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1 year ago

14

Describe how sound signals can be transmitted from a radio station and received on a radio at your home. Create a labeled diagra

m that shows this process

2 answers:

vlada-n [284]1 year ago

6 0

A radio wave is generated by a transmitter and then detected by a receiver. An antenna allows a radio transmitter to send energy into space and a receiver to pick up energy from space. Transmitters and receivers are typically designed to operate over a limited range of frequencies

MrRissso [65]1 year ago

4 0

Answer:

The person above is right.

Explanation:

You just need to type in sound signals transmitted from a radio station and received on a radio at your home to find a diagram https://flypaper.soundfly.com/wp-content/uploads/2019/03/am.gif

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A monkey weighs 6.00 x 102 N and swings from vine to vine. As the monkey grabs a new vine, both vines make an angle of 35.0° wit

zmey [24]

Answer:

$T=366.23\ N$

Explanation:

Given:

mass of monkey, $w=600\ N$

angle of vine from the vertical, $\theta=35^{\circ}$

Now follow the schematic to understand the symmetry and solution via Lami's theorem.

<u>The weight of the monkey will be balanced equally by the tension in both the vines:</u>

Using Lami's Theorem:

$\frac{w}{sin\ 70^{\circ}} =\frac{T}{sin\ 145^{\circ}}$

$\frac{600}{sin\ 70^{\circ}} =\frac{T}{sin\ 145^{\circ} }$

$T=366.23\ N$

4 0

1 year ago

Find earth's approximate mass from the fact that the moon orbits earth in an average time of 27.3 days at an average distance of

Aleks [24]

We can solve the problem by using Kepler's third law, which states:

$\frac{4 \pi^2}{T^2}=\frac{GM}{r^3}$

where T is the period of revolution of the Moon around the Earth, G is the gravitational constant, M the Earth's mass and r the average distance between Earth and Moon.

Using the data of the problem:

$T=27.3 d \cdot 24 \cdot 60 \cdot 60 = 2358720 s=2.36 \cdot 10^6 s$

$r=384000 km=3.84 \cdot 10^8 m$

We can re-arrange the equation and find the Earth's mass:

$M=\frac{4 \pi^2 r^3}{GT^2}=\frac{4 \pi^2 (3.84 \cdot 10^8 m)^3}{(6.67 \cdot 10^{-11})(2.36 \cdot 10^6 s)^2}=6.0 \cdot 10^{24} kg$

4 0

1 year ago

Suppose that the mirror described in Part A is initially at rest a distance R away from the sun. What is the critical value of a

Sati [7]

Ans;

6.25m²/Kd

Explanation: see attached file

5 0

1 year ago

A 6.0-Ï‰ and a 12-Ï‰ resistor are connected in parallel across an ideal 36-v battery. What power is dissipated by the 6.0-Ï‰ res

Elena L [17]

Answer:

P = 216 Watts

Explanation:

Given that,

Resistance 1, $R_1=6\ \Omega$

Resistance 2, $R_2=12\ \Omega$

Voltage, V = 36 V

We need to find the power dissipated by the 6 ohms resistor. In case of parallel circuit, voltage throughout the circuit is same while the current is different.

Firstly, lets find current of the 6 ohms resistor such that,

V = IR

$I=\dfrac{V}{R}\\\\I=\dfrac{36}{6}\\\\I=6\ A$

The power dissipated by 6 ohms resistor is given by :

$P=I^2R$

$P=6^2\times 6\\\\P=216\ W$

So, the power dissipated across 6 ohm resistor is 126 watts.

8 0

1 year ago

A light ray just grazes the surface of the Earth (M = 6.0 × 10 24 kg, R = 6.4×10 6 m). Through what angle α is the light ray ben

grandymaker [24]

Answer:

(a). The deflection angle is $2.77\times10^{-9}\ rad$

(b). The deflection angle is $3.95\times10^{-4}\ rad$

(c). The deflection angle is $7.41\times10^{-1}\ rad$

Explanation:

Given that,

Mass of earth $M_{e}=6.0\times10^{24}\ kg$

Radius of earth $R_{e}=6.4\times10^{6}\ m$

Mass of white dwarf $M=2.0\times10^{30}\ kg$

Radius of white dwarf $R=1.5\times10^{7}\ m$

Mass of Neutron $M=3.0\times10^{30}\ kg$

Radius of neutron $R=1.2\times10^{4}\ m$

We need to calculate the deflection angle for earth

Using formula of angle

$\alpha=\dfrac{4G M}{c^2R}$

Where, R = radius

G = gravitational constant

M = mass

c = speed of light

Put the value into the formula

$\alpha=\dfrac{4\times6.67\times10^{-11}\times6.0\times10^{24}}{(3\times10^{8})^2\times6.4\times10^{6}}$

$\alpha=2.77\times10^{-9}\ rad$

The deflection angle is $2.77\times10^{-9}\ rad$

We need to calculate the deflection angle for white dwarf

Using formula of angle

$\alpha=\dfrac{4G M}{c^2R}$

Put the value into the formula

$\alpha=\dfrac{4\times6.67\times10^{-11}\times2.0\times10^{30}}{(3\times10^{8})^2\times1.5\times10^{7}}$

$\alpha=3.95\times10^{-4}\ rad$

The deflection angle is $3.95\times10^{-4}\ rad$

We need to calculate the deflection angle for neutron star

Using formula of angle

$\alpha=\dfrac{4G M}{c^2R}$

Put the value into the formula

$\alpha=\dfrac{4\times6.67\times10^{-11}\times3.0\times10^{30}}{(3\times10^{8})^2\times1.2\times10^{4}}$

$\alpha=7.41\times10^{-1}\ rad$

The deflection angle is $7.41\times10^{-1}\ rad$

Hence, This is the required solution.

3 0

2 years ago