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3 years ago

The Milky Way and the Andromeda are both________.

Physics

1 answer:

Igoryamba3 years ago

6 0

They are both spiral galaxies

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How would a chart of mid-17th century religious beliefs differ from this chart?

tino4ka555 [31]

Answer:

The correct answer is B) Fewer people would identify as an atheist because people were not willing to share alternative religious beliefs publically.

Explanation:

A chart of mid-17th-century religious beliefs differs from this chart in that Fewer people would identify as an atheist because people were not willing to share alternative religious beliefs publically.

In the 1600s, people did not have total freedom of speech if they had any. The church had a tremendous influence in the life of people and religious beliefs defined societies and families. The Church exerted its power and influence in many aspects of the people's lives and something out of the purview of the church or different to the religious beliefs of the Church was considered to be sacrilegious. The Church prosecuted people for being against the Church, so people of that time preferred to say that they were religious people supporting the church. Being an atheist was not really an option in the 17th century.

Hope this helps!!

4 0

3 years ago

Read 2 more answers

Simplicity expression

Volgvan

They have expression, modesty, sex appeal, simplicity.

7 0

3 years ago

25 mL of water is poured into a graduated cylinder. Several coins are dropped into the water, and the new volume is measured to

Arisa [49]

5 cubic cm is the answer

4 0

2 years ago

Read 2 more answers

Ablok slides on ahorizonted sur fors which has been lubricated with heavy oil such that the blok soffers a viscous resistance th

svetoff [14.1K]

Answer:

$v = (v_0^{\frac{3}{2}}-\frac{3cx}{2m})^\frac{2}{3}$

$x = \frac{2m}{c}*v_0^{\frac{3}{2}}$

Explanation:

Given

$f(v) =- cv^\frac{1}{2}$

To start with, we begin with

$F = ma$

Substitute the expression for F

$-cv^\frac{1}{2} = ma$

$-ma = cv^\frac{1}{2}$

Acceleration (a) is:

$a = \frac{dv}{dt}$

So, the expression becomes:

$m\frac{dv}{dt} = -cv^\frac{1}{2}$

-----------------------------------------------------------------------------------------

Velocity (v) is:

$v = \frac{dx}{dt}$ --- distance/time

$v * \frac{dv}{dx}= \frac{dx}{dt}* \frac{dv}{dx}$

$v * \frac{dv}{dx}= \frac{dv}{dt}$

$\frac{dv}{dt} = v * \frac{dv}{dx}$

--------------------------------------------------------------------------------------------

So, we have:

$m\frac{dv}{dt} = -cv^\frac{1}{2}$

$mv * \frac{dv}{dx} = -cv^\frac{1}{2}$

Divide both sides by $v^\frac{1}{2}$

$mv^{1-\frac{1}{2}} * \frac{dv}{dx} = -c$

$mv^{\frac{1}{2}} * \frac{dv}{dx} = -c$

Divide both sides by m

$v^{\frac{1}{2}} * \frac{dv}{dx} = -\frac{c}{m}$

$v^{\frac{1}{2}} * dv = -\frac{c}{m} * dx$

Integrate:

$\int\limits^v_{v_0} {v^{\frac{1}{2}}} \, dv = -\frac{c}{m}\int\limits^x_0 {}} \, dx$

$\frac{2}{3}v^{\frac{3}{2}}|\limits^v_{v_0} = -\frac{c}{m}x|\limits^x_0$

$\frac{2}{3}(v^{\frac{3}{2}} - v_0^{\frac{3}{2}} ) = -\frac{cx}{m}$

$v^{\frac{3}{2}} - v_0^{\frac{3}{2}} = -\frac{3cx}{2m}$

$v^{\frac{3}{2}} = v_0^{\frac{3}{2}}-\frac{3cx}{2m}$

$v = (v_0^{\frac{3}{2}}-\frac{3cx}{2m})^\frac{2}{3}$

Next, is to get the maximum velocity by distance x.

To do this, we find the derivation by x

$\frac{dv}{dx} = 0$

$\frac{2}{3}(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{2}{3} - 1} * -\frac{3c}{2m} = 0$

$\frac{2}{3}(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{2-3}{3}} * -\frac{3c}{2m} = 0$

$\frac{2}{3}(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{-1}{3}} * -\frac{3c}{2m} = 0$

$(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{-1}{3}} * -\frac{c}{m} = 0$

Divide both sides by $-\frac{c}{m}$

$(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{-1}{3}} = 0$

Take cube roots of both sides

$(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{-1} = 0$

$v_0^{\frac{3}{2}}-\frac{3cx}{2m} = 0$

$\frac{3cx}{2m} = v_0^{\frac{3}{2}}$

$x = \frac{2m}{c}*v_0^{\frac{3}{2}}$

7 0

3 years ago

The mine car and its contents have a total mass of 6.0 Mg and a center of gravity at G. If the coefficient of static friction be

ohaa [14]

Answer:

The mine car does not move

Explanation:

Attached below is a comprehensive explanation on how I arrived at my answer.

Cheers

3 0

3 years ago