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Mamont248 [21]
3 years ago
6

A 23.0 kg child is riding a playground merry-go-round that is rotating at 30.0 rpm. What centripetal force, in N, must she exert

to stay on if she is 2.00 m from its center
Physics
1 answer:
sammy [17]3 years ago
5 0

Answer:

The value is  F  =  454 \  N

Explanation:

From the question we are told that

   The mass of the child is  m  =  23.0 \  kg

   The angular velocity is  w =  30 \ rpm  = \frac{2 \pi  *  30 }{60} = 3.142  \ rad/s

   The distance from the center is  r =  2.0 \  m

Generally the centripetal force is mathematically represented as

       F  =  m *  w^2 *  r

=>   F  =  23 *  3.142 ^2 *   2    

=>   F  =  454 \  N  

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Which of the following statements are true?Check all that apply.
MaRussiya [10]

Answer:

1.The Sun is located at one of the foci of the planets' elliptical orbits.

2.The path of the planets around the Sun is elliptical in shape.

Explanation:

As per Kepler's law of planet motion we know that all planets revolve around the sun in elliptical path in such a way that position of Sun must be at one of the focii of the path

So all planets are in elliptical path always

Position of sun is always at one of the focus

so correct answer will be

1.The Sun is located at one of the foci of the planets' elliptical orbits.

2.The path of the planets around the Sun is elliptical in shape.

4 0
3 years ago
A meteor moving 468 km per minute traveling in a south-to-north direction passed near Earth in 2013. Does this statement describ
Thepotemich [5.8K]

Answer:

this statement describes meteor's velocity,

because velocity is a vector quantity which has both magnitude as well as a specific direction and here the meteor's direction is specified in the statement hence we conclude that this statement describes meteor's velocity as well as speed too.

3 0
3 years ago
As a car drives with its tires rolling freely without any slippage, the type of friction acting between the tires and the road i
Darya [45]
<span>As a car drives with its tires rolling freely without any slippage, the type of friction acting between the tires and the road is kinetic friction.

We exert force to move the object from rest and in this case, static friction works. But, when the object comes in motion, then kinetic friction works. Here, since the car is driving without slipping means, kinetic friction acts on it. Its also called sliding or dynamic friction.</span>
5 0
3 years ago
A wheel has a radius of r = 2.0 m and it rolls down a smooth incline. The height of the incline is h = 8.0 m . What is the angul
zepelin [54]

The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.

For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.

As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.

i.e.

P.E = K.E + R.K.E

\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}

\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}

\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 +  \omega^2}

\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2}  }

\mathbf{\omega^2=\dfrac{39.24 }{2}}

\mathbf{\omega=\sqrt{19.62 } \ rad/sec}

\mathbf{\omega=4.429 \ rad/sec}

Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

Learn more about angular velocity here:

brainly.com/question/1452612

4 0
2 years ago
Read 2 more answers
The car travels 52m in 21 seconds before it comes to a complete stop after the driver applies the brakes. What was the car's ini
prisoha [69]

Answer:

2.48 m/s

Explanation:

D=vt

52 = V 21

V= 52/21=2.47m/s

5 0
3 years ago
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