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-BARSIC- [3]
1 year ago
10

A mass m is attached to an ideal massless spring with spring constant k. In experiment 1 the mass oscillates with amplitude a, a

nd period t. A student grabs the mass and brings it to rest before starting experiment 2. In experiment 2, the mass is set to oscillate with a larger amplitude of 3a. What is the period of the oscillation in experiment 2?.
Physics
1 answer:
Alina [70]1 year ago
4 0

The answer is time period of spring mass system does not depend on its amplitude.

<h3>Does the oscillation's amplitude affect the period?</h3>

The period does not depend on the Amplitude. The greater the amplitude, the more the distance to be covered, but it will do so more quickly. The period won't change since the effects of distance and speed cancel each other out. As the amplitude rises, the mass moves farther with each cycle. Therefore, increasing the amplitude has no overall impact on the oscillation's period.

<h3>What is oscillation?</h3>

A kind of energy called sound is created when objects vibrate. For it to spread, a medium is needed. As a result, sound cannot travel through a vacuum because there is nothing for the sound waves to travel through. The object's back and forth movement creates the sound. Sound vibration is the name for this. Additionally called oscillatory motion. Oscillation describes the regular rhythmic back-and-forth movement.

<h3>What is amplitude and time period of oscillation?</h3>

The time required to complete a task is referred to as a period. An occurrence is referred to as periodic if it happens frequently. The period is the length of time it takes for a periodic event to repeat itself. The particle's time period is determined by how long it takes it to complete one vibration cycle.

A sound wave's amplitude measures the height of the wave. When a sound is made, the maximum displacement of the medium's vibrating particles from their mean position determines the amplitude of the sound wave. It is the separation between the wave's crest or trough and its mean location.

To know more about Amplitude and Time period:

brainly.com/question/20885248

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An electric field of 710,000 N/C points due west at a certain spot. What is the magnitude of the force that acts on a charge of
Anni [7]

Answer:

The magnitude of force is 4.26\times 10^{- 6} N

Solution:

As per the question:

The strength of Electric field due west at a certain point, \vec{E_{w}} = 710,000 N/C

Charge, Q = - 6 C

Now, the force acting on the charge Q in the electric field is given by:

\vec{F} = Q\vec{E_{w}}

\vec{F} = -6\times 710,000 = - 4.26\times 10^{- 6} N

Here, the negative sign indicates that the force acting is opposite in direction.

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3 years ago
What is the process of making acid rain? beginning from burning of fossil fuels...to gaseous products released..to nitrogen oxid
elena55 [62]
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4 0
4 years ago
Car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver not
musickatia [10]

Answer:

Part A: t = v_0/a_0

Part B: t = v_0/a_0

Part C: v_0^2/a_0

Explanation:

Part A:

We will use the following kinematics equation:

v = v_0 + at\\0 = v_0 - a_0t\\t = \frac{v_0}{a_0}

Part B:

We will use the same kinematics equation:

v = v_0 + at\\v_0 = 0 + a_0t\\t = \frac{v_0}{a_0}

Part C:

The total time takes is 2t.

So the train moves a distance of

x = v_0(2t) = 2v_0(\frac{v_0}{a_0}) = \frac{2v_0^2}{a_0}

And the car moves a distance in Part A and in Part B:

d_A = v_0t + \frac{1}{2}at^2 = v_0(\frac{v_0}{a_0}) - \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{a_0} - \frac{v_0^2}{2a_0} = \frac{v_0^2}{2a_0}\\d_B = v_0t + \frac{1}{2}at^2 = 0 + \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{2a_0}

So the total distance that the car traveled is d = \frac{v_0^2}{a_0}

The difference between the train and the car is

x - d = \frac{2v_0^2}{a_0} - \frac{v_0^2}{a_0} = \frac{v_0^2}{a_0}

8 0
3 years ago
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 22 ft/s2. What is the dist
makvit [3.9K]

Answer:

The correct solution is "122.2211".

Explanation:

Given:

deceleration,

a = 22 ft/sec²

Initial velocity,

V_i=50 \ m/h

Now,

V_i=50 \ m/h\times 5280 \ ft/m\times hr/3600 \ s

    =73.333 \ ft/sec

Now,

Final velocity,

V_f=0

Initial velocity,

V_{initial} = 73.333 \ ft/sec

hence,

⇒ V_f^2=V_i^2+2aD

By putting the values, we get

      0=(73.333)^2+2\times( -22) D

  44D=(73.333)^2

      D=\frac{(73.333)^2}{44}

          =122.2211

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Communication definition​
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Answer:

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Explanation:

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