PHYSICS COLLEGE

A bicycle wheel of radius 0.70 m is rolling without slipping on a horizontal surface with an angular speed of 2.0 rev/s when the cyclist begins to uniformly apply the brakes. the bicycle stops in 5.0 s. how far did the bicycle travel during the 5.0 seconds of braking?

Answers

Answer 1
Answer:

Distance traveled by the bicycle during the 5 seconds of braking is 22m

Explanation:

initial angular velocity= 2 rev/s

final angular velocity= 0 rev/s

Angular displacement Ф=((wi+wf)/(2) )t

Ф=((0+2)/(2) )5=5 rev

so the distance travelled= 5(2πr)

distance=5(2π*0.7)

distance=22m
Answer 2
Answer:

The bicycle traveled about 22 m during the 5.0 seconds of braking

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Further explanation

Centripetal Acceleration can be formulated as follows:

\large {\boxed {a = \frac{ v^2 } { R } }

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

\texttt{ }

Centripetal Force can be formulated as follows:

\large {\boxed {F = m \frac{ v^2 } { R } }

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

\texttt{ }

Given:

radius of wheel = R = 0.70 m

initial angular speed = ω = 2.0 rev/s = 4π rad/s

final angular speed = ωo = 0 rad/s

time taken = t = 5.0 s

Asked:

distance covered = d = ?

Solution:

d = \theta R

d = (\omega + \omega_o)(1)/(2)t R

d = ( 4 \pi + 0 ) (1)/(2)(5.0)( 0.70 )

d = 4\pi (1.75)

d = 7\pi \texttt{ m}

d \approx 22 \texttt{ m}

\texttt{ }

Learn more

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Answer details

Grade: High School

Subject: Physics

Chapter: Circular Motion

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A 13.0-Ω resistor, 13.5-mH inductor, and 50.0-µF capacitor are connected in series to a 55.0-V (rms) source having variable frequency. If the operating frequency is twice the resonance frequency, find the energy delivered to the circuit during one period.

Answers

Answer:

E = 0.13 J

Explanation:

At resonance condition we have

\omega = \sqrt{(1)/(LC)}

\omega = \sqrt{(1)/((13.5 * 10^(-3)){50* 10^(-6))}}

\omega = 1217.2 rad/s

now if the frequency is double that of resonance condition then we have

x_L = 2\omega L

x_L = 2(1217.2)(13.5* 10^(-3))

x_L = 32.86 ohm

now we have

x_c = (1)/(2(1217)(50* 10^(-6)))

x_c = 8.22 ohm

now average power is given as

P = i_(rms)V_(rms)* (R)/(z)

P = (55)/(√((32.86 - 8.22)^2 + 13^2)))(55)* (13)/(√((32.86 - 8.22)^2 + 13^2))

P_(avg) = 50.67 Watt

Now time period is given as

T = (2\pi)/(\omega)

so total energy consumed is given as

E_(avg) = 50.67((2\pi)/(2(1217.2)))

E = 0.13 J

Please show steps as to how to solve this problem
Thank you!

Answers

Answer:

Torques must balance

F1 * X1 = F2 * Y2

or M1 g X1 = M2 g X2

X2 = M1 / M2 * X1 = 130.4 / 62.3 * 10.7

X2 = 22.4 cm

Torque = F1  * X2 =

62.3 gm* 980 cm/sec^2  * 22.4 cm = 137,000 gm cm^2 / sec^2

Normally x cross y   will be out of the page

r X F  for F1 will be into the page so the torque must be negative

How long would it take a 500. W electric motor to do 15010 J of work?

Answers

time = energy / power = 15010 / 500 = .... seconds

Give some reasons for our knowledge of the solar system has increased considerably in the past few years. Support your response with at least 3 reasons with details regarding concepts from the units learned in this course.

Answers

Answer:

Improvement in observational, and exploratory technology

Rapid increase in knowledge

International collaboration

Explanation:

Our knowledge of the solar system has increased greatly in the past few years due to to some factors which are listed below.

Improvement in observational, and exploratory technology: In recent years, developments in technology has led to the invention of advanced observational instruments and probes, that are used to study the solar system. Also more exploratory units are now developed to go out into the solar system and gather useful data which is then further processed to yield more results about our solar system.

Rapid increase in knowledge: The past few years has seen an increased number of theories proposed to explain phenomena in the solar system. Some of these theories have been seen to be accurate under experimentation, leading to newer and fresher insights into our solar system. Also, new experiments and research are carried out, all these leading to an exponential growth in our knowledge of the solar system.

International Collaboration: The sharing of knowledge by scientists all over has led to a better, quick understanding of the solar system. Also, scientists from  different countries, working together on different experiment and data sharing regarding our solar system now allows our knowledge of the solar system to deepen faster.

An effect analogous to two-slit interference can occur with sound waves, instead of light. In an open field, two speakers placed 1.30 m apart are powered by a single function generator producing sine waves at 1200-Hz frequency. A student walks along a line 12.5 m away and parallel to the line between the speakers. She hears an alternating pattern of loud and quiet, due to constructive and destructive interference. What is : (a) the wavelength of this sound and (b) the distance between the central maximum and the first maximum (loud) position along this line

Answers

Answer:

2.72 m

Explanation:

wavelength of sound λ = velocity / frequency

= 340 / 1200

= .2833 m

Distance of point of  first constructive interference

= λ D / d ( D is distance of the screen and d is distance between source of sound.

Here D = 12.5 m

d = 1.3 m

λ D / d= ( .2833 x 12.5) / 1.3

= 2.72 m

Distance of point of  first constructive interference = 2.72 m

Final answer:

The wavelength of the produced sound is approximately 0.29 m. Constructive interference occurs when the path difference between the two waves is a multiple of this wavelength, allowing you to calculate the distance between the central maximum and first maximum loud position.

Explanation:

For part (a) of the question, we need to calculate the wavelength of the sound wave. The wave speed (v) is given by the multiplication of frequency (f) and wavelength (λ). The speed of sound in air is approximately 343 m/s and given that the frequency produced by the function generator is 1200 Hz, the wavelength can be calculated using the formula λ = v / f = 343 / 1200 ≈ 0.29 m.

For part (b) the distance between the central maximum (loud) position and the first maximum along this line requires understanding of sound wave interference and constructive interference. For constructive interference to occur, the path difference between the two waves needs to be a multiple of the wavelength. Thus, in the first constructive interference position (first maximum loud position), the path difference equals one wavelength (0.29m). Since the student is walking 12.5 m away and parallel to the line between the speakers (which is the hypotenuse of a right triangle stakeout, with one side being 0.65m), we can use Pythagorean theorem to find out the distance.

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What If? Fluoride ions (which have the same charge as an electron) are initially moving with the same speed as the electrons from part (a) through a different uniform electric field. The ions come to a stop in the same distance d. Let the mass of an ion be M and the mass of an electron be m. Find the ratio of the magnitude of electric field the ions travel through to the magnitude of the electric field found in part (a). (Use the following as necessary: d, K, m, M, and e for the charge of the electron.)

Answers

Answer:

E₁ / E₂ = M / m

Explanation:

Let the electric field be E₁ and E₂ for ions and electrons respectively .

Force on ions = E₁ e where e is charge on ions .

Acceleration on ions a = E₁ e / M . Let initial velocity of both be u . Final velocity v = 0

v² = u² - 2as

0 = u² - 2 x E₁ e d  / M  

u² = 2 x E₁ e d  / M

Similarly for electrons

u² = 2 x E₂ e d  / m

Hence

2 x E₁ e d  / M =  2 x E₂ e d  / m

E₁ / E₂ = M / m

Final answer:

The ratio of the magnitude of the electric field the ions travel through to the magnitude of the electric field found in part (a) is M/m.

Explanation:

The ratio of the magnitude of the electric field the ions travel through to the magnitude of the electric field found in part (a) can be determined using the concept of mechanical energy conservation. Since the ions come to a stop, their initial kinetic energy must be equal to the work done by the electric field on them. The work done is given by the equation:

Work = Change in kinetic energy

The change in kinetic energy can be calculated using the formula:

Change in kinetic energy = (1/2)Mv2 - (1/2)mv2

where M and m are the masses of the ions and electrons respectively, and v is their initial speed. Solving for the ratio, we get:

Ratio = (1/2)M/(1/2)m = M/m

So, the ratio of the magnitude of electric field the ions travel through to the magnitude of the electric field found in part (a) is M/m.

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