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irina [24]
3 years ago
6

Una rueda que tiene 15 cm de radio, realiza 64 vueltas en 16 seg. Calcula: Periodo Frecuencia Velocidad angular Velocidad lineal

Physics
1 answer:
liraira [26]3 years ago
7 0

Answer:

i) El período de la rueda es de 0,25 segundos.

ii) La frecuencia de la rueda es 4 Hertz

iii) La velocidad angular es aproximadamente 25.133

iv) La velocidad lineal es de aproximadamente 3,77 m / s

Explanation:

El radio de la rueda, r = 15 cm = 0,15 m

El número de vueltas que hace la rueda = 64 vueltas

El tiempo que tarda el volante en dar 64 vueltas = 16 segundos

i) El período = El tiempo que tarda la rueda en dar 1 vuelta

∴ El período de la rueda, T = 16 segundos/(64 vueltas) = 0,25 segundos

El período de la rueda, T = 0,25 segundos

ii) La frecuencia = El número de vueltas por segundo

∴ La frecuencia de la rueda, f = 64 vueltas /(16 segundos) = 4 Hertz

1 vuelta = 2 · π radianes

La frecuencia de la rueda, f = 4 Hertz

iii) Velocidad angular = La medida del ángulo girado por segundo

∴ La velocidad angular, ω = 64 × 2 × π/16 segundos ≈ 8 · π rad/segundos ≈ 25.133 rad/seg

La velocidad angular, ω ≈ 25.133

iv) La velocidad lineal, v = r × ω

∴ v = 0,15 m × 8 · π rad / segundos ≈ 3,77 m/s

La velocidad lineal, v ≈ 3.77 m/s

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8 0
3 years ago
A 12.8-kg monkey hangs from a cord suspended from the ceiling of an elevator. The cord can withstand a tension of 156N and break
Varvara68 [4.7K]

Answer:

2.4 m/s^2

Downward

Explanation:

We are given that

Mass of monkey=12.8 kg

Tension=156 N

We have to find the magnitude of the elevator's minimum acceleration.

T=m(g+a)

Where g=9.8 m/s^2

Substitute the values

156=12.8(a+9.8)

9.8+a=\frac{156}{12.8}=12.2

a=12.2-9.8=2.4 m/s^2

Hence, the acceleration =a=2.4 m/s^2

Direction of the elevator's minimum acceleration is downward because the elevator moves downwards.

8 0
3 years ago
What happens to the force between two charges when each charge is doubled and the distance between them is 1/4 its original
DIA [1.3K]

Answer:

F' = 64 F

Explanation:

The electric force between charges is given by :

F=\dfrac{kq_1q_2}{r^2}

Where

q₁ and q₂ are charges

r is the distance between charges

When  each charge is doubled and the distance between them is 1/4 its original magnitude such that,

q₁' = 2q₁, q₂' = 2q₂ and r' = (r/4)

New force,

F'=\dfrac{kq_1'q_2'}{r'^2}

Apply new values,

F'=\dfrac{k\times 2q_1\times 2q_2}{(\dfrac{r}{4})^2}\\\\=\dfrac{k\times 4q_1q_2}{\dfrac{r^2}{16}}\\\\=64\times \dfrac{kq_1q_2}{r^2}\\\\=64F

So, the new force becomes 64 times the initial force.

7 0
3 years ago
5. A construction worker on a high-rise building is on a platform suspended between two cables as illustrated below. The constru
Natalka [10]

Answer:

a) Tc = 750 [N]  ;b) See the explanation below.

Explanation:

To solve this problem, we first need a graphical explanation of this, as well as knowing the corresponding questions. Therefore, a search was carried out in google, in the attached image we will find a graphical description of the problem.

b)

The solution of this type of problem corresponds to the use of Newton's third law, applying static which tells us that the sum of the forces in a system in equilibrium without movement must be equal to zero.

a)

In this way we can find by means of a sum of forces on the y axis equal to zero:

- 850 - 450 + 550 + Tc = 0

Tc = 750 [N]

4 0
3 years ago
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