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Masteriza [31]
3 years ago
9

Why might some people not have believed Galileo's discoveries?

Physics
2 answers:
Mademuasel [1]3 years ago
8 0
They did not believed Galileo's discoveries because religiouse reasons the preast said that all the bible is true but Galileo despised it.
tankabanditka [31]3 years ago
6 0

Answer:

Explanation:

Many people did not believed in the discoveries made by Galileo especially the concept of heliocentrism because of a conflict between Catholic church and Galileo.

There were many controversies between the idea of Heliocentrism that Catholic churches have threatened  Galileo to abolish it.

There was many other astronomers as well who did not believed in his theory because of absence of stellar parallax and considered this theory as foolish and absurd.

Dan Brown in his novels has very well explained the difference in ideas of Galileo and Catholic church.

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A cosmic ray (an electron or nucleus moving ar speeds close to the speed of light) travels across the Milky Way at a speed of 0.
Fiesta28 [93]

Answer:

Cosmic ray's frame of reference: 99,875 years

Stationary frame of reference: 501,891 years

Explanation:

First of all, we convert the distance from parsec into metres:

d=30,000 pc =9.26\cdot 10^{20} m

The speed of the cosmic ray is

v=0.98 c

where

c=3.0 \cdot 10^8 m/s is the speed of light. Substituting,

v=(0.98)(3.0\cdot 10^8)=2.94\cdot 10^8 m/s

And so, the time taken to complete the journey in the cosmic's ray frame of reference (called proper time) is:

T_0 = \frac{d}{v}=\frac{9.26\cdot 10^{20}}{2.94\cdot 10^8}=3.15\cdot 10^{12} s

Converting into years,

T_0 = \frac{3.15\cdot 10^{12}}{(365\cdot 24\cdot 60 \cdot 60}=99,875 years

Instead, the time elapsed in the stationary frame of reference is given by Lorentz transformation:

T=\frac{T_0}{\sqrt{1-(\frac{v}{c^2})^2}}

And substituting v = 0.98c, we find:

T=\frac{99,875}{\sqrt{1-(\frac{0.98c}{c})^2}}=501,891 years

3 0
3 years ago
A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of
34kurt

Answer:

The centripetal acceleration of the runner is 1.73\ m/s^2.

Explanation:

Given that,

A runner completes the 200 m dash in 24.0 s and runs at constant speed throughout the race. We need to find the centripetal acceleration as he runs the curved portion of the track. We know that the centripetal acceleration is given by :

a=\dfrac{v^2}{r}

v is the velocity of runner

v=\dfrac{200\ m}{24\ s}\\\\v=8.34\ m/s

Centripetal acceleration,

a=\dfrac{(8.34)^2}{40}\\\\a=1.73\ m/s^2

So, the centripetal acceleration of the runner is 1.73\ m/s^2. Hence, this is the required solution.

5 0
3 years ago
A capacitor with an initial potential difference of 185 V is discharged through a resistor when a switch between them is closed
GrogVix [38]

Answer:

  • a. \tau =  2.1161 s
  • b. V(18.8 \ s) = 0.0256 \ V

Explanation:

<h3>a.</h3>

The equation for the voltage V of  discharging capacitor in an RC circuit at time t is:

V(t) = V_0 e^{(- \frac{t}{\tau}) }

where V_0 is the initial voltage, and \tau is the time constant.

For our problem, we know

V_0 = 185 \ V

and

V(10 \ s) = V_0 e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V

So

185 \ V \ e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V

e^{(- \frac{10 \ s}{\tau}) } = \frac{1.64 \ V}{ 185 \ V }

ln (e^{(- \frac{10 \ s}{\tau}) } ) = ln (\frac{1.64 \ V}{ 185 \ V })

- \frac{10 \ s}{\tau}  = ln (\frac{1.64 \ V}{ 185 \ V })

\tau =  \frac{-10 \s}{ln (\frac{1.64 \ V}{ 185 \ V }) }

This gives us

\tau =  2.1161 s

and this is the time constant.

<h3>b.</h3>

At t = 18.8 s we got:

V(18.8 \ s) = 185 \ V  \ e^{(- \frac{18.8 \ s}{2.1161 s}) }

V(18.8 \ s) = 185 \ V \ e^{(- \frac{18.8 \ s}{2.1161 s}) }

V(18.8 \ s) = 0.0256 \ V

4 0
3 years ago
Free fall differs from terminal velocity, in that during free fall
viktelen [127]

Answer:

B

Explanation:

8 0
2 years ago
You are explaining to friends why astronauts feel weightless orbiting in the space shuttle, and they respond that they thought g
bezimeni [28]

Answer:

It's only 1.11 m/s2 weaker at 400 km above surface of Earth

Explanation:

Let Earth radius be 6371 km, or 6371000 m. At 400km above the Earth surface would be 6371 + 400 = 6771 km, or 6771000 m

We can use Newton's gravitational law to calculate difference in gravitational acceleration between point A (Earth surface) and point B (400km above Earth surface):

g = G\frac{M}{r^2}

where G is the gravitational constant, M is the mass of Earth and r is the distance form the center of Earth to the object

\frac{g_B}{g_A} = \frac{GM/r^2_B}{GM/r^2_A}

\frac{g_B}{g_A} = \left(\frac{r_A}{r_B}\right)^2

\frac{g_B}{g_A} = \left(\frac{6371000}{6771000}\right)^2

\frac{g_B}{g_A} = 0.94^2 = 0.885

g_B = 0.885 g_A

So the gravitational acceleration at 400km above surface is only 0.885 the gravitational energy at the surface, or 0.885*9.81 = 8.7 m/s2, a difference of (9.81 - 8.7) = 1.11 m/s2.

6 0
3 years ago
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