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anzhelika [568]

3 years ago

14

An ideal horizontal spring-mass system oscillates with a period of 0.60 seconds. The mass is 3 kg but the spring constant is not

known. When this same spring and mass are arranged to hang vertically, the frequency of oscillations will be: a) 0.40 Hz b) 1.2 Hz c) 2.5 Hz d) none of the above are correct e) the spring constant is needed to determine the frequency

1 answer:

VashaNatasha [74]3 years ago

4 0

Explanation:

Given that,

An ideal horizontal spring-mass system oscillates with a period of 0.60 seconds.

Mass is 3 kg but the spring constant is not known. We need to find the frequency of oscillations when this same spring and mass are arranged to hang vertically. We know that the relation between frequency and time period is inverse. Frequency is given by :

$f=\dfrac{1}{T}\\\\f=\dfrac{1}{0.6\ s}\\\\f=1.67\ Hz$

Hence, the correct option is (d) "none of the above".

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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).

6 0

4 years ago

You have a battery marked " 6.00 V 6.00 V ." When you draw a current of 0.383 A 0.383 A from it, the potential difference betwee

Archy [21]

Answer:

V = 4.81 V

Explanation:

As the potential difference between the battery terminals, is less than the rated value of the battery, this means that there is some loss in the internal resistance of the battery.
We can calculate this loss, applying Ohm's law to the internal resistance, as follows:

$V_{rint} = I* r_{int}$

The value of the potential difference between the terminals of the battery, is just the voltage of the battery, minus the loss in the internal resistance, as follows:

$V = V_{b} - V_{rint} = 5.03 V = 6.0 V - 0.383 A* r_{int}$

We can solve for rint, as follows:

$r_{int} = \frac{V_{b}-V}{I} =\frac{6.0V-5.03V}{0.383A} = 2.53 \Omega$

When the circuit draws from battery a current I of 0.469A, we can find the potential difference between the terminals of the battery, as follows:

$V = V_{b} - V_{rint} = 6.0 V - 0.469 A* 2.53 \Omega= 6.0 V - 1.19 V = 4.81 V$

As the current draw is larger, the loss in the internal resistance will be larger too, so the potential difference between the terminals of the battery will be lower.

5 0

3 years ago

Sally and Sam are in a spaceship that comes to within 14,000 km of the asteroid Ceres. Determine the force Sally experiences, in

tatiyna

Sam and Sally are traveling aboard a spacecraft that approaches the asteroid Ceres within 14,000 kilometers. Sally will experience 1.989 × 10⁻¹¹ N of force.

<h3>What is the gravitational force?</h3>

Newton's law of gravity states that each particle having mass in the universe attracts each other particle with a force known as the gravitational force.

The gravitational force is proportional to the product of the masses of the two bodies and inversely proportional to the square of their distance.

Given data

Mass of asteroid ,m₁ = 8.7 1020 kg

Mass of sally,m₂ = 67 kg

Gravitational constant,G = 6.6 × 10⁻¹¹ kg⁻² m²

Distance of seperation,R = 14,000 km

$\rm F = G\frac{m_1m_2}{R^ 2} \\\\ F = 6.6 \times 10^{-11 }\times \frac{8.71020 \times 67 }{(14,000)^2} \\\\F = 1.989 \times 10^{-11 } \ N$

Hence, the force Sally experiences will be 1.989 × 10⁻¹¹ N.

To learn more about the gravitational force, refer to the link;

brainly.com/question/24783651

#SPJ1

4 0

2 years ago

katrin2010 [14]

The particles of the medium (slinky in this case) move up and down (choice #2) in a transverse wave scenario.

This is the defining characteristic of transverse waves, like particles on the surface of water while a wave travels on it, or like particles in a slack rope when someone sends a wave through by giving it a jolt.

The other kind of waves is longitudinal, where the particles of the medium move "left-and-right" along the direction of the wave propagation. In the case of the slinky, this would be achieved by giving a tensioned slinky an "inward" jolt. You would see that such a jolt would give rise to a longitudinal wave traveling along the length of the tensioned slinky. Another example of longitudinal waves are sound waves.

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4 years ago

Which are parts of the middle ear? Check all that apply.

stellarik [79]

Answer:

middle ear has three bones! the hammer, anvil, stirrup, and ear drum

Explanation:

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3 years ago