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3 years ago

people used to think that the sun moved around earth galileo proved this was wrong why is this important

Physics

1 answer:

tamaranim1 [39]3 years ago

6 0

To approve to people that the Catholic Church was wrong but they threatened Galileo and he stop

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For metalloids on the periodic table, how do the group number and the period number relate?

lapo4ka [179]

Im guessing it's (a) since the numbers go in chronological order and you read the periodic table left to right

3 0

3 years ago

Read 2 more answers

If the mass of a material is 42 grams and the volume of the material is 15 cm^3, what would the density of the material be?

Fantom [35]

Density = mass / volume

Density = 42g / 15cm^3

Density = 2.8g/cm^3

3 0

3 years ago

A vinyl record is played by rotating the record so that an approximately circular groove in the vinyl slides under a stylus. Bum

AleksandrR [38]

Answer:

Hits per second=199 hit/s

Explanation:

#Given the angular velocity, $\omega=33\frac{1}{3} rev/min$ , radius of the record $r=0.1m$ and the distance between any two successive bumps on the groove as $d=1.75mm$ .

The linear speed of the record in meters per second is:

$v=\omega r=33\frac{1}{3}\times\frac{2\pi}{60}\times 10\times 10^{_2}\\\\=0.3843m/s\\$

#From $v$ above, if the bumps are uniformly separated by 1m, then the rate at which they hit the stylus is:

$Hits/second=v/d \ \ \ \ d=1.75mm\\\\=0.3483/0.000175\\\\=199.0385714\approx 199$

Hence the bumps hit the stylus at around 199hit/s

8 0

3 years ago

During a new moon, the moon is between the

mylen [45]

The moon is between the sun and earth.

The side where the light from the sun hits the moon is facing away from earth.

5 0

3 years ago

7.22 Ignoring reflection at the air–water boundary, if the amplitude of a 1 GHz incident wave in air is 20 V/m at the water surf

Serga [27]

Answer:

z = 0.8 (approx)

Explanation:

given,

Amplitude of 1 GHz incident wave in air = 20 V/m

Water has,

μr = 1

at 1 GHz, r = 80 and σ = 1 S/m.

depth of water when amplitude is down to 1 μV/m

Intrinsic impedance of air = 120 π Ω

Intrinsic impedance of water = $\dfrac{120\pi}{\epsilon_r}$

Using equation to solve the problem

$E(z) = E_0 e^{-\alpha\ z}$

E(z) is the amplitude under water at z depth

E_o is the amplitude of wave on the surface of water

z is the depth under water

$\alpha = \dfrac{\sigma}{2}\sqrt{\dfrac{(120\pi)^2}{\Epsilon_r}}$

$\alpha = \dfrac{1}{2}\sqrt{\dfrac{(120\pi)^2}{80}}$

$\alpha =21.07\ Np/m$

now ,

$1 \times 10^{-6} = 20 e^{-21.07\times z}$

$e^{21.07\times z}= 20\times 10^{6}$

taking ln both side

21.07 x z = 16.81

z = 0.797

z = 0.8 (approx)

5 0

3 years ago