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dimaraw [331]
3 years ago
7

When the displacement in SHM is equal to 1/2 of the amplitude xm, what fraction of the total energy is (a) kinetic energy and (b

) potential energy? (c) At what displacement, in terms of the amplitude, is the energy of the system half kinetic energy and half potential energy? (Give the ratio of the answer to the amplitude)
Physics
1 answer:
nikitadnepr [17]3 years ago
3 0

Answer:

Explanation:

Let x₀ be the amplitude , ω be the angular velocity

velocity at displacement  x ,

v = ω \sqrt{(x_0^2-x^2)}

at x =\frac{x_0}{2}

v = ω [ \sqrt{(x_0^2-\frac{x_0}{4} ^2)}

v² =  ω² x\frac{3x_0^2}{4}

1/2 m v²

= 3/8 m x ω²x₀²

Total energy = 1/2 m ω²x₀²

Kinetic energy as fraction of total energy

= 3/8 m x ω²x₀² x 2 /  m ω²x₀²

= 3 / 4

Fraction of potential energy

= 1 - 3/4

1/4

c )

Half kinetic = half potential energy

=  1/4 m ω²x₀²

1/2 m ω²( x₀²- x ² )

x₀²/2  = x₀²- x ²

x² = x₀²/2

x = x₀/√2

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UNO [17]

Answer:

international air transport association

7 0
3 years ago
when the armature of an ac generatr rotates at 15.0 rad/s, the amplitude of the induced emf is 27.0 V. What is the amplitude of
nata0808 [166]

To solve this problem we will apply the concepts related to the electric field. This is defined as the product between the angular frequency, the number of turns of the body (solenoid in this case) the magnetic field and the sine of the angular frequency and time. Mathematically this can be described as

E = \omega NBA |sin \omega t|

Here,

\omega = Angular frequency

N = Number of turns

B = Magnetic field

The emf has its maximum value when sin \omega t = \pm 1

Thus the amplitude of the emf is

E = \omega NBA

When number of turns of armature, area and applied magnetic field remains constant, induced emf is proportional to angular speed.

E \propto \omega

Further it can be written as follows,

\frac{E_1}{E_2} \propto \frac{\omega_1}{\omega_2}

E_2 = \frac{\omega_2}{\omega_1}E_1

E_2 = \frac{10rad/s}{15rad/s}(27.0V)

E_2 = 18V

Therefore the maximum amplitude of induced emf when armature rotates at 10.0rad/s is 18V

5 0
3 years ago
A container of gas is at a pressure of 3.7 x 10^5 Pa. How much work is done by the gas if its volume expands by 1.6 m^3 ?
Dmitriy789 [7]

Answer:

592000 J

Explanation:

We'll begin by converting 3.7×10⁵ Pa to Kg/ms². This can be obtained as follow:

1 Pa = 1 Kg/ms²

Therefore,

3.7×10⁵ Pa = 3.7×10⁵ Kg/ms²

Next, we shall determine the workdone.

Workdone is given by the following equation:

Workdone (Wd) = pressure (P) × change in volume (ΔV)

Wd = PΔV

With the above formula, the work done can be obtained as follow:

Pressure (P) = 3.7×10⁵ Kg/ms²

Change in volume (ΔV) = 1.6 m³

Workdone (Wd) =?

Wd = PΔV

Wd = 3.7×10⁵ × 1.6

Wd = 592000 Kgm²/s²

Finally, we shall convert 592000 Kgm²/s² to Joule (J). This can be obtained as follow:

1 Kgm²/s² = 1 J

Therefore,

592000 Kgm²/s² = 592000 J

Therefore, the Workdone is 592000 J.

6 0
3 years ago
The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 2.0 rev/s in 10.0 s. At th
Arte-miy333 [17]

Answer:

22 revolutions

Explanation:

2 rev/s = 2*(2π rad/rev) = 12.57 rad/s

The angular acceleration when it starting

\alpha_a = \frac{\Delta \omega}{\Delta t} = \frac{12.57}{10} = 1.257 rad/s^2

The angular acceleration when it stopping:

\alpha_o = \frac{\Delta \omega}{\Delta t} = \frac{-12.57}{12} = -1.05 rad/s^2

The angular distance it covers when starting from rest:

\omega^2 - 0^2 = 2\alpha_a\theta_a

\theta_a = \frac{\omega^2}{2\alpha_a} = \frac{12.57^2}{2*1.257} = 62.8 rad

The angular distance it covers when coming to complete stop:

0 - \omega^2 = 2\alpha_o\theta_o

\theta_o = \frac{-\omega^2}{2\alpha_o} = \frac{-12.57^2}{2*(-1.05)} = 75.4 rad

So the total angular distance it covers within 22 s is 62.8 + 75.4 = 138.23 rad or 138.23 / (2π) = 22 revolutions

6 0
3 years ago
Convertir 500 f a grado celsios
Allushta [10]

Answer:

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Explanation:

si

5 0
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