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sveta [45]
3 years ago
15

A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangent

ially to the merry-go-round. Find the kinetic energy of the merry-goround after 3.47 s. The acceleration of gravity is 9.8 m/s 2 . Assume the merry-go-round is a solid cylinder. Answer in units of J. 028 10.0
Physics
1 answer:
Julli [10]3 years ago
5 0

Answer:

The kinetic energy of the merry-go-round is \bf{475.47~J}.

Explanation:

Given:

Weight of the merry-go-round, W_{g} = 826~N

Radius of the merry-go-round, r = 1.17~m

the force on the merry-go-round, F = 57.8~N

Acceleration due to gravity, g= 9.8~m.s^{-2}

Time given, t=3.47~s

Mass of the merry-go-round is given by

m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg

Moment of inertial of the merry-go-round is given by

I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}

Torque on the merry-go-round is given by

\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m

The angular acceleration is given by

\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}

The angular velocity is given by

\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}

The kinetic energy of the merry-go-round is given by

E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J

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Andreyy89

Answer:

27.95[kW*min]

Explanation:

We must remember that the power can be determined by the product of the current by the voltage.

P=V*I

where:

P = power [W]

V = voltage [volt]

I = amperage [Amp]

Now replacing:

P=110*8.47\\P=931.7[W]

Now the energy consumed can be obtained mediate the multiplication of the power by the amount of time in operation, we must obtain an amount in Kw per hour [kW-min]

Energy = 931.7[kW]*30[days]*10[\frac{min}{1day} ]=279510[W*min]or 27.95[kW*min]

3 0
3 years ago
3) A driver in a 1000-kg car traveling at 24 m/s slams on the brakes and skids to a stop. If the coefticient of friction between
Natali5045456 [20]

Answer:

A) 37 m

Explanation:

The car is moving of uniformly accelerated motion, so the distance it covers can be calculated by using the following SUVAT equation:

v^2 -u^2 = 2ad (1)

where

v = 0 m/s is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d is the length of the skid

We need to find the acceleration first. We know that the force responsible for the (de)celeration is the force of friction, so:

F=ma=-\mu mg

where

m = 1000 kg is the mass of the car

\mu = 0.80 is the coefficient of friction

a is the deceleration of the car

g = 9.8 m/s^2 is the acceleration due to gravity

The negative sign is due to the fact that the force of friction is against the motion of the car, so the sign of the acceleration will be negative because the car is slowing down. From this equation, we find:

a=-\mu g

And we can substitute it into eq.(1) to find d:

v^2 -u^2 = 2(\-mu g) d\\d= \frac{v^2-u^2}{-2 \mu g}=\frac{0-(24 m/s)^2}{-2(0.80)(9.8 m/s^2)}=36.7 m \sim 37 m

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3 years ago
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Maksim231197 [3]

Answer:

The Kinetic Energy INCREASE in the following situations:

a)A block slides down a frictionless incline:

e)A merry go round rotates faster due to the push by a person.

Explanation:

The energy kinetics is proportional to the square of the velocity:

E_k=1/2*mv^2

We study the cases:

a)A block slides down a frictionless incline:

The gravity made a positive work and the box get a extra velocity, then the kinetics energy INCREASE

b)A box is pulled across a rough floor at constant speed:

The magnitude of the velocity does not change, so the kinetics energy does not change as well

c)A stone at the end of a string is whirled in a horizontal circle at constant speed:

The magnitude of the velocity does not change, so the kinetics energy does not change as well

d)A projectile approaches its maximum height:

If the projectile approaches its maximum height, its velocity approaches to zero. Then the kinetics Energy DECREASE

e)A merry go round rotates faster due to the push by a person.

Thanks to the push, the magnitude of the velocity INCREASE, so the kinetics energy INCREASE as well

6 0
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PSYW - Please Show Your Work
Ad libitum [116K]

Answer:

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Explanation:

7 0
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Can you help quick in science please
nikdorinn [45]
Uhhhh...you should have paid attention in class, just saying...
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3 years ago
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