A pendulum on a planet, where gravitational acceleration is unknown, oscillates with a time period 5 sec. If the mass is increased six times, what is the time period of the pendulum?

Answer:

* air resistance.

*the direction of the rotation of the Earth

rotation of the thrown body

Explanation:

The projectile launch is described by the expressions

x-axis         x = v₀ₓ t

y-axis         y = [tex]v_{oy}[/tex] t - ½ gt²

When the things that affect this movement are analyzed, in order of importance we have:

* air resistance. This significantly changes the body's horizontal position, so it introduces a horizontal acceleration that is not contained in the equations.

* air resistance. At the height that the body reaches, since air resistance has the same direction as the gravity of gravity and therefore the relationship is more challenging.

* to a lesser extent the direction of launch, in the direction of the rotation of the Earth against. Since this creates an operational on the x and y axis that changes the initial assumption

* The possible rotation of the thrown body, since this rotation creates a lift that is not taken in the equations, this value is more noticeable the lighter the body, this effect has to keep the body longer in the air achieving more reach and height

Answer:

2500 N/m²

Explanation:

Pressure: This can be defined as the force acting normally on a surface per unit area.

The expression for pressure is give as

P = F/A...................... Equation 1

Where P = pressure (N/m²), F = force (N), A = Contact area (m²)

Given: F = 500 N, A = 2000 cm² = (2000/10000) m = 0.2 m.

Substitute into equation  1

P = 500/0.2

P = 2500 N/m²

Hence the pressure exerted to the surface is 2500 N/m²

Answer:b

Explanation:I’m just trynna get more money dude

Answer:

Increases

Explanation:

Due to an increase in temperature, molecules within the medium will vibrate more vigorously, meaning that the rate of chemical reactions generally increases with temperature due to an increase in kinetic energy. Because sound is a form of kinetic energy, it is safe to assume that the speed of sound waves increases with temperature.

Answer:

A- increases because The particles bump into each other more often.

Explanation:

Just took the test

Answer:

  0.365 cm³

Explanation:

The change in volume is found by multiplying the coefficient of expansion by the volume and the temperature change. The temperature change is in °F, but the expansion coefficient is per °C, so we need to convert the temperature scale in the computation.

  ΔV = V·Ce·ΔT

  = (π/6·d³)(1.5×10⁻⁶/°C)((5 °C)/(9 °F))(305 °F)

  = (1436.76 cm³)(1.5×10⁻⁶/°C)(169.44 °C)

  = 0.365 cm³ . . . . increase in volume

Answer:

a) The spring constant of the spring is [tex]28.381\,\frac{N}{m}[/tex], b) The angular frequency of the block is [tex]7.038\,\frac{rad}{s}[/tex].

Explanation:

This question is incomplete and complete version will be presented herein:

A spring is hung from the ceiling. A 0.573-kg block is then attached to the free end of the spring. When released from rest, the block drops 0.198 m before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring (b) Find the angular frequency of the block 's vibrations.

a) Since spring is hung from the ceiling and is stretched by action of gravity on 0.573 kilogram block. According to the Hooke's Law, force experimented by the spring is directly proportional to elongation. An expression describing the phenomenon is presented and described below: (System at equilibrium - Newton's Second Law)

[tex]m\cdot g = k\cdot \Delta x[/tex]

Where:

[tex]m[/tex] - Mass, measured in kilograms.

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

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

[tex]\Delta x[/tex] - Spring linear deformation, measured in meters.

Now, the spring constant is cleared in this equation and outcome is computed: ([tex]m = 0.573\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\Delta x = 0.198\,m[/tex])

[tex]k = \frac{m\cdot g}{\Delta x}[/tex]

[tex]k = \frac{(0.573\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.198\,m}[/tex]

[tex]k = 28.381\,\frac{N}{m}[/tex]

The spring constant of the spring is [tex]28.381\,\frac{N}{m}[/tex].

b) Let suppose that mass-spring system is experimenting a simple harmonic motion, so that angular frequency is equal to:

[tex]\omega = \sqrt{\frac{k}{m} }[/tex]

Given that [tex]k = 28.381\,\frac{N}{m}[/tex] and [tex]m = 0.573\,kg[/tex], the angular frequency, measured in radians per second, of the block is:

[tex]\omega = \sqrt{\frac{28.381\,\frac{N}{m} }{0.573\,kg} }[/tex]

[tex]\omega = 7.038\,\frac{rad}{s}[/tex]

The angular frequency of the block is [tex]7.038\,\frac{rad}{s}[/tex].

Answer:

India is a peninsula.

Explanation:

India is called as Indian Peninsula because it is surrounded by the Indian ocean on the south, the Arabian sea on the west and the Bay of Bengal on the east.

Answer:

Explanation:

According to newton's secomd law, ∑F = ma

∑F is the summation of the force acting on the body

m is the mass of the body

a is the acceleration

Given the normal force when the elevator starts N1 = 592N

Normal force after the elevator stopped N2 = 400N

When the elevator starts, its moves upward, the sum of force ∑F = Normal (N)force on the elevator - weight of the person( Fg)

When moving up;

N1 - Fg = ma

N1 = ma + Fg ...(1)

Stopping motion of the elevator occurs after the elevator has accelerates down. The sum of forces in this case will give;

N2 - Fg = -ma

N2 = -ma+Fg ...(2)

Adding equation 1 and 2 we will have;

N1+N2 = 2Fg

592N + 400N = 2Fg

992N 2Fg

Fg = 992/2

Fg = 496N

The weight of the person is 496N

\b) To get the person mass, we will use the relationship Fg = mg

g = 9.81m/s

496 = 9.81m

mass m = 496/9.81

mass = 50.56kg

c) To get the magnitude of acceleration of the elevator, we will subtract equation 1 from 2 to have;

N1-N2 = 2ma

592-400 = 2(50.56)a

192 = 101.12a

a = 192/101.12

a = 1.90m/s²

Answer:

The height of the tower is 8.96 m.

Explanation:

We have, a visitor to a lighthouse wishes to determine the height of the tower. She ties a spool of thread to a small rock to make a simple pendulum, which she hangs down the center of a spiral staircase of the tower. The period of oscillation is 6.01 s.

It is required to find the height of the tower. Let it is l. The time period of a simple pendulum is given by :

[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]

l is length of pendulum, or height of tower

[tex]l=\dfrac{T^2g}{4\pi^2}\\\\l=\dfrac{(6.01)^2\times 9.8}{4\pi^2}\\\\l=8.96\ m[/tex]

So, the height of the tower is 8.96 m.

Answer:

☟

Explanation:

Fuels made from oil mixtures containing large hydrocarbon molecules are not efficient as they do not flow easily and are difficult to ignite. Crude oil often contains too many large hydrocarbon molecules and not enough small hydrocarbon molecules to meet demand. This is where cracking comes in.

Cracking allows large hydrocarbon molecules to be broken down into smaller, more useful hydrocarbon molecules. Fractions containing large hydrocarbon molecules are heated to vaporise them. They are then either:

heated to 600-700°C

passed over a catalyst of silica or alumina

These processes break covalent bonds in the molecules, causing thermal decompositionreactions. Cracking produces smaller alkanesand alkenes (hydrocarbons that contain carbon-carbon double bonds). For example:

hexane → butane + ethene

C6H14 → C4H10 + C2H4

Some of the smaller hydrocarbons formed by cracking are used as fuels, and the alkenes are used to make polymers in plastics manufacture. Sometimes, hydrogen is also produced during cracking.

Fractional distillation of crude oil

Fractional distillation separates a mixture into a number of different parts, called fractions.

A tall fractionating column is fitted above the mixture, with several condensers coming off at different heights. The column is hot at the bottom and cool at the top. Substances with high boiling points condense at the bottom and substances with lower boiling points condense on the way to the top.

Crude oil is a mixture of hydrocarbons. The crude oil is evaporated and its vapours condense at different temperatures in the fractionating column. Each fraction contains hydrocarbon molecules with a similar number of carbon atoms and a similar range of boiling points.

Oil fractions

The diagram below summarises the main fractions from crude oil and their uses, and the trends in properties. Note that the gases leave at the top of the column, the liquids condense in the middle and the solids stay at the bottom.

As you go up the fractionating column, the hydrocarbons have:

lower boiling points

lower viscosity (they flow more easily)

higher flammability (they ignite more easily).

Other fossil fuels

Crude oil is not the only fossil fuel.

Natural gas mainly consists of methane. It is used in domestic boilers, cookers and Bunsen burners, as well as in some power stations.

Coal was formed from the remains of ancient forests. It can be burned in power stations. Coal is mainly carbon but it may also contain sulfur compounds, which produce sulfur dioxide when the coal is burned. This gas is a cause of acid rain. Also, as all fossil fuels contain carbon, the burning of any fossil fuel will contribute to global warming due to the production of carbon dioxide.

In fractional distillation, the crude oil is added to the chamber and heated. The components with the highest boiling point will condense in the lower part of the column and the components with the lower boiling point will condense at the top of the column. Petrol with a low boiling point is collected from the top of the column.

Fractional distillation can be described as the separation of a mixture into its component fractions. The chemical compound is separated by heating them to a temperature at which fractions of the mixture will vaporize.

Generally, the components have boiling points that differ by less than 25 °C  from each other under one atmosphere. When the mixture is heated, the component with the lower boiling point boils and changes to vapours.

The more volatile component remains in a vapour state and repeated distillations are used in the process, and the mixture is separated into component parts.

Therefore, the petrol from the crude oil can easily be separated as it has a boiling point of about 25-60°C.

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Answer:

4.93 m

Explanation:

According to the question, the computation of the height is shown below:

But before that first we need to find out the speed which is shown below:

As we know that

[tex]Speed = \frac{Distance}{Time}[/tex]

[tex]Speed = \frac{5}{0.504}[/tex]

= 9.92 m/s

Now

[tex]v^2 - u^2 = 2\times g\times h[/tex]

[tex]9.92^2 = 2\times 9.98 \times h[/tex]

98.4064 = 19.96 × height

So, the height is 4.93 m

We simply applied the above formulas so that the height i.e H could arrive

Answer:

correct answer is c

v = -0.8 m / s

Explanation:

This is a problem of quantity of movement, for this we must define a system formed by the two cars, so that the forces during the collision are internal and therefore the quantity of movement is conserved

initial

    p₀ = m₁ v₁ - m₂ v₂

final

    = (m₁ + m₂) v

We have taken the direction to the right as positive

    p₀ =p_{f}

    m₁ v₁ - m₂ v₂ = (m₁ + m₂) v

    v = (m₁ v₁ - m₂ v₂) / (m₁ + m₂)

we calculate

    v = (2  4 - 8  2) / (2 + 8)

    v = (8 -16) / 10

     v = -0.8 m / s

the negative sign indicates that the set is moving to the left

correct answer is c

Answer:

a. 900 J

b. 0.383

Explanation:

According to the question, the given data is as follows

Horizontal force = 150 N

Packing crate = 40.0 kg

Distance = 6.00 m

Based on the above information

a. The work done by the 150-N force is

[tex]W = F x = \mu N x = \mu\ m\ g\ x[/tex]

[tex]W = 150 \times 6[/tex]

= 900 J

b. Now the coefficient of kinetic friction between the crate and surface is

[tex]\mu = \frac {F}{m\timesg}[/tex]

[tex]= \frac{150}{40\times 9.8}[/tex]

= .383

We simply applied the above formulas so that each one part could calculate

We want to find the work and kinetic friction for the given situation. The solutions are:

Here we have a horizontal force of 150N pushing a 40.0 kg packing crate a distance of 6.00m at a constant speed.

a) First we want to find the work, it is given by the force applied times the distance moved, so the work is just:

W = 150N*6.00m = 900 N*m

b) Now we want to find the coefficient of kinetic friction, it must be such that the kinetic friction force is equal to the pushing force, in this way there is no net force, and then there is no acceleration.

Remember that the friction force is:

F = m*g*μ

Where:

Then we must solve:

150N = 40kg*(9.8 m/s^2)*μ = 392N*μ

150N/392N = 0.38

So the coefficient of kinetic friction between the crate and the surface is 0.38

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Answer:

25m

Explanation:

Let's assume the Jeep attains a velocity of 36km/h ; a constant speed same with that of the car.

While the Jeep is accelerating to that speed, the car with that speed passes it.

Now we can calculate the time taken for the Jeep to attain the velocity of 36km/h on her constant acceleration.

This time is t = v/a; from Newton's Law of Motion:

a = V-U / t ; a-acceleration

V is final velocity = 36km/h

U is initial velocity 0 since the body starts from rest.

Hence t = 36000/3600 ÷ 4 = 2.5s

Note conversting from km/h to m/s we multiply by 1000/3600.

But the distance covered by the car while the Jeep just accelerates is

S = U × t = 10× 2.5 = 25m.

Note From Newton's law of Motion, distance for constant speed is defined as: U × t

Hence the Car would be 25m off the starting point just as the Jeep accelerates. It would overtake the Jeep when it just covers 25m from the Jeep starting point.

Answer:

The answer is D)Theory

Explanation:

This is due because a theory is a scientific term and is a testable model that scientists seek to explain a phenomenon. You can also find out the answer by the process of elimination it can't be data because that would be something they already know and something they use to prove not explain. It can't be law because it isn't testable but can be used to explain. So that leaves you with two answers hypothesis and theory which are very similar but it isn't hypothesis because it isn't used to explain it to help the scientists come up with a theory and accumulate what might happen.

Answer:

Silver.

Explanation:

To determine the identity of the metal, we need to calculate the density of the metal. This is illustrated below:

Mass of metal (m) = 18.15g

Length (L)= 1.2cm

Volume (V) = L³ = 1.2³ = 1.728cm³

Density =.?

The density of a substance is simply defined as the mass of the substance per unit volume of the substance. Mathematically, it is expressed as:

Density = Mass /volume

With the above formula, we can obtain the density of the metal as follow:

Mass = 18.15g

Volume = 1.728cm³

Density =.?

Density = Mass /volume

Density = 18.15g/1.728cm³

Density of the metal = 10.50g/cm³

Comparing the density of metal obtained with the densities given in the table above, we can see that the density of the metal is the same with that of silver.

Therefore, the metal is silver.

Explanation:

a scientific theory is a well substantiated explanation of some aspect of the nature world, based on a body of facts that have been repeatedly confirmed through observation and experiment. search fact-supported theories are not "guesses" but reliable account of the real world .

Answer:

9.05 m/s ,   -14.72°  (respect to x axis)

Explanation:

To find the final velocity of the bowling ball you take into account the conservation of the momentum for both x and y component of the total momentum. Then, you have:

[tex]p_{xi}=p_{xf}\\\\p_{yi}=p_{yf}\\\\[/tex]

[tex]m_1v_{1xi}+m_2v_{2xi}=m_1v_1cos\theta+m_2v_{2}cos\phi\\\\0=m_1v_1sin\theta-m_2v_2sin\phi[/tex]

m1: mass of the bowling ball = 5.50 kg

m2: mass of the bowling pin = 0.850 kg

v1xi: initial velocity of the bowling ball = 9.0 m/s

v2xi: initial velocity of bowling pin = 0m/s

v1: final velocity of bowling ball = ?

v2: final velocity of bowling pin = 15.0 m/s

θ: angle of the scattered bowling pin = ?

Φ: angle of the scattered bowling ball = 85.0°

Where you have used that before the bowling ball hits the pin, the y component of the total momentum is zero.

First you solve for v1cosθ in the equation for the x component of the momentum:

[tex]v_1cos\theta=\frac{m_1v_{1xi}-m_2v_2cos\phi}{m_1}\\\\v_1cos\theta=\frac{(5.50kg)(9.0m/s)-(0.850kg)(15.0m/s)cos85.0\°}{5.50kg}\\\\v_1cos\theta=8.79m/s[/tex]

and also you solve for v1sinθ in the equation for the y component of the momentum:

[tex]v_1sin\theta=\frac{(0.850kg)(15.0m/s)sin(85.0\°)}{5.50kg}\\\\v_1sin\theta=2.3m/s[/tex]

Next, you divide v1cosθ and v1sinθ:

[tex]\frac{v_1sin\theta}{v_1cos\theta}=tan\theta=\frac{2.3}{8.79}=0.26\\\\\theta=tan^{-1}(0.26)=14.72[/tex]

the direction of the bawling ball is -14.72° respect to the x axis

The final velocity of the bawling ball is:

[tex]v_1=\frac{2.3m/s}{sin\theta}=\frac{2.3}{sin(14.72\°)}=9.05\frac{m}{s}[/tex]

hence, the final velocity of the bawling ball is 9.05 m/s

Answer:

a. FALSE

b. FALSE

c. TRUTH

d. FALSE

e. FALSE

Explanation:

To determine which statements are truth or false you focus in the following formula, for the electric potential generated by a conducting sphere:

[tex]V=\frac{Q}{4\pi \epsilon_o R}[/tex]      inside the sphere

[tex]V'=\frac{Q}{4\pi \epsilon_o r}[/tex]      for r > R (outside the sphere)

R: radius of the sphere

ε0: dielectric permittivity of vacuum

Q: charge of the sphere

As you can notice, inside the sphere the potential is constant. Inside the sphere, the potential is the same. Outside the surface the potential decreases as 1/r, being r the distance to the center of the sphere.

Hence, you can conclude:

a. The potential at the center of the sphere is zero. FALSE

b.The potential is lowest, but not zero, at the center of the sphere. FALSE

c. The potential at the center of the sphere is the same as the potential at the surface. TRUTH

d. The potential at the center is the same as the potential at infinity. FALSE

e. The potential at the surface is higher than the potential at the center. FALSE

Answer:

Explanation:

Moment of inertia of the rod = 1/12 m L²

m is mass of the rod and L is its length

= 1/2 x 2.3 x 2 x 2

= 4.6 kg m²

Moment of inertia of masses attached with the rod

= m₁ d² + m₂ d²

m₁ and m₂ are masses attached , and d is their distance from the axis of rotation

= 5.3 x 1² + 3.5 x 1²

= 8.8 kg m²

Total moment of inertia = 13.4 kg m²

B )

Rotational kinetic energy = 1/2 I ω²

I is total moment of inertia and ω is angular velocity

= .5 x 13.4 x 2²

= 26.8 J .

C )

when mass of rod is negligible , moment of inertia will be due to masses only

Total moment of inertia of masses

= 8.8 kg m²

D )

kinetic energy of the system

= .5 x 8.8 x 2²

= 17.6 J .

(A) Total moment of inertia is 13.4 kgm²

(B) Total kinetic energy is 26.8J

(C) Moment of inertia is  8.8 kgm²
(D) Kinetic energy is 17.6J

(A) The moment of inertia of the rod is given by:

I = 1/12 mL²

where m is the mass of the rod

and L is the length

So,

I = (1/12) × 2.3 × 2²

I = 4.6 kgm²

Now, the moment of inertia of masses attached to the rod is given by:

I' = m₁ d² + m₂d²

where m₁ and m₂ are masses

and d is their distance from the axis of rotation

I' = 5.3 × 1² + 3.5 × 1²

I' = 8.8 kgm²

The total moment of inertia of the system is given by:

I(tot) = I + I'

I(tot) = 13.4 kgm²

(B) The rotational kinetic energy of an object with a moment of inertia I and angular velocity ω is given by:

KE = 1/2 I(tot)ω²

KE = 0.5 × 13.4 × 2²

KE = 26.8J

(C) If the mass of the rod is negligible, then the moment of inertia of the rod will be zero. So the total moment of inertia will be

I(tot) = I' = 8.8 kgm²

(D) the kinetic energy of the system when the mass of the rod is negligible and the angular speed is 2 rad/s is given by:

KE = 1/2 I'ω²

KE = 0.5 × 8.8 × 2²

KE = 17.6J

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Answer:

normal force = 10 N

Explanation:

Given data

frictional force = 0.400 N

coefficient of kinetic friction = 0.04

Solution

we get here normal force that is express as

normal force = [tex]\frac{Frictional\ force}{coefficient\ of\ friction}[/tex]        ............1

put here value and we will get value

normal force = [tex]\frac{0.400}{0.04}[/tex]  

solve it we get

normal force = 10 N

Answer:

D. Velocity of the car

Explanation:

The centripetal force acting on the car is given by the following formula:

[tex]F_c=ma_c=m\frac{v^2}{r}[/tex]    (1)

m: mass of the car  = 888 kg

v: tangential speed of the car = 7 m/s

r: radius of the flat circular track = 59 m

By the form of the equation (1) you can notice that the greatest change in the centripetal force is obtained when the velocity v is twice. In fact, you have:

[tex]F_c=m\frac{(2v)^2}{r}=4m\frac{v^2}{r}=4F_c[/tex]

Then, the greatest values of the centripetal force is:

[tex]F_c=4(888kg)\frac{(7m/s)^2}{59m}=2949.96N[/tex]

The greatest change in Fc is obtained by changing the value of the speed

answer

D. Velocity of the car

Answer:

About 3.06 seconds

Explanation:

[tex]v_f=v_o+at[/tex]

Since at the peak of its trajectory, the ball will have no velocity, you can write the following equation:

[tex]0=30+(-9.81)t\\\\-30=-9.81t\\\\t\approx 3.06s[/tex]

Hope this helps!

Answer:

FInal speed (v) = 0.509 m/s (Approx)

Explanation:

Given:

Mass of ant (m) = 12 mg

Force (f) = 47 N

Time taken (t) = 0.13 ms

Find:

FInal speed (v) = ?

Computation:

Initial velocity (u) = 0

Impulse = change in momentum

Force × TIme = change in momentum

47 × 0.13 = mv - mu

6.11 = 12 (V)

FInal speed (v) = 0.509 m/s (Approx)

Answer:

E) True. The girl has a larger tangential acceleration than the boy.

Explanation:

In this exercise they do not ask us to say which statement is correct, for this we propose the solution to the problem.

Angular and linear quantities are related

          v = w r

          a = α r

the boy's radius is r₁ = 1.2m the girl's radius is r₂ = 1.8m

as the merry-go-round rotates at a constant angular velocity this is the same for both, but the tangential velocity is different

          v₁ = w 1,2 (boy)

          v₂ = w 1.8 (girl)

whereby

          v₂> v₁

reviewing the claims we have

          a₁ = α 1,2

          a₂ = α 1.8

          a₂> a₁

A) False. Tangential velocity is different from zero

B) False angular acceleration is the same for both

C) False. It is the opposite, according to the previous analysis

D) False. Angular acceleration is equal

E) True. You agree with the analysis above,

Answer:

The number of molecules displaced in a vibration makes the amplitude of a sound.

A line of best fit expresses the relationship between the points.

Explanation:

It does not go through all the points but goes through most of them and it is like a hardrawn curve

Answer:

4.437 m/s

Explanation:

Diameter of rotation d is 1.7 m

Radius of rotation = d/2 = 1.7/2 = 0.85 m

If he takes 1.2 sec to complete one revolution, then his angular speed is 1/1.2 = 0.83 rev/s

We convert to rad/s

Angular speed = 2 x pi x 0.83

= 2 x 3.142 x 0.83 = 5.22 rad/s

Speed is equal to the angular speed times the radius of rotation

Speed = 5.22 x 0.85 = 4.437 m/s

In the given case, the speed of the discus at release, If the thrower takes 1.2s to complete one revolution, starting from rest would be - 8.90 m/s.

Given:

radius f the circle would be = 1.7/2 = 0.85 m

This rotation exercise can be treated using the rotation kinematics.

Angular acceleration:

θ = w₀ t + ½ α t²

t = 1.2 s to give a revolution (T = 2π rad) and with part of the rest the initial angular velocity is zero (wo = 0)

 =>  θ = 0 + ½ α t²

 => α = 2θ / t²

=>  α= 2 × 2π / 1.2²

 => α = 4π = 8.7266 rad / s²

Let's calculate the angular velocity:

=> w = wo + α t

=>  w = 0 + α t

=> w = 8.7266 × 1.2

=> w = 10.47192 rad / s

The relationship between linear and angular velocity is

=> r = d / 2

=> r = 1.7 / 2 = 0.85 m

=> v = w r

=> v = 10.47192 × 0.85  

=> v = 8.90 m / s

Thus, the correct speed would be - 8.90 m/s

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Answer:

The centripetal acceleration of the traveler is [tex]40.98\ m/s^2[/tex]

Explanation:

It is given that, A carousel has a diameter of 6.0-m and completes one rotation every 1.7 s.

We need to find the centripetal acceleration of the traveler. It is given by the formula as follows :

[tex]a=\dfrac{v^2}{r}[/tex]

r is radius of carousel

[tex]v=\dfrac{2\pi r}{T}[/tex]

So,

[tex]a=\dfrac{4\pi ^2r}{T^2}[/tex]

Plugging all the values we get :

[tex]a=\dfrac{4\pi ^2\times 3}{(1.7)^2}\\\\a=40.98\ m/s^2[/tex]

So, the centripetal acceleration of the traveler is [tex]40.98\ m/s^2[/tex].