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Flauer [41]
2 years ago
11

. A blowing ball with a 12 cm radius has an initial angular velocity of 5.0 rad/s. Sometime later, after rotating through 5.5 ra

dians, the bowling ball has an angular velocity of 1.5 rad/s. What is the average angular velocity of the bowling ball
Physics
1 answer:
Anna35 [415]2 years ago
7 0

Answer:

3.3 rad/sec.

Explanation:

The term '' average angular velocity'' can be defined as the rate at which the angular velocity changes. Kindly note that the rate in which the angular velocity changes is in radian per seconds[rad/sec]. So, without mincing words let's dive straight into the solution to the question above.

The radius of the blowing ball = 12cm, the initial angular velocity = 5.0 rad/s, the final angular velocity =  1.5 rad/s and the angle of rotation = 5.5 radians.

Therefore, from the definition of average angular velocity given above, the mathematical representation can be written as given below:

The average angular velocity  = [ the  initial angular velocity] + [the final angular velocity] / 2.

Thus, the average angular velocity = [ 5.0 rad/s + 1.5 rad/s] / 2 = 3.3 rad/s.

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Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

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3 years ago
Which would deliver a greater change in momentum to an opponent’s body - a dodgeball that traveled at 10m/s and rebounded with a
Mazyrski [523]
2nd sentence seems bout right ?? i think
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3 years ago
A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0
otez555 [7]

The given question is incomplete. The complete question is as follows.

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.

Explanation:

The given data is as follows.

    F_{1} = 20 N, F_{2} = 25 N, a = -0.9 m/s^{2}

             W = 83 N

         m = \frac{83}{9.81}

             = 8.46

Now, we will balance the forces along the y-component as follows.

       N = W + F_{2}

           = 83 + 25 = 108 N

Now, balancing the forces along the x component as follows.

       F_{1} - F_{r} = ma

        20 - F_{r} = 8.46 \times (-0.9)

             F_{r} = 7.614 N

Also, we know that relation between force and coefficient of friction is as follows.

             F_{r} = \mu \times N

          \mu = \frac{F_{r}}{N}

                    = \frac{7.614}{108}

                    = 0.0705

Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.

7 0
3 years ago
Read 2 more answers
A satellite orbits earth with a mean altitude of 361 km. If the orbit is circular, what are the satellite's time period and spee
Advocard [28]

Answer:

v = 7.69 x 10³ m/s = 7690 m/s

T = 5500 s = 91.67 min = 1.53 h

Explanation:

In order for the satellite to orbit the earth, the force of gravitation on satellite must be equal to the centripetal force acting on it:

F_{gravitation}= F_{centripetal}\\\\\frac{GM_{s} M_{E}}{r^2}  = \frac{M_{s} v^2}{r}\\\\\frac{GM_{E}}{r} = v^2\\\\v = \sqrt{\frac{GM_{E}}{r} } \\\\

where,

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Me = Mass of Earth = 5.97 x 10²⁴ kg

r = distance between the center of Earth and Satellite = Radius of Earth + Altitude = 6.371 x 10⁶ m + 0.361 x 10⁶ m = 6.732 x 10⁶ m

v = orbital speed = ?

Therefore,

v = \sqrt{\frac{(6.67 x 10^{-11}N.m^2/kg^2)(5.97 x 10^{24} kg)}{6.732 x 10^6 m} }\\\\

<u>v = 7.69 x 10³ m/s</u>

For time period satellite completes one revolution around the earth. It means that the distance covered by satellite is equal to circumference of circle at the given altitude.

So, its orbital speed can be given as:

v = \frac{Circumference of Circle at Given Altitude}{T}\\\\v =  \frac{2\pi r}{T}\\\\

where,

T = Time Period of Satellite = ?

Therefore,

T = \frac{2\pi r}{v}\\\\T = \frac{(2\pi )(6.732 x 10^6 m}{7.69 x 10^3 m/s}\\\\

<u>T = 5500 s = 91.67 min = 1.53 h</u>

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