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Zielflug [23.3K]
2 years ago
14

While traveling along a highway a driver slows from 31 m/s to 15 m/s in 8 seconds. What is the automobile’s acceleration? (Remem

ber that a negative value indicates a slowing down or deceleration.)​
Physics
1 answer:
MaRussiya [10]2 years ago
8 0

Answer:

The automobile's acceleration in that time interval is -2 m/s^2

Explanation:

The acceleration is defined as the rate of change of the velocity.

The average acceleration in a given lapse of time is calculated as:

A = (final velocity - initial velocity)/time.

In this case, we have:

initial velocity = 31 m/s

final velocity = 15 m/s

time = 8 seconds.

Then the average acceleration is:

A = (15m/s - 31m/s)/8s = -2 m/s^2

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3. The velocity of sound is 332 m/s. Answer the following questions:
Angelina_Jolie [31]

Answer:

20 Hz, 20000 Hz

0.0166 m, 16.6 m

Explanation:

The minimum frequency that a human ear can hear is 20 Hz

The maximum frequency that a human ear can hear is 20000 Hz.

v = Velocity of sound = 332 m/s

Wavelength is given by

\lambda=\dfrac{v}{f}\\\Rightarrow \lambda=\dfrac{332}{20}\\\Rightarrow \lambda=16.6\ \text{m}

The longest wavelength that can be heard by the human ear is 16.6 m

\lambda=\dfrac{332}{20000}\\\Rightarrow \lambda=0.0166\ \text{m}

The shortest wavelength that can be heard by the human ear is 0.0166 m.

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2 years ago
Undersea mountain ranges in the middle of the ocean floors are known as d͟e͟e͟p͟-͟o͟c͟e͟a͟n͟ ͟t͟r͟e͟n͟c͟h͟e͟s.
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False

mid-ocean ridge

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3 years ago
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Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by
Ulleksa [173]

A) v=\sqrt{\frac{2qV}{m}}

B) r=\frac{mv}{qB}

C) T=\frac{2\pi m}{qB}

D) \omega=\frac{qB}{m}

E) r=\frac{\sqrt{2mK}}{qB}

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

\Delta U = qV

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

\Delta K=\frac{1}{2}mv^2

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

qV=\frac{1}{2}mv^2

And solving for v, we find the speed v at which the particle enters the cyclotron:

v=\sqrt{\frac{2qV}{m}}

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

F=qvB

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

F=m\frac{v^2}{r}

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

qvB=m\frac{v^2}{r}

And solving for r, we find the radius of the orbit:

r=\frac{mv}{qB} (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference (2\pi r) and the velocity of the particle (v):

T=\frac{2\pi r}{v} (2)

From eq.(1), we can rewrite the velocity of the particle as

v=\frac{qBr}{m}

Substituting into(2), we can rewrite the period of revolution of the particle as:

T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

\omega=\frac{2\pi}{T} (3)

where T is the period.

The period has been found in part C:

T=\frac{2\pi m}{qB}

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

v=\frac{qBr}{m}

which can be rewritten as

r=\frac{mv}{qB} (4)

The kinetic energy of the particle is written as

K=\frac{1}{2}mv^2

So, from this we can find another expression for the velocity:

v=\sqrt{\frac{2K}{m}}

And substitutin into (4), we find:

r=\frac{\sqrt{2mK}}{qB}

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

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The conventional signal used by sailboats in conditions of reduced visibility such as heavy fog is one long blast followed by two short blasts.

The blasts help other boat operators locate one another's vessel in a condition where it is not easy to see. This signal is repeated in order to not only let others know of the vessel's position, but also help them know which way it is traveling. For example, if the blasts start to become distant, then the sailboat is travelling away from you.
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3 years ago
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Interactions of current carrying wires
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