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3 years ago

Throughout the problem, take the speed of sound in air to be 343 .

Physics

1 answer:

anyanavicka [17]3 years ago

8 0

Answer:

A pipe has a length of 1.15 m.

a) Determine the frequency of the first harmonic if the pipe is open at each end. The velocity of sound in air is 343 m/s. Answer in units of Hz

b) What is the frequency of the first harmonic if the pipe is closed at one end?

Answer in units of Hz

Explanation:

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A brick sits on the top of a hill with a gravitational potential energy of 245 J. To determine the gravitational potential of th

Bogdan [553]

Answer:

The mass of the object, its acceleration due to gravity and the distance between the top of the hill and the ground level.

Explanation:

gravitational potential energy is the energy possessed by a body under influence of gravitational force by virtue of its position.

In order to determine the gravitational potential energy of the brick, we must know the mass (m) of the brick, its acceleration due to gravity (g) since it is acting under the influence of gravitational force and the distance between the top of the hill and the ground level. (The height).

Potential energy of a body is calculated as mass × acceleration due to gravity × height.

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3 years ago

Q.01 When charging a secondary cell, energy is stored within a dielectric material using an electric field. True or False

Nitella [24]

True, when charging a secondary cell, energy can be stored within a dielectric material using an electric field.

<h3>Relationship between dielectric material and electric field</h3>

The electric field in a capacitor separates the negative and positive charges in the dielectric material, this causes an attractive force between each plate and the dielectric.

The dielectric material can store electric energy due to its polarization in the presence of external electric field, which causes the positive charge to store on one electrode and negative charge on the other.

Thus, when charging a secondary cell, energy can be stored within a dielectric material using an electric field.

Learn more about dielectric material here: brainly.com/question/17090590

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3 years ago

A uniform solid sphere rolls down an incline without slipping. if the linear acceleration of the center of mass of the sphere is

Ulleksa [173]

A uniform solid sphere rolls down an incline without slipping<span>. </span>If the linear acceleration of the center of mass of the sphere is 0.19g<span>, </span>then what is the angle the incline makes with the horizontal<span>?</span>

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3 years ago

Two carts, cart AA (mass 4.00 kgkg) and cart BB (mass 6.50 kgkg) move on a frictionless, horizontal track. Initially, cart BB is

tankabanditka [31]

Answer:

$V{_a}'=-0.95m/s$

Explanation:

From the question we are told that:

Mass of cart A $m_a=4kg$

Mass of cart B $m_a=6.50kg$

Speed of cart AA $V-{a}=4.00$

Generally the equation for velocity of A after collision $V{_a}'$ is mathematically given by

$V_{a}'=\frac{M_a-M_b}{M_a+M+b} V_a$

$V{_a}'=\frac{4-6.5}{4+M6.5} *4$

$V{_a}'=-0.95m/s$

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3 years ago

The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that were obser

Dafna1 [17]

A) Orbital speed: $v=\sqrt{\frac{GM}{R}}$

B) Kinetic energy: $K= \frac{GmM}{2R}$

D) The orbital period is $T=\frac{2\pi}{\sqrt{GM}}R^{3/2}$

F) The angular momentum is $L=m\sqrt{GMR}$

G) Exponent of radial dependence:

Speed: -1/2

Kinetic energy: -1

Orbital period: 3/2

Angular momentum: 1/2

Explanation:

We know that for a satellite in circular orbit around a planet of mass M, the gravitational force between the satellite and the planet is

$F=G\frac{mM}{R^2}$

where m is the mass of the satellite.

This force provides the centripetal force needed for the circular motion, which is

$F=m\frac{v^2}{R}$

where v is the orbital speed.

Since the gravitational force provides the centripetal force, we can equate the two expressions:

$G\frac{mM}{R^2}=m\frac{v^2}{r}$

And solving for v, we find

$v=\sqrt{\frac{GM}{R}}$

The kinetic energy of an object is given by

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

where

m is the mass of the object

v is its speed

In this problem,

m is the mass of the satellite

$v=\sqrt{\frac{GM}{R}}$ is the speed of the satellite (found in part A)

Substituting, we find an expression for the kinetic energy of the satellite:

$K=\frac{1}{2}m(\sqrt{\frac{GM}{R}})^2 = \frac{GmM}{2R}$

The orbital speed of the satellite can be rewritten as the ratio between the distance covered during one orbit (the circumference of the orbit) divided by the period of revolution:

$v=\frac{2\pi R}{T}$