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2 years ago

Imagine a mass hanging from a spring that behaves ideally. Which factor will not affect the period of this mass-spring system?

Physics

2 answers:

natulia [17]2 years ago

0 0

Answer:

varying the displacement of the mass from its equilibrium position

increasing the amount of time the oscillating system is observed

keeping track of the precise position of the mass through time

Explanation:

The period of a mass-spring system is given by:

$T=2 \pi \sqrt{\frac{m}{k}}$

where

m is the mass hanging from the spring

k is the spring constant

As we can see from the formula, the period depends only on these two factors: mass and spring constant, and nothing else. Therefore, the following factors do not affect the period of the system:

varying the displacement of the mass from its equilibrium position

increasing the amount of time the oscillating system is observed

keeping track of the precise position of the mass through time

insens350 [35]2 years ago

0 0

$T=2\pi \sqrt{\frac{m}{k} }$

Where T is the period, m is the mass of the object, and k is the spring constant. It is important to note that the amplitude of the oscillation has no effect on the period.
Which means the question should in fact say factors - the displacement will have no effect, nor will observing it in any way - the only two things there that will affect the period of the spring is changing the weight and the spring constant.

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An office window has dimensions 3.1 m by 2.1 m. As a result of the passage of a storm, the outside air pressure drops to 0.954 a

Virty [35]

Answer:

Net forces which pushes the window is 30342.78 N.

Explanation:

Given:

Dimension of the office window.

Length of the window = $3.1$ m

Width of the window = $2.1$ m

Area of the window = $(3.1\times 2.1) = 6.51\ m^2$

Difference in air pressure = Inside pressure - Outside pressure

= $(1.0-0.954)$ atm = $0.046$ atm

Conversion of the pressure in its SI unit.

⇒ $1$ atm = $101325$ Pa

⇒ $0.046$ atm = $0.046\times 101325 =4660.95$ Pa

We have to find the net force.

We know,

⇒ Pressure = Force/Area

⇒ $Pressure=\frac{Force }{Area}$

⇒ $Force =Pressure\times Area$

⇒ Plugging the values.

⇒ $Force =4660.95\times 6.51$

⇒ $Force=30342.78$ Newton (N)

So,

The net forces which pushes the window is 30342.78 N.

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3 years ago

What type of energy does a pv cell create ?

Alex_Xolod [135]

Answer: Solar Photovoltaic (PV) cells generate electricity by absorbing sunlight and using that light energy to create an electrical current. There are many PV cells within a single solar panel, and the current created by all of the cells together adds up to enough electricity to help power your school, home and businesses.

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1 year ago

Read 2 more answers

What is matter in regards to energy?

svp [43]

Answer:

Energy always involves. motion in some form. Matter is an object that that up space so it is. measured in units of space

Explanation:

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2 years ago

A spherical shell is rolling without slipping at constant speed on a level floor. What percentage of the shell's total kinetic e

IgorC [24]

Answer:

41.667 per cent of the total kinetic energy is translational kinetic energy.

Explanation:

As the spherical shell is rolling without slipping at constant speed, the system can be considered as conservative due to the absence of non-conservative forces (i.e. drag, friction) and energy equation can be expressed only by the Principle of Energy Conservation, whose total energy is equal to the sum of rotational and translational kinetic energies. That is to say:

$E = K_{t} + K_{r}$

Where:

$E$ - Total energy, measured in joules.

$K_{r}$ - Rotational kinetic energy, measured in joules.

$K_{t}$ - Translational kinetic energy, measured in joules.

The spherical shell can be considered as a rigid body, since there is no information of any deformation due to the motion. Then, rotational and translational components of kinetic energy are described by the following equations:

Rotational kinetic energy

$K_{r} = \frac{1}{2}\cdot I_{g}\cdot \omega^{2}$

Translational kinetic energy

$K_{t} = \frac{1}{2}\cdot m \cdot R^{2}\cdot \omega^{2}$

Where:

$I_{g}$ - Moment of inertia of the spherical shell with respect to its center of mass, measured in $kg\cdot m^{2}$ .

$\omega$ - Angular speed of the spherical shell, measured in radians per second.

$R$ - Radius of the spherical shell, measured in meters.

After replacing each component and simplifying algebraically, the total energy of the spherical shell is equal to:

$E = \frac{1}{2}\cdot (I_{g} + m\cdot R^{2})\cdot \omega^{2}$

In addition, the moment of inertia of a spherical shell is equal to:

$I_{g} = \frac{2}{3}\cdot m\cdot R^{2}$

Then, total energy is reduced to this expression:

$E = \frac{5}{6}\cdot m \cdot R^{2}\cdot \omega^{2}$

The fraction of the total kinetic energy that is translational in percentage is given by the following expression:

$\%K_{t} = \frac{K_{t}}{E}\times 100\,\%$

$\%K_{t} = \frac{\frac{1}{2}\cdot m \cdot R^{2}\cdot \omega^{2} }{\frac{5}{6}\cdot m \cdot R^{2}\cdot \omega^{2} } \times 100\,\%$

$\%K_{t} = \frac{5}{12}\times 100\,\%$

$\%K_{t} = 41.667\,\%$

41.667 per cent of the total kinetic energy is translational kinetic energy.

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3 years ago

A resistor, an inductor, and a switch are all connected in series to an ideal battery of constant terminal voltage. Suppose at f

Airida [17]

Answer:

c. The steady-state value of the current depends on the resistance of the resistor.

Explanation:

Since all the components are connected in series, when the switch is at first open, current will not flow round the circuit. As current needs to flow through from the positive terminal of the battery through the resistor, inductor, and switch to the negative terminal of the battery.

But the moment the switch is closed, at the initial time t = 0, the current flow through from the positive terminal of the battery through the resistor, inductor, and switch to the negative terminal of the battery. It then begins to increase at a rate that depends upon the value of the inductance of the inductor.

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3 years ago