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Setler [38]
3 years ago
8

A merry-go-round at a playground is a circular platform that is mounted parallel to the ground and can rotate about an axis that

is perpendicular to the platform at its center. the angular speed of the merry-go-round is constant, and a child at a distance of 1.4 m from the axis has a tangential speed of 2.2 m/s. what is the tangential speed of another child, who is located at a distance of 2.1 m from the axis?
Physics
1 answer:
julsineya [31]3 years ago
3 0

Since angular speed is persistent, v/r is constant.
Where:

 v is tangential speed; and

 r is distance from axis.

 

Then equate v/r in both cases to get v in the second case.

Hence, speed = 2.2 x 2.1 / 1.4 meters

= 3.3 meters / seconds

 

Alternative solution would be:

w = 2.2 / 1.4 = 1.57

v = rw = 2.1 x 1.57 = 3.3 meters / seconds 

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As a pendulum swings back and forth_____________.
Nookie1986 [14]

Answer:e

Explanation:

As the Pendulum swings back and forth its kinetic energy is converted into potential and Potential into kinetic .

At highest point Pendulum is momentarily at rest and thus possess all the energy in the form of Potential Energy which is converted into kinetic energy as the pendulum moves downward.

The kinetic energy is maximum at mean position i.e. at lowest point.

Thus all of the given options are true

8 0
4 years ago
Frank and Lisa are analyzing the chart, which shows the speed at which light travels through different media. Frank says that li
Lelechka [254]
The correct answer is
c) Neither person is correct because light does not change frequency when it travels through different media

In fact, light changes both wavelength and speed when it travels through different media, but the frequency remains the same.
4 0
3 years ago
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An open pipe is 1.42 m long
lora16 [44]

Answer:

the fundamental frequency produced by the open pipe is 120.78 Hz

Explanation:

Given;

length of the open pipe, L = 1.42 m

speed of sound in air, v = 343 m/s

The length of the open pipe for the fundamental frequency is equivalent to half of wavelength;

L = \frac{\lambda}{2} \\\\\lambda = 2L

The fundamental frequency produced by the open pipe is calculated as;

f_o = \frac{v}{\lambda} \\\\f_o = \frac{v}{2L} \\\\f_o = \frac{343}{2 \times 1.42} \\\\f_o = 120.78 \ Hz

Therefore, the fundamental frequency produced by the open pipe is 120.78 Hz

6 0
3 years ago
The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C. If the current is supplied by
Artemon [7]

Answer:

E = 20.03 J

Explanation:

Given that,

The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C,

Voltage, V = 12 V

We need to find the energy delivered to the lightbulb filament during 2.00 s.

The energy delivered is given by :

E=I^2Rt. ....(1)

As,

I=\dfrac{q}{t}\\\\I=\dfrac{1.67}{2}\\\\I=0.835\ A

As per Ohm's law, V = IR

R=\dfrac{V}{I}\\\\R=\dfrac{12}{0.835}\\\\R=14.37\ \Omega

Using formula (1).

E=0.835^2\times 14.37\times 2\\\\=20.03\ J

So, the energy delivered to the lightbulb filament is 20.03 J.

6 0
3 years ago
An object accelerates 12.0 m/s² when a force of 6.0 newtons is applied to it.
Nimfa-mama [501]

Answer:

<h2>0.5 kg</h2>

Explanation:

The mass of the object can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{6}{12}  =  \frac{1}{2}  \\

We have the final answer as

<h3>0.5 kg</h3>

Hope this helps you

7 0
3 years ago
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