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3 years ago

7

Kepler's____ of planetary motion states that the planets have an elliptical orbit, with the Sun at one focal point of the ellips

e. A. 1st law B. 2nd law C. 3rd law

1 answer:

pentagon [3]3 years ago

6 0

A.Kepler's first law of planetry motion.

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1. Most waves are caused by vibrations. What motion in a transmitting antenna causes the vibrations that produce radio waves?

morpeh [17]

The answer 1 is
The best-known use of radio waves is for communication; television, cellphones and radios all receive radio waves and convert them to mechanical vibrations in the speaker to create sound waves that can be heard.Electromagnetic radiation is transmitted inwaves or particles at different wavelengths and frequencies.

the answer 2 is

Radio Waves: Instant Communication
Microwaves: Data and Heat

i think the answer number 3 is
It is all about wavelength versus tunnel diameter. The wavelength of GPS is about 20cm it would happily propagate in any normal tunnel if it could get in but the earth and other structures absorb it. AM radio (600kHz - 1500kHz) cannot propagate in any normal tunnel because the wavelength is too long (500m-200m) relative to the diameter, and thus gets reflected at the entrance. FM (100MHz ~ 3m) would propagate and it does for a while but then it suffers reflections

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3 years ago

Read 2 more answers

At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet

expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

also the PE is given as function of position as

$PE = \frac{1}{2}m\omega^2x^2$

now it is given that

KE = PE

now we will have

$\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2$

$A^2 - x^2 = x^2$

$2x^2 = A^2$

$x = \frac{A}{\sqrt2}$

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

now to find the fraction of kinetic energy

$f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}$

$f = \frac{A^2 - (\frac{A}{3})^2}{A^2}$

$f_k = \frac{8}{9}$

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

$f_{PE} = 1 - f_k$

$f_{PE} = 1 - (8/9) = 1/9$

7 0

3 years ago

A model rocket has an acceleration of 12 m/s2. A net force of 18 N is acting on the rocket. What is the mass of the rocket? its

larisa86 [58]

Answer:

C.) 1.5 kg

Explanation:

Start with the equation:

$F_n_e_t=ma$

Plug in what you know, and solve:

$18=m(12)\\m=1.5kg$

Find matching soluation:

C.) 1.5 kg

5 0

2 years ago

Explain how fluids exert pressure

Veseljchak [2.6K]

In a fluid, all the forces exerted by the individual particles combine to make up the pressure exerted by the fluid
Due to fundamental nature of fluids, a fluid cannot remain at rest under the presence of shear stress. However, fluids can exert pressure normal to any contacting surface. If a point in the fluid is thought of as a small cube, then it follows from the principles of equilibrium that the pressure on every side of this unit of fluid must be equal. but if this were not a case, the fluid would move in the directions of the resulting force, So the pressure on a fluid at rest is isotropic.
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4 0

3 years ago

A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she

lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

$f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L}$

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}} }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = 1.3225 \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz = 740.6 \ Hz$

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, $L_{new}$ = 2/3·L

The value of the fundamental frequency will then be given as follows;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \times \dfrac{2 \times L}{3} } =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz = 840 \ Hz$

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0

3 years ago