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soldier1979 [14.2K]
3 years ago
5

A thin hoop is supported in a vertical plane by a nail. What should the radius of the hoop be in order for it to have a period o

f oscillation of 1.00 s
Physics
1 answer:
maxonik [38]3 years ago
4 0

Answer:

0.124 m

Explanation:

the period of a simple pendulum with a small amplitude is given as

T = 2π [√(I/mgd)]

From the above stated formula,

I = moment of inertia

m = mass of the pendulum

g = acceleration due to gravity, 9.8 m/s²

d = distance from rotation axis due to center of gravity

Also, moment of Inertia, I = 2mr², if we substitute this in the above formula, we have

T = 2π [√(2mr²/mgd)]

If we assume that our r = d, then we have

T = 2π [√(2r/g)]

If we make r the subject of the formula in the above equation, we get

r = gT² / 8π²

r = (9.8 * 1²) / 8 * π²

r = 9.8 / 78.98

r = 0.124 m

Thus, the radius of the hoop is 0.124 m

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A body weighs 10newton in air and 9.5 in water. How will it weigh in alchohol if density 0.8gcm-³​
Katen [24]

Answer:

The weight of object in alcohol is, W₀ = 9.6 N

Explanation:

Given,

The weight of body in air, Wₐ = 10 N

The weight of the body in water, Wₓ = 9.5 N

The density of the alcohol, ρ = 0.8 g/cm³

                                                = 800 kg/m³

The weight of body in water = weight of object in air - weight of water displaced

                        9.5 N = 10 N - weight of water displaced

                       Weight of water displaced = 0.5 N

The mass of the displaced water, m = 0.5 / 9.8

                                                            = 0.051 kg

Therefore the volume of the water displaced,

                                              V = m / ρ

                                                  = 0.051 / 1000

                                                   = 5.1 x 10⁻⁵ m³

<em>The volume of water displaced is equal to the volume of alcohol displaced</em>

Therefore, the mass of the alcohol displaced, = V x ρ

                                                                             = 5.1 x 10⁻⁵ x 800

                                                                             = 0.0405 kg

The weight of the alcohol displaced, w = 0.0405 x 9.8

                                                                    = 0.4 N

Therefore,

The weight of body in alcohol = weight of object in air - weight of alcohol displaced

                             W₀ = W - w

                                    = 10 N - 0.4 N

                                     = 9.6 N

Hence, the weight of object in alcohol is, W₀ = 9.6 N

4 0
3 years ago
Science is made up of two things, what is it
rodikova [14]

Answer:

Natural science is concerned with the description, prediction, and understanding of natural phenomena based on empirical evidence from observation and experimentation. It can be divided into two main branches: life science (or biological science) and physical science.

Explanation:

4 0
3 years ago
A pendulum has 294 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom o
baherus [9]

Newton's law of conservation states that energy of an isolated system remains a constant. It can neither be created nor destroyed but can be transformed from one form to the other.


Implying the above law of conservation of energy in the case of pendulum we can conclude that at the bottom of the swing the entire potential energy gets converted to kinetic energy. Also the potential energy is zero at this point.


Mathematically also potential energy is represented as


Potential energy= mgh


Where m is the mass of the pendulum.


g is the acceleration due to gravity


h is the height from the bottom z the ground.


At the bottom of the swing,the height is zero, hence the potential energy is also zero.


The kinetic energy is represented mathematically as


Kinetic energy= 1/2 mv^2


Where m is the mass of the pendulum


v is the velocity of the pendulum


At the bottom the pendulum has the maximum velocity. Hence the kinetic energy is maximum at the bottom.


Energy can neither be created e destroyed. It can only be transferred from one form to another. Implying this law and the above explainations we conclude that at the bottom of the pendulum,the potential energy=0 and the kinetic energy=294J as the entire potential energy is converted to kinetic energy at the bottom.



6 0
4 years ago
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
3 years ago
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