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aleksklad [387]
3 years ago
5

1. Karuzela ma średnicę 3 m. Dziecko siedzące na karuzeli potrzebuje 3 s na wykonanie jednego pełnego obrotu. Oblicz częstotliwo

ść obrotów karuzeli
oraz prędkość dziecka na karuzeli.
2. Wskazówka godzinowa zegara ma długość 1,2 m a minutowa posiada długość 2,8 m. Wyznacz stosunek częstotliwości obrotów tych wskazówek. Ile
razy większą prędkość ma punkt leżący na końcu wskazówki minutowej od punktu leżącego na końcu wskazówki godzinowej.
Physics
1 answer:
DochEvi [55]3 years ago
7 0

First, figure out what the angular speed of the merry-go-around is.

v ϖ = Using the formula for linear speed Since the angular speed is constant, there is no angular acceleration. Tangential acceleration is t a Radial acceleration is r a Thus the total acceleration is a = rω =1.2m×1.6rad / s =1.9m/ s = rα =1.2m×0rad / s2 = 0m/ s2 =rϖ2 =1.2m×(1.6rad / s)2 =3.1m/ s2 2 2 t r = a +a = 0+(3.1)2 =3.1m/ s2 1 4.0 rev

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Explain what is meant by “field” and compare the properties of gravitational, electric, and magnetic forces in terms of particle
notsponge [240]
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6 0
3 years ago
Part 1 :
dybincka [34]

Answer:

1. 218.55 N

2. 30.96^{o}

3. 2.1 m/s^{2}

Explanation:

Part 1;

Net force F=mg sin \theta where m is mass, g is gravitational force and \theta is the angle of inclination

F= 46*9.8*sin 29^{o}= 218.55N

Frictional force, F_{r} is given by

F_{r} = \mu_{s}mg cos \theta where \mu_{s} is the coefficient of static friction

F_{r} = 0.6*46*9.8 cos 29

F_{r}=236.57N

Since F_{r}>F, therefore, the block doesn’t slip and the frictional force acting is mgh=218.55N

Part 2.

Using the relationship that

Frictional force F_{s} = \mu_{s} mg cos \theta

mg sin \theta =\mu_{s} mg cos \theta

\mu_{s}= \frac {sin \theta}{cos \theta}

\mu_{s}= tan \theta

The maximum angle of inclination \theta = tan^{-1} \mu_{s}

\theta = tan^{-1} (0.6)

\theta= 30.96^{o}  

        

Part 3:

Net force on the object is given by

ma = mg sin 38 - \mu_{k} mg cos 38 where \mu_{k} is the coefficient of kinetic friction

 a = g ( sin 38 - \mu_{k} cos 38 )

                 = 9.8 ( sin 38 - (0.51) cos 38 )

                = 2.1m/s^{2}

7 0
4 years ago
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