What is the toy's total energy at any point of its motion? Express your answer with the appropriate units.]

Complete question:

The length of nylon rope from which a mountain climber is suspended has an effective force constant of 1.40 ×10⁴ N/m.

What is the frequency (in Hz) at which he bounces, given that his mass plus the mass of his equipment is 84.0 kg?

Answer:

The frequency (in Hz) at which he bounce is 2.054 Hz

Explanation:

Given;

effective force constant, K = 1.40 ×10⁴ N/m.

The total mass = his mass plus the mass of his equipment, m = 84 kg

The frequency (in Hz) at which he bounce is given by;

[tex]f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\\\\f = \frac{1}{2\pi} \sqrt{\frac{1.4*10^4}{84}}\\\\f = 2.054 \ Hz[/tex]

Therefore, the frequency (in Hz) at which he bounce is 2.054 Hz

Answer:

TRICARE for Life (TFL), a program for Medicare-eligible military retirees and their dependents, acts as a supplement to Medicare.

Explanation:

Answer:

1) 5.52 cm , C) 5.5 cm

Explanation:

When a measurement is carried out, in addition to the value of the magnitude, the error or uncertainty of the measurement must occur, in a direct measurement with an instrument the uncertainty is equal to the appreciation of the instrument.

Uniform see the errors by the number of significant figures days, in this cases they are two decimals for which the appreciation of the instrument ± - 0.01

now we can analyze the measurements made

1) 5.52 cm. Validate. It is a valid measurement is within the uncertainty range

2) 6.63 cm. It does not validate. It is out of the error range

3) 5.5 cm Valid. It is within the given error range,

4) 5.93 cm Not Valid. It is out of the error range.

Answer:

6.63 and 5.93

Explanation:

Answer:

a

     [tex]l_o  =52 \  m[/tex]

b

      [tex]l = 37.13 \ LY[/tex]

Explanation:

From the question we are told that

    The  speed of the spaceship is  [tex]v  =  0.800c[/tex]

    Here  c is the speed of light with value  [tex]c =  3.0*10^{8} \ m/s[/tex]

    The  length is  [tex]l = 31.2 \  m[/tex]

     The  distance of the star for earth is [tex]d = 145 \  light \  years[/tex]

     The  speed is [tex]v_s = 2.90 *10^{8}[/tex]

Generally the from the length contraction equation we have that

       [tex]l  =  l_o  \sqrt{1 -[\frac{v}{c } ]}[/tex]

Now the when at rest the length is  [tex]l_o[/tex]

So  

      [tex]l_o =\frac{l}{\sqrt{ 1 - \frac{v^2}{c^2 } } }[/tex]

      [tex]l_o =\frac{ 31.2 }{ \sqrt{1 - \frac{(0.800c ) ^2}{c^2} } }[/tex]

      [tex]l_o=52 \  m[/tex]

Considering b  

  Applying above equation

            [tex]l  =l_o \sqrt{1 -  [\frac{v}{c } ]}[/tex]

Here [tex]l_o  =145 \  LY(light \ years )[/tex]

So

           [tex]l=145 *  \sqrt{1 -  \frac{v_s^2}{c^2 } }[/tex]

            [tex]l =145 *  \sqrt{ 1 - \frac{2.9 *10^{8}}{3.0*10^{8}} }[/tex]

            [tex]l = 37.13 \ LY[/tex]

Complete Question

The human eye can readily detect wavelengths from about 400 nm to 700 nm. If white light illuminates a diffraction grating having 710 lines/mm, over what range of angles does the visible m = 1 spectrum extend

Answer:

The  value  [tex]\theta = 16.5 ^ o[/tex]

Explanation:

From the question we are told that

   The  diffraction grating haves [tex]a =  710 \  lines /mm[/tex]

Generally the separation of the slit is mathematically represented  as

      [tex]d = \frac{1}{710} \ mm =0.001408 \ mm =   1.408 *10^{-6} \  m[/tex]

 Generally the condition for constructive interference is mathematically represented as  

           [tex]dsin(\theta ) = n \lambda[/tex]

So

          [tex]\theta =  sin ^{-1} [\frac{n * \lambda }{d} ][/tex]

=>       [tex]\theta  =  sin^{-1}[\frac{1 *  400 *10^{-9}}{ 1.408*10^{-6}} ][/tex]

=>         [tex]\theta = 16.5 ^ o[/tex]

Answer:

force applied by red = 50 N

force applied by blue = - 63 N (since it is in the opposite direction)

net force = force by red + force by blue

net force = 50 + (- 63)

net force = - 13 N  

Answer:

They should be connected in parallel

Explanation:

Because In Parallel circuit, voltage across all lamps are same and current will flow independent each other such that a malfunction of one will nor affect the rest unlike in series connection

Complete Question

A wave is described by y(x,t) = 0.1 sin(3x + 10t), where x is in meters, y is in centimetres and t is in seconds. The angular wave  frequency is

Answer:

The  value is [tex]w =  10 \ rad /s[/tex]

Explanation:

From the question we are told that  

    The equation describing the wave is y(x,t) = 0.1 sin(3x + 10t)

Generally the sinusoidal equation representing the motion of a wave is mathematically represented as

         [tex]y(x,t) =  Asin(kx + wt )[/tex]

Where  w  is the  angular frequency

Now comparing this equation  with that given we see that

       [tex]w =  10 \ rad /s[/tex]

Answer:

height of cliff (h) = 112.38m

Explanation:

The time 5.12 s is the total time it takes for the rock to fall, and for the soundwave to travel back to the top of the cliff before it is heard.

[tex]5.12 = t_f\ +\ t_s - - - - -(1)\\where:\\t_f = time\ of\ fall\ of\ the\ piece\ of\ rock\\t_s = time\ travelled\ by\ the\ return\ sound[/tex]

Let h be the height of the cliff in meters, the time taken for the rock to fall is given by:

[tex]t_f=\sqrt{\frac{2h}{g} } \\where:\\t_f = time\ of\ fall\\h = height\ of\ cliff\\g= acceleration\ due\ to\ gravity= 9.8 m \slash s^2[/tex]

[tex]\therefore t_f = \sqrt{\frac{2h}{9.8}} \\squaring\ both\ sides\\(t_f)^2 = \frac{2h}{9.8}\\ 2h = 9.8 \timess\ (t_f)^2\\h = \frac{9.8 \timess\ (t_f)^2}{2} \\h= 4.9(t_f)^2 - - - - - (2)[/tex]

Next, let us calculate the time taken fot the sound to return

[tex]t_s = \frac{h}{v} \\where:\\t_s = time\ for\ sound\ to\ travel\ up\ the\ cliff\\h= distance\ tavelled\ = height\ of\ cliff\\v= speed\ = 340m \slash s\ (speed\ of\ sound)\\\therefore t_s = \frac{h}{340} - - - - - (3)\\[/tex]

now putting the values of  h from equation 2 into equation (3)

[tex]t_s = \frac{4.9(t_f)^2}{340}[/tex]

Putting the value of [tex]t_s[/tex] into equation (1)

[tex]5.12 = t_f +\frac{4.9(t_f)^2}{340} \\[/tex]

multiplying through by 340

[tex]1740.8 = 340(t_f) + 4.9(t_f)^2\\4.9(t_f)^2 + 340 (t_f) - 1740.8 = 0[/tex]

now let us solve the quadratic equationsss;

[tex]Let\ (t_f) = x[/tex]

[tex]4.9x^2 + 340x - 1740.8 = 0\\using\ quadratic\ formula\\x = \frac{-b \pm\sqrt{b^2 - 4ac} }{2a} \\x = \frac{-340 \pm\sqrt{(340)^2 -\ 4 \times4.9 \times(-1740.8)} }{2\times4.9}\\x = \frac{-340\ \pm\ 386.936}{9.8} \\x =\frac{386.938 - 340}{9.8} \\x = \frac{46.936}{9.8}\\ x = 4.789\\x = t_f\\t_f=4.789s[/tex]

note, time cannot be negative, so we ignored the negative answer

putting the value of  [tex]t_f[/tex]  into equation (2) to find height of cliff (h)

[tex]h= 4.9(t_f)^2\\h = 4.9 \times(4.789)^2\\h = 112.38m[/tex]

Therefore, height of cliff (h) = 112.38m

Answer:

The antenna which is a transmitting and receiving device emits energy from current as radio waves, it does this by attracting the radio waves which are a form of EMWs and converts it to small voltages which are amplified to the final voltage signal which hear

Answer:

Zero

Explanation:

Because using

Deta X= dsinစ x n(lambda)

But we know that for central maxima

n is zero

So after substituting

Deta x = 0

Answer:

2000Hz and 1500Hz

Explanation:

Using

a) f = f0((c+vr)/(c+vs))

=>>> f0((c)/(c-0.5c))

=>>>1000/0.5 = 2000Hz

b) f = f0((c+vr)/(c+vs))

=>>>f0((c+0.5c)/(c))

=>>>>1000 x 1.5 = 1500Hz

Here we have a problem referring to the Doppler effect, the solutions are:

The Doppler effect is an effect that explains how the perception of waves changes as the source moves or as the listener moves.

The formula, for sound, is:

[tex]f = \frac{v + v_0}{v - v_1}*f_0[/tex]

where:

a) First we have that the source moves towards a stationary listener, then we have:

[tex]f = \frac{v }{v - v/2}*1.0 kHz\\\\f = \frac{v }{v/2}*1.0 kHz = 2.0 kHz[/tex]

b) in this case, the listener moves towards the source, so we have:

[tex]f = \frac{v + v/2 }{v }*1.0 kHz\\\\f = \frac{ (3/2)*v }{v}*1.0 kHz = 1.5 kHz[/tex]

So in this case the perceived frequency is smaller than in the point a.

This is because the waves will move at a fixed rate in air, in one case, the successive waves are emitted from different points in space (each time closer to the listener) while in the other case the waves are emitted from a fixed point, and the listener moves towards them, thus feels that the waves move faster.

If you want to learn more about the Doppler effect, you can read:

https://brainly.com/question/3826119

Answer:

1km/h^2 = 5/648 cm/s^2 [about 7.7 x 10^(-3) /s^2]

Explanation:

1km = 10^3 m = 10^5 cm

1h = 60x60 = 3600s

--> 1km/h^2 = 10^5 cm / (3600^2) s^2

<=> 1km/h^2 = 5/648 cm/s^2 [about 7.7 x 10^(-3) /s^2]

Question:  Initially, the car travels along a straight road with a speed of 35 m/s. If the brakes are applied and the speed of the car is reduced to 13 m/s in 17 s, determine the constant deceleration of the car.

Answer:

1.29 m/s²

Explanation:

From the question,

a = (v-u)/t............................ Equation 1

Where a = deceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.

Given: v = 13 m/s, u = 35 m/s, t = 17 s.

a = (13-35)/17

a = -22/17

a = -1.29 m/s²

Hence the deceleration of the car is 1.29 m/s²

Answer:

The final speed of small metal sphere is 12.6 m/s.

Explanation:

Given that,

Distance = 0.4 m

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

We need to calculate the final speed of small metal sphere

Using conservation of energy

[tex]\dfrac{1}{2}mv_{1}^2+\dfrac{kq_{1}q_{2}}{r_{1}}=\dfrac{1}{2}mv_{2}^2+\dfrac{kq_{1}q_{2}}{r_{2}}[/tex]

[tex]\dfrac{1}{2}m(v_{1}^2-v_{2}^2)=kq_{1}q_{2}(\dfrac{1}{r_{2}}-\dfrac{1}{r_{1}})[/tex]

Put the value into the formula

[tex]\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{2}^2)=9\times10^{9}\times(-2)\times10^{-6}\times(-8)\times10^{-6}(\dfrac{1}{0.4}-\dfrac{1}{0.8})[/tex]

[tex]400-v_{2}^2=240[/tex]

[tex]-v_{2}^2=240-400[/tex]

[tex]v_{2}=\sqrt{160}\ m/s[/tex]

[tex]v_{2}=12.6\ m/s[/tex]

Hence, The final speed of small metal sphere is 12.6 m/s.

Answer:

The Sun is known to emit almost all wavelengths of electromagnetic radiation but 99% of the radiation emitted by the sun lie in the ultraviolet, visible, and infrared regions.

Ultraviolet (UV) is a form of electromagnetic radiation with wavelength from 10 nm to 400 nm (750 THz). The wavelength is shorter than that of visible light but longer than X-rays. UV rays make up about 10% of the e-m waves from the sun. UV radiation is carcinogenic to the skin, and is absorbed by the melanin pigment in the skin.

Visible light is the only e-m wave that our eyes can pick up, i.e the only e-m wave we can see. The frequency of this spectrum corresponds to a band in the vicinity of 405–790 THz. It can further be separated into different colors. It makes up a large portion of the e-m waves coming from the sun.

Infrared wave is an e-m radiation whose wavelengths longer than those of visible light. It is is generally invisible to the human eye. The wavelength of an infrared wave extend from the red edge of the visible spectrum at 700 nanometers (frequency 430 THz), to 1 millimeter (300 GHz). Most of the thermal radiation emitted by objects near room temperature is infrared. Most of the heat from the sun reach us as infrared radiation. As with all e-m radiation, infrared radiation carries radiant energy and behaves both like a wave and like a photon.

Answer:

t = 0.25 seconds

Explanation:

Given that,

Initial speed of a bullet, v = 800 m/s

Distance from the target is 200 m

We need to find the time required for the bullet to reach the target. Time is simply calculated by the definition of velocity i.e.

[tex]t=\dfrac{d}{v}\\\\t=\dfrac{200\ m}{800\ m/s}\\\\t=0.25\ s[/tex]

So, it will take 0.25 seconds to reach the target.

Answer:

The number of photons of light produced by the laser in 1 hour is

1 Photon / hour

Explanation:

Number of photons of light produced is given by

[tex]Number of photons = \frac{Power}{Energy}[/tex]

From the question,

Power = 5 mW (milliwatts) = 5 × 10⁻³ W

Since 1 Watt = 1 Js⁻¹

Then, 5 × 10⁻³ W = 5 × 10⁻³ Js⁻¹

For the Energy,

As given from the question

Energy=(Power)x(time)

Time = 1 hour = (1 × 60 × 60) s = 3600 s

∴ Energy = 5 × 10⁻³ Js⁻¹ x 1 x 3600s

Energy =18.0 J

Now for the Number of photons produced,

[tex]Number of photons = \frac{Power}{Energy}[/tex]

Power = 5 × 10⁻³ Js⁻¹ = 0.005 Js⁻¹

[tex]Number of photons = \frac{0.005}{18}[/tex]

Number of photons = 2.78 × 10⁻⁴ Photons / sec

This is the number of photons produced in 1 second.

For the number of photons produced in 1 hour, we will multiply the result by 3600

(NOTE: 1 sec = [tex]\frac{1}{3600}[/tex] hour)

Number of photons = 2.78 × 10⁻⁴ Photons / sec

= 2.78 × 10⁻⁴ × 3600 Photons / hour

= 1.0008 Photons / hour

≅ 1 Photon / hour

Hence, the number of photons of light produced by the laser in 1 hour is

1 Photon / hour

Answer:

2.7%

Explanation:

Given:

Uncertainty of the speedometer (u)= 2.5km/h

Speed measured at that uncertainty (v) = 92km/h

Percent uncertainty (p) is given as the ratio of the uncertainty to the speed measured then multiplied by 100%. i.e

p = [tex]\frac{u}{v} * 100[/tex]%

p = [tex]\frac{2.5}{92} * 100[/tex]%

p = 2.7%

Therefore, the percent uncertainty is 2.7%

Answer:

The force is  [tex]F = 423.04 \ N[/tex]

Explanation:

From the question we are told that

   The weight is  [tex]W = 35 .0 \ N[/tex]

    The  force constant is  [tex]k = 220 \ N/m[/tex]

    The frequency is  [tex]f = 10.5 \ Hz[/tex]

     The amplitude is  [tex]A = 3.00 \ cm = 0.03 \ m[/tex]

Generally the maximum driving force is mathematically represented as

     [tex]F = m * w^2 A[/tex]

Here  m is the mass of the weight which is  mathematically represented as  

       [tex]m = \frac{ W }{g}[/tex]

=>     [tex]m = \frac{ 35 }{9.8 }[/tex]

=>     [tex]m = 3.571 \ kg[/tex]

  Also   [tex]w[/tex] is the angular frequency of the weight which is mathematically represented as

         [tex]w = 2 \pi * f[/tex]

         [tex]w = 2* 3.142 * 10[/tex]

=>      [tex]w = 62.84 \ rad/s[/tex]

So

    [tex]F = 3.571 * 62.84^2 * 0.03[/tex]

    [tex]F = 423.04 \ N[/tex]

Answer:

soccer ball

Explanation:

Because it's round in shape and therefore cannot be stopped as both moves at same speed. There's no edges or corners to bounce up and slow the ball.

Answer:

They will both be since they are going the same speed

Hope it helps please mark brainliest

Answer:

8×10⁻¹⁷ N

Explanation:

from the question, Electric force is given as

F = QV/r.............. Equation 1

Where F = Electric Force,  Q = Charge, V = Electric potential, r = distance.

Given: Q = 8×10⁻¹⁹ C, V = 5 kV = 5000 V, r = 2 cm = 0.02 m.

Substitute into equation 1

F = 8×10⁻¹⁹(5000)(0.02)

F = 8×10⁻¹⁷ N

Hence the electric force on the dust particle is  8×10⁻¹⁷ N

A cyclist must lean into a turn to prevent tipping over in the other direction.The frictional force provides the centripetal force necessary to turn the cyclist to the left.But the frictional force also produces a clockwise torque that will cause the rider and the bicycle to tip clockwise to the right.The force is provided by the friction of the tires.

Answer:

to become stable

Explanation:

when it bends it body, it moves closer to the center of gravity which makes the bicycle stable and hence the turn can be taken easily

Answer:

1)   α = 2.2 rad / s² , 2)   t = 7.068 s , 3) in this interval   s = 23.096 m

total distance    s = 57.4 m

Explanation:

For this exercise we use the angular kinematic relations, before starting the problem we reduce all the magnitude to the SI system

        v₀ = 88 km / h (1000 m / 1km) 1h / 3600s) = 24.44 m / s

        v = 56.0 km / h = 15.55 m / s

        θ = 90 rev (2pi rad / 1 rev) = 565,487 rad

        d = 0.84 m

          r = d / 2

         r = 0.84 / 2

         r = 0.42 m

1) ask for angular acceleration

        w² = w₀² - 2 α Δθ

        α = (w₀² -w²) / 2 Δθ

To find the angular velocities we use the acceleration between the linear and angular velocity

         v = w r

         w = v / r

         w₀ = 24.44 / 0.42

         w₀ = 58.20 rad / s

         w = 15.55 / 0.42

         w = 30.037 rad / s

we calculate

        α = (58.20² - 30.037²) / (2   565.487)

        α = 2.2 rad / s²

2) how much longer does it take to stop

         w₂ = 15.55 rad / s

         w = 0

         w = w₂ - α t

         t = (w₂ -0) / α

         t = 15.55 / 2.2

         t = 7.068 s

3) the distance that the car travels from the beginning of the movement, we can find it by looking for the number of revolutions until it stops and then using the relationship between the angular and linear variable

         w² = w₀² - 2 α θ

at the end of the movement speed is zero

         0 = w₀² - 2 α θ

         θ = w₀² / 2 α

         θ = 24.44² / (2 2.2)

          θ = 135.80 rad

If the angles are measured in radians, we can apply the relation

         θ = s / R

         s = R ttea

         s = 0.42 135

         s = 57.4 m

this is the distance from when the movement starts

the distance for the final part of the movement is

          w = 15.55 rad / s

          θ = w² / 2 α

          θ = 15.55 2 / (2 2.2)

          θ = 54.99 rad

the distance in this interval is

          s = 0.42 54.99

          s = 23.096 m

Answer:

The electric field due to charge at distance r is [tex](-0.8169\times10^{6},0.8169\times10^{6}, 2.800\times10^{6})\ N/C[/tex]

Explanation:

Given that,

Initial position of particle [tex]r_{1}=4.00\times10^{-9}i-5.00\times10^{-9}j-2.00\times10^{-9}k\ m[/tex]

Final position of particle [tex]r_{2}=-2.00\times10^{-9}i+4.00\times10^{-9}j+ 3.00\times10^{-9}k\ m[/tex]

We need to calculate the magnitude of the position vector

Using formula of position vector

[tex]\vec{r}=\vec{r_{2}}-\vec{r_{1}}[/tex]

Put the value into the formula

[tex]\vec{r}=-2.00\times10^{-9}i+4.00\times10^{-9}j+ 3.00\times10^{-9}k-(4.00\times10^{-9}i-5.00\times10^{-9}j-2.00\times10^{-9}k)[/tex]

[tex]\vec{r}=-9\times10^{-9}i+9\times10^{-9}j+5\times10^{-9}k[/tex]

[tex]\vec{|r|}=\sqrt{(9^2+9^2+5^2)\times10^{-18}}[/tex]

[tex]\vec{|r|}=13.6\times10^{-9}\ m[/tex]

We need to calculate the unit vector along electric field direction

Using formula of unit vector

[tex]\hat{r}=\dfrac{\vec{r}}{|\vec{r}|}[/tex]

Put the value into the formula

[tex]\hat{r}=\dfrac{-9\times10^{-9}i+9\times10^{-9}j+5\times10^{-9}k}{13.6\times10^{-9}}[/tex]

[tex]\hat{r}=-0.66i+0.66j+0.36k[/tex]

We need to calculate the electric field due to charge at distance r

Using formula of electric field

[tex]E=\dfrac{kq}{r^2}\hat{r}[/tex]

Put the value into the formula

[tex]E=\dfrac{9\times10^{9}\times1.6\times10^{-19}}{(13.6\times10^{-9})^2}\times(-0.66i+0.66j+0.36k)[/tex]

[tex]E=7.78\times10^{6}\times(-0.66i+0.66j+0.36k)[/tex]

[tex]E=-0.8169\times10^{6}i+0.8169\times10^{6}j+2.800\times10^{6}k\ N/C[/tex]

[tex]E=(-0.8169\times10^{6},0.8169\times10^{6}, 2.800\times10^{6})\ N/C[/tex]

Hence, The electric field due to charge at distance r is [tex](-0.8169\times10^{6},0.8169\times10^{6}, 2.800\times10^{6})\ N/C[/tex]

In each case, the larger unit is the one that represents a greater quantity or magnitude compared to the other unit. For example, a centimeter is larger than a millimeter because one centimeter is equal to 10 millimeters. Similarly, a megagram is larger than a kilogram as one megagram is equivalent to 1,000 kilograms. The pattern continues for each pair, where the larger unit represents a higher order of magnitude or a greater number of the smaller units.

To know more about magnitude, here

brainly.com/question/31022175

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Explanation:

Assuming the circular path is horizontal, the sum of forces in the centripetal direction is:

∑F = ma

T = mv²/r

T = (1.0000 kg) (5.00 m/s)² / (0.500 m)

T = 50.0 N

Answer:

The geocentric model of Ptolemy in which the sun and other planets were believed to move round the Earth.

Explanation:

The Copernicus heliocentric model of the solar system was put forward by  Nicolaus Copernicus and was published in 1543. The model proposed by Copernicus puts the Sun near the center of the Universe, motionless, with Earth and the other planets orbiting around the sun in circular paths, modified by epicycles, and at uniform speed. This is contrary to Ptolemy's geocentric solar system that puts the Earth at the center of the solar system, in which the sun and other planets were believed to move round the Earth.

Answer:

1670 kg

Explanation:

Convert the dimensions to cm.

12 ft × (12 in/ft) × (2.54 cm/in) = 365.76 cm

32 in × (2.54 cm/in) = 81.28 cm

Calculate the volume.

V = πr²h

V = π (81.28 cm / 2)² (365.76 cm)

V = 1,897,813.27 cm³

V = 1,897,813.27 mL

V = 1,897.81 L

Density of benzene is 0.88 kg/L.

m = (0.88 kg/L) (1,897.81 L)

m = 1670 kg

Answer:

The angular acceleration of the top is 0.18 rad/s²

Explanation:

Given;

angular velocity of the top, ω = 16 rad/s

time it takes to come to a complete stop, t = 88.9s

The angular acceleration of the top, is calculated as;

[tex]\alpha = \frac{\omega}{t}[/tex]

α is the angular acceleration

ω is angular acceleration

t is the time

[tex]\alpha = \frac{\omega}{t}\\\\\alpha = \frac{16}{88.9}\\\\\alpha =0.18 \ rad/s^2[/tex]

Therefore, the angular acceleration of the top is 0.18 rad/s²