)

f = 7.07 Hz

Therefore, the frequency of the coil is 7.07 Hz

Answer:

His angular velocity will increase.

Explanation:

According to the conservation of rotational momentum, the initial angular momentum of a system must be equal to the final angular momentum of the system.

The angular momentum of a system = [tex]I[/tex]'ω'

where

[tex]I[/tex]' is the initial rotational inertia

ω' is the initial angular velocity

the rotational inertia = [tex]mr'^{2}[/tex]

where m is the mass of the system

and r' is the initial radius of rotation

Note that the professor does not change his position about the axis of rotation, so we are working relative to the dumbbells.

we can see that with the mass of the dumbbells remaining constant, if we reduce the radius of rotation of the dumbbells to r, the rotational inertia will reduce to [tex]I[/tex].

From

[tex]I[/tex]'ω' = [tex]I[/tex]ω

since [tex]I[/tex] is now reduced, ω will be greater than ω'

therefore, the angular velocity increases.