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3 years ago

A model train traveling at a constant speed around a circular track has a constant velocity

1 answer:

3 0

A model train traveling at a constant speed around a circular track has a constant velocity. FALSE.
Hope this helps you!

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Please Help!!!!

Len [333]

Answer:

v = 0.92 c

Explanation:

Here, we will use the time dilation formula from Einstein's theory of relativity to find the speed of traveling of the friend:

$t =\frac{t_o}{\sqrt{1-\frac{v^2}{c^2}}} \\\\\\\sqrt{1-\frac{v^2}{c^2}}=\frac{t_o}{t}\\\\$

where,

v = speed of traveling = ?

c = speed of light

t = time of return = 10 years

t₀ = time passed on earth = 4 years

Therefore,

$\sqrt{1-\frac{v^2}{c^2}} = \frac{4\ years}{10\ years}\\\\ 1-\frac{v^2}{c^2}=(\frac{2}{5})^2\\\\\frac{v^2}{c^2} = 1-\frac{4}{25}\\\\\frac{v^2}{c^2} = \frac{21}{25}\\\\v^2 = 0.84c^2\\\\$

8 0

2 years ago

The metal ions in iron are held together by

Sidana [21]

Metals are giant structures of atoms held together by metallic bonds. “Giant” implies that large but variable numbers of Atoms are involved - depending on the size of the bits of metal. most metals are close packed - that is, they fit as many items as possible into the available volume.

5 0

3 years ago

How did Galileo increase public support for Copernicus’s model?

miskamm [114]

I think by using data collected by Tycho Brahe

8 0

3 years ago

Read 2 more answers

What is night vision pls tell fast

mario62 [17]

Answer:

Night vision is when you can see in the dark or at night like owls

Explanation:

3 0

3 years ago

Read 2 more answers

Each of the following diagrams shows a spaceship somewhere along the way between Earth and the Moon (not to scale); the midpoint

djyliett [7]

Answer:

$F_5 >F_4>F_1 >F_2>F_3$

Where $F_i$ represent the force for each of the 5 cases $i -1,2,3,4,5$ presented on the figure attached.

Explanation:

For this case the figure attached shows the illustration for the problem

We have an inverse square law with distance for the force, so then the force of gravity between Earth and the spaceship is lower when the spaceship is far away from Earth.

Th formula is given by:

$F = G \frac{m_{Earth} m_{Spaceship}}{r^2}$

Where G is a constant $G = 6.674 x10^{-11} m^2/ (ks s^2)$

$m_{Earth}$ represent the mass for the earth

$m_{spaceship}$ represent the mass for the spaceship

$r$ represent the radius between the earth and the spaceship

For this reason when the distance between the Earth and the Spaceship increases the Force of gravity needs to decrease since are inversely proportional the force and the radius, and for the other case when the Earth and the spaceship are near then the radius decrease and the Force increase.

Based on this case we can create the following rank:

$F_5 >F_4>F_1 >F_2>F_3$

Where $F_i$ represent the force for each of the 5 cases $i -1,2,3,4,5$ presented on the figure attached.

6 0

3 years ago