# Movimiento Circular Uniforme

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• 1.

### Una esfera hace 5 / π revoluciones por segundo, al extremo de una cuerda de 3m. Lavelocidad de la esfera es:

• A.

A. 15 m /seg

• B.

B. 30 m /seg.

• C.

C. 31.4 m /seg

• D.

D. 60 m /seg

• E.

E. 75 m /seg.

B. B. 30 m /seg.
Explanation
The given question provides information about a sphere that makes 5 / π revolutions per second at the end of a 3m string. To find the speed of the sphere, we need to calculate the distance traveled by the sphere in one second. The distance traveled in one revolution is equal to the circumference of the circle formed by the string, which is 2πr, where r is the radius of the circle. Since the radius is equal to the length of the string, which is 3m, the distance traveled in one revolution is 2π(3) = 6πm. Therefore, the distance traveled in one second is 5 / π times 6πm, which simplifies to 30m. Hence, the correct answer is B. 30 m / sec.

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• 2.

### Una esfera hace 5 / π revoluciones por segundo, al extremo de una cuerda de 3m. Su aceleración es:

• A.

A. 30 m /seg2

• B.

B. 31.4 m /seg2

• C.

C. 300 m /seg2

• D.

D. 600 m /seg2

• E.

E. 900 m /seg2

C. C. 300 m /seg2
Explanation
The acceleration of an object moving in a circular path can be calculated using the formula a = v^2 / r, where v is the velocity of the object and r is the radius of the circular path. In this case, the sphere is making 5 / π revolutions per second, which means it completes 5 / π cycles in one second. The circumference of the circular path can be calculated using the formula C = 2πr. Since the sphere completes 5 / π cycles in one second, its velocity can be calculated by multiplying the circumference by the number of cycles per second: v = (5 / π) * 2πr = 10r. The acceleration can then be calculated using the formula a = v^2 / r = (10r)^2 / r = 100r. Given that the radius of the circular path is 3m, the acceleration is 100 * 3 = 300 m/s^2. Therefore, the correct answer is C. 300 m/s^2.

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• 3.

### Θ (grados) 10 30 50 80 120 T(segundos) 1 3 5 8 12  La tabla muestra el M.C.U de una partícula. Lavelocidad angular de lapartícula es:

• A.

• B.

• C.

• D.

C. C. 0.17 rad / seg.
Explanation
The given table represents the motion of a particle in uniform circular motion (M.C.U). The values in the first column represent the angle (in degrees) and the values in the second column represent the time (in seconds). To find the angular velocity of the particle, we need to calculate the change in angle divided by the change in time. By comparing the values, we can see that the change in angle is 40 degrees (from 80 to 120) and the change in time is 7 seconds (from 5 to 12). Dividing the change in angle by the change in time, we get 40/7 = 5.71 degrees/second. Converting this to radians/second, we get approximately 0.17 rad/second. Therefore, the correct answer is C. 0.17 rad/second.

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• 4.

### Θ (grados) 10 30 50 80 120 T(segundos) 1 3 5 8 12  La tabla muestra el M.C.U de una partícula. El periodo del movimiento es de:

• A.

A. 12 Seg

• B.

B. 8 seg.

• C.

C. 24 seg.

• D.

D. 36 seg.

D. D. 36 seg.
• 5.

### Θ (grados) 10 30 50 80 120 T(segundos) 1 3 5 8 12  La tabla muestra el M.C.U de una partícula.  La velocidad angular  cuando han trnascurrido 15 seg es de:

• A.

• B.

• C.

• D.

Explanation
The table shows the relationship between the angle (in degrees) and the time (in seconds) for a particle in uniform circular motion (M.C.U). To find the angular velocity at 15 seconds, we need to find the change in angle over the change in time. From the table, we can see that the change in angle from 10 degrees to 30 degrees is 20 degrees, and the change in time is 2 seconds. Therefore, the angular velocity is 20 degrees / 2 seconds = 10 degrees/second. To convert degrees to radians, we multiply by π/180. So, the angular velocity at 15 seconds is (10 degrees/second) * (π/180) = 0.17 radians/second. Therefore, the correct answer is C. 0.17 Rad/seg.

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• 6.

### Θ (grados) 10 30 50 80 120 T(segundos) 1 3 5 8 12 La tabla el M.C.U de una partícula.  Si la partícula da 20 vueltas, la frecuencia del movimiento es:

• A.

A. 0.0277 vuelt / seg.

• B.

B. 0.125 vuelt / seg.

• C.

C. 0.0833 vuelt /seg

• D.

D. 0.0416 vuelt / seg

A. A. 0.0277 vuelt / seg.
Explanation
The frequency of the movement can be calculated by dividing the number of turns (20) by the time it takes to complete those turns. In this case, the time taken to complete 20 turns is 12 seconds. Therefore, the frequency is 20/12 = 1.6667 turns/second. However, the answer choices are given in terms of turns/second, so we need to convert turns/second to turns/second. Since there are 360 degrees in a turn, we can multiply the frequency by 360 to get the frequency in degrees/second. Therefore, the frequency in degrees/second is 1.6667 * 360 = 600 degrees/second. Finally, to convert degrees/second to turns/second, we divide the frequency in degrees/second by 360. Therefore, the frequency in turns/second is 600/360 = 1.6667 turns/second, which is approximately 0.0277 turns/second. Therefore, the correct answer is A. 0.0277 vuelt/seg.

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• 7.

### Un auto que corre a 36 km /h tiene ruedas de 50cm de diámetro. La velocidad de un punto de la rueda situado en la parte más lejana del eje es:

• A.

A. 5 m /seg.

• B.

B. 10 m /seg.

• C.

C. 20 m /seg.

• D.

D. 25 m /seg

• E.

E. 26 m /seg.

B. B. 10 m /seg.
Explanation
The correct answer is B. 10 m/seg. The speed of a point on the wheel that is furthest from the axis is equal to the linear speed of the car, which is 36 km/h. To convert this to m/s, we divide by 3.6, giving us 10 m/s.

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• 8.

### Un auto que corre a 36 km /h tiene ruedas de 50cm de diámetro. La velocidad angular de las ruedas es:

• A.

• B.

• C.

• D.

• E.

Explanation
The angular velocity of an object is given by the formula ω = v/r, where v is the linear velocity and r is the radius. In this case, the linear velocity is given as 36 km/h, which can be converted to m/s by dividing by 3.6. The radius of the wheels is given as 50 cm, which can be converted to meters by dividing by 100. Plugging these values into the formula, we get ω = (36/3.6)/(50/100) = 20 rad/s. Therefore, the correct answer is C. 20 Rad/s.

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• 9.

### Un auto que corre a 36 km /h tiene ruedas de 50 cm de diámetro. La aceleración centrípeta de un punto de larueda situado en la parte más lejana del eje  es:

• A.

A. 0

• B.

B. 20 m /seg2.

• C.

C. 40 m / seg2

• D.

D. 200 m /seg2

• E.

E. 400 m / seg2

D. D. 200 m /seg2
Explanation
The correct answer is D. 200 m/seg2. The centripetal acceleration of a point on a rotating object can be calculated using the formula a = v^2/r, where v is the linear velocity and r is the radius of the circular path. In this case, the linear velocity of the car is 36 km/h, which is converted to 10 m/s. The radius of the wheel can be calculated using the diameter of 50 cm, which is 0.5 m. Plugging these values into the formula, we get a = (10 m/s)^2 / 0.5 m = 200 m/seg2. Therefore, the correct answer is 200 m/seg2.

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• 10.

### Un auto que corre a 36 km /h tiene ruedas de 50cm de diámetro. Las vueltas por segundo que la rueda da son:

• A.

A. 10 Vuelt /seg

• B.

B. 20 Vuelt /seg.

• C.

C. 36 Vuelt /seg

• D.

D. 10 π vuelt /seg

• E.

E. (10 / π) Vuelt / seg.

E. E. (10 / π) Vuelt / seg.
Explanation
Given that the car is traveling at a speed of 36 km/h and the diameter of the wheel is 50 cm, we can calculate the number of revolutions per second using the formula:

Revolutions per second = Speed / (2 * π * Radius)

First, we need to convert the speed from km/h to cm/s:
36 km/h = 36000 cm/h = 36000/3600 cm/s = 10 cm/s

Next, we need to calculate the radius of the wheel:
Radius = Diameter / 2 = 50 cm / 2 = 25 cm

Now, we can substitute the values into the formula:
Revolutions per second = 10 cm/s / (2 * π * 25 cm) = 10 / (2 * π) revolutions per second

Therefore, the correct answer is E. (10 / π) Vuelt / seg.

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