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ohaa [14]
3 years ago
6

You observe that a mass suspended by a spring takes 0.25 s to make a full oscillation. What is the frequency of this oscillation

? What is its period? What is the oscillation's angular frequency?
Physics
1 answer:
Katarina [22]3 years ago
5 0

Answer:

Frequency of oscillation, f = 4 Hz

time period, T = 0.25 s

Angular frequency, \omega = 25.13 rad/s

Given:

Time taken to make one oscillation, T = 0.25 s

Solution:

Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:

f = \frac{1}{T}

f = \frac{1}{0.25}

Frequency of oscillation, f = 4 Hz

The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.

Therefore, time period, T = 0.25 s

Angular frequency of oscillation is given by:

\omega = 2\pi \times f

\omega = 2\pi \times 4

\omega = 25.13 rad/s

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A traves de una manguera de 1 in de diámetro fluye gasolina con una velocidad media de 5ft/s ¿cuál es el gasto?
jonny [76]

Answer:

El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.

Explanation:

El gasto es el flujo volumétrico de gasolina (Q), medido en pies cúbicos por segundo, que sale de la manguera. Asumiendo que la velocidad de salida es constante, tenemos que el gasto a través de la manguera es:

Q = \frac{\pi}{4}\cdot D^{2}\cdot v (1)

Donde:

D - Diámetro de la manguera, medido en pies.

v - Velocidad medida de salida, medida en pies por segundo.

Si sabemos que D = \frac{1}{12}\,ft y v = 5\,\frac{ft}{s }, entonces el gasto de gasolina es:

Q = \frac{\pi}{4}\cdot \left(\frac{1}{12}\,ft \right)^{2} \cdot \left(5\,\frac{ft}{s} \right)

Q \approx 0.0273\,\frac{ft^{3}}{s}

El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.

6 0
3 years ago
a catcher "gives" with the ball when he catches a 0.196 kg baseball moving at 31 m/s. if he moves his glove a distance of 5.32 c
Aloiza [94]

Answer:

3540.5N

Explanation:

Step one:

given data

mass m= 0.196kg

speed  v= 31m/s

distance r= 5.32cm = 0.0532m

Step two

The expression relating force, mass, velocity and distance is

F= mv^2/r

substitute we have

F=0.196*31^2/0.0532

F=0.196*961/0.0532

F=188.356/0.0532

F=3540.5N

6 0
3 years ago
A car moving at a speed of 36 km/h reaches the foot of a smooth
boyakko [2]

Answer:

d = 10.2 m

Explanation:

When the car travels up the inclined plane, its kinetic energy will be used to do the work in climbing up. So according to the law of conservation of energy, we can write that:

Kinetic\ Energy\ of\ the \ Car = Work\ Done\ while\ moving\ up\ the\ plane\\\frac{1}{2}mv^{2} = Fd

where,

m = mass of car

v = speed of car at the start of plane = (36 km/h)(1000 m/1 km)(1 h/3600 s)

v = 10 m/s

F = force on the car in direction of inclination = W Sin θ

W = weight of car = mg

θ = Angle of inclinition = 30°

d = distance covered up the ramp = ?

Therefore,

\frac{1}{2}mv^{2} = mgdSin\theta\\\frac{1}{2}v^{2} = gdSin\theta\\\frac{1}{2}(10\ m/s)^{2} = d(9.81\ m/s^{2}) Sin\ 30^{0}

<u>d = 10.2 m</u>

4 0
2 years ago
1000 kg of water initially at 6 m/s runs through a hydro-generator. If the water leaves the generator at velocity of 4 m/s, and
r-ruslan [8.4K]

Answer:

8.00 kJ

Explanation:

The first thing is to determine what quantities are there.

the mass of water = 1 000 kg

initial velocity, u = 6 m/s

final velocity, v = 4 m/s

the generator is operating at 100 % efficiency, so there is no energy loss.

The kinetic energy, Ek is converted to electrical energy, therefore Ek = electrical energy.

The kinetic energy is calculated as follows:

Ek = 1/2 mv²

    = 1/2×(1 000)× (4)²

    = 8 000 J/s

    = 8.00 kJ  Ans

4 0
3 years ago
A 1400-kg rocket has a net propulsion force of 20 kN (kiloNewtons). Over a short time period, it uniformly speeds up from an ini
zubka84 [21]

Answer:

  W = 8.92 10² kJ

Explanation:

For this exercise they give us the strength, we must calculate the distance traveled, for this we need the rocket acceleration let's use Newton's second law

        F = m a

        a = F / m

        a = 20 103/1400

        a = 14.29 m/s²

With kinematics we can find the distance traveled

      v_{f}² = v₀² + 2 a x

      x = ( v_{f}²-v₀²) / 2 a

      x = (50² -35²) / 2 14.29

      x = 1275 / 28.58

      x = 44.61 m

Let's calculate the work

      W = F.d

The bold is vector; as indicated by the force is in the direction of movement the scalar product is reduced to the ordinary product

     W = F d

     W = 20 10³ 44.61

     W = 8.92 10⁵ J

     W = 8.92 10² kJ

6 0
3 years ago
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