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svet-max [94.6K]
3 years ago
6

Determine the total moment of inertia of a merry-go round with 5 children sitting on it. Of the five children, four are seated a

t the edge of the merry-go-round, and the fifth child is sitting exactly at the center. The children have the same mass of 30.0 kg. You may consider the merry-go-round as a cylinder of mass 150 kg and a radius 2.0 meters.
Physics
1 answer:
BaLLatris [955]3 years ago
7 0

Answer:

Explanation:

Given that,

We have five children.

Each of mass m =30kg

They sit on a merry go round

Mass of Merry go round M= 150kg

Radius of Merry go round is r =2m

Four children sit at the edge of the merry go round but one child sit at the centre.

The four child that sit at the edge are 2m from the centre of the merry go round but the one at the centre is 0m from the centre

Moment of inertia?

Moment of inertia is given as

I=Σmi•ri²

For the question, the moment of inertia is the combination of inertial of child and the merry go round

I= I(merry go round) + I(four child)+ I(last child)

The merry go round is assumed to be a solid cylinder, so it is going to have the moment of inertia of solid cylinder

Then,

I(merry go round ) =½ Mr²

Also, Four of the child has the same moment of inertia, they are 2m form the centre of the merry go round why the last child has no moment of inertia

I= I(merry go round) + I(four child) +I(last child )

I= ½Mr² + 4mr² + mr'²

I = ½ × 150 ×2² + 4×30×2² + 30×0²

I = 300 +480+0

I = 780 kgm²

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Answer:

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A 65-kg skier grips a moving rope that is powered by an engine and is pulled at a constant speed to the top of a 230 hill. The s
Sveta_85 [38]

Answer:

The required power by the engine is 33.0 hp

Explanation:

Solution

Newton's second law says that, the net force Fnet on an object of mass m will accelerates the object

Where

Fnet = ma

a = acceleration

θ = angle of incline,

m = mass of the 30 skiers,

f = frictional force

N = normal force

mg sinθ, mg cos θ are components of weight skier

F = the force applied by engine

Now,

The skier mass  is 65 kg

We calculate the mass of the 30 skier

m = 30 (65kg) = 1950 kg

Calculate the net force acting on the skiers along the x-axis

Fnet, x=ma

Now,

F-mg sin θ - f = 0

F= mg sin θ + f -----(1)

The kinetic frictional force is denoted by

f = μk N ------(2)

μk = The coefficient of the kinetic friction

We now, calculate the net force acting on the skiers along y axis

Fnet, y = ma

N- mg cos θ = 0

so,

N = mg cos θ

This value is  substituted in equation (2)

f = μk mg cos θ

we substitute the value for equation (1)

F = mg sin θ + μk mg cos θ

mg =  sinθ + μk cos θ)-----(3)

The next step is to calculate the work done by the engine in pulling the skiers, the incline top by applying the equation 3

W = Fx

= mg ( sinθ + μk cos θ)x

x = the displacement

we now substitute 1950 kg for m, 23° for θ, 0.10 for μk and 320m for x

so,

W = mg ( sinθ + μk cos θ)x

= (1950 kg) (9.81 m/s²) (sin 23° + (0.10) cos 23°) (320 m)

= 2.99 * 10 ^6 J

Then,

The time from minute to s is converted

t =(2.0min) ( 60sec/1.0min) = 120 sec

Now we calculate the power needed by the engine to pull the skiers at the incline top

Thus,

P = W/t

we substitute  2.955 * 10 ^6 J for W and  120 s for t

we have,

P = 2.955 * 10 ^ 6 J/ 120 s

= ( 2.4625 * 10 ^ 4 W) (1.0 hp/746 W)

= 33.0 hp

In conclusion the required power by the engine is 33.0 hp

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3 years ago
What is the equation for horizontal velocity
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3 years ago
A 110 V power line is protected by a 15 A fuse. What is the maximum number of 400 W lamps that can be simultaneously operated in
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Answer:

Total number of lamps will be 4            

Explanation:

We have given power of the lamp W = 400 watt

Potential difference across the lamp V=110 volt

We know that power is equal to P=VI

So 400=110\times I

I=3.636A

Total current is given 15 A

As it is given that lamps are connected in parallel so total current is the sum of current through each lamp

So number of lamp will be n=\frac{15}{3.636}=4.125

As the lamp can not be in negative

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5 0
3 years ago
Sound enters the ear, travels through the auditory canal, and reaches the eardrum. The auditory canal is approximately a tube op
Juliette [100K]

Answer:

The fundamental frequency of can is 2.7 kHz.                          

Explanation:

Given that,

A typical length for the auditory canal in an adult is about 3.1 cm, l = 3.1 cm

The speed of sound is, v = 336 m/s

We need to find the fundamental frequency of the canal. For a tube open at only one end, the fundamental frequency is given by :

f=\dfrac{v}{4l}\\\\f=\dfrac{336}{4\times 3.1\times 10^{-2}}\\\\f=2709.67\ Hz\\\\f=2.7\ kHz

So, the fundamental frequency of can is 2.7 kHz. Hence, this is the required solution.

7 0
2 years ago
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