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2 years ago

If the pendulum took longer to complete one oscillation, how would the graph change?

Physics

2 answers:

Sindrei [870]2 years ago

7 0

Answer:

took longer to complete one oscillation, that means its PERIOD increased, and the distance between the peaks of the graph would be longer.

line would be less. the period of oscillation would have any effect on the graph

Komok [63]2 years ago

6 0

Answer: The distance between peaks would be longer.

Explanation:

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When the chemical name of a compound is being written, the subscripts will determine the symbols. elements. prefixes. suffixes.W

Xelga [282]

Answer:

C. Prefixes

Explanation:

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2 years ago

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A bicycle tire is spinning counterclockwise at 2.70 rad/s. During a time period Δt = 1.50 s, the tire is stopped and spun in the

Andru [333]

Answer:

Δω = -5.4 rad/s

αav = -3.6 rad/s²

Explanation:

Given:

Initial angular velocity = ωi = 2.70 rad/s

Final angular velocity = ωf = -2.70 rad/s (negative sign is

due to the movement in opposite direction)

Change in time period = Δt = 1.50 s

Required:

Change in angular velocity = Δω = ?

Average angular acceleration = αav = ?

Solution:

Angular velocity (Δω):

Δω = ωf - ωi

Δω = -2.70 - 2.70

Δω = -5.4 rad/s.

Average angular acceleration (αav):

αav = Δω/Δt

αav = -5.4/1.50

αav = -3.6 rad/s²

Since, the angular velocity is decreasing from 2.70 rad/s (in counter clockwise direction) to rest and then to -2.70 rad/s (in clockwise direction) so, the change in angular velocity is negative.

7 0

2 years ago

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A capacitor is connected across an ac source. Suppose the frequency of the source is doubled. What happens to the capacitive rea

fgiga [73]

The capacitive reactance is reduced by a factor of 2.

<h3>Calculation:</h3>

We know the capacitive reactance is given as,

$Xc = \frac{1}{2\pi fC}$

where,

$Xc\\$ = capacitive reactance

f = frequency

C = capacitance

It is given that frequency is doubled, i.e.,

f' = 2f

To find,

$Xc$ =?

$Xc' = \frac{1}{2\pi f'C}$

$= \frac{1}{2\pi 2f C}$

$= \frac{1}{2} (\frac{1}{2\pi fC} )\\$

$Xc' = \frac{1}{2} Xc$

Therefore, the capacitive reactance is reduced by a factor of 2.

I understand the question you are looking for is this:

A capacitor is connected across an AC source. Suppose the frequency of the source is doubled. What happens to the capacitive reactant of the inductor?

The capacitive reactance is doubled.
The capacitive reactance is traduced by a factor of 4.
The capacitive reactance remains constant.
The capacitive reactance is quadrupled.
The capacitive reactance is reduced by a factor of 2.

Learn more about capacitive reactance here:

brainly.com/question/23427243

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1 year ago

The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit

Margaret [11]

Let us consider two bodies having masses m and m' respectively.

Let they are separated by a distance of r from each other.

As per the Newtons law of gravitation ,the gravitational force between two bodies is given as - $F = G\frac{mm'}{r^{2} }$ where G is the gravitational force constant.