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3 years ago

Where on the Earth would you have to be to see the North Celestial Pole exactly halfway between your horizon and zenith?

Physics

1 answer:

mario62 [17]3 years ago

4 0

Answer:

The point straight overhead on the celestial sphere for any observer is called the zenith and is always 90 degrees from the horizon. The arc that goes through the north point on the horizon, zenith, and south point on the horizon is called the meridian.

From any location on Earth you see only half of the celestial sphere, the half above the horizon.

If you stood at the North Pole of Earth, for example, you would see the north celestial pole overhead, at your zenith. The celestial equator, 90° from the celestial poles, would lie along your horizon.

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What does this equation mean FeDe=FrDr

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3 years ago

Why couldnt mendeleev organize the entire table during his research

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3 years ago

Read 2 more answers

Each of the space shuttle's main engines is fed liquid hydrogen bya high-pressure pump. Turbine blades inside the pump rotateat

kvasek [131]

Answer:

a) $a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}$

b) $k = \frac{9.8}{9.74}=1.006$

Explanation:

Part a

For this case we can begin finding the period like this:

$T= \frac{1}{w} =\frac{1}{617 rad/s}=0.00162 s$

Then we know that the centripetal acceleration is given by:

$a= \frac{v^2}{r}$

And the velocity is given by:

$v=\frac{2\pi r}{T}$

If we replace this into the acceleration we got:

$a = \frac{(\frac{2\pi r}{T})^2}{r}= \frac{4 \pi^2 r}{T^2}$

And we can replace the values and we got:

$a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}$

Part b

For this case we want to find a value of k such that:

$a= k 9.8$

Where a = 9.74, so then we can solve for k like this:

$k = \frac{9.8}{9.74}=1.006$

8 0

3 years ago

Find the minimum radius of a helium balloon that will lift her off the ground. the density of helium gas is 0.178 kg/m3, and the

marysya [2.9K]

$Volume of balloon = \frac{4}{3} \pi r^{3}$

Where r is the radius of balloon.

Here mass of woman = 68 kg

Mass of air displaced by a balloon with volume V = 1.29*V

Mass of helium inside balloon = 0.178*V

Total mass to be lifted by balloon = 68 +0.178*V

Buoyant force = 1.29V-0.178V=1.112V

So we have 1.112 V = 68+ 0.178*V

0.934 V = 68

V = 72.81 $m^3$

\frac{4}{3} \pi r^{3}[/tex]= 72.81

r = 2.59 m

So radius of helium balloon = 2.59 m

4 0

3 years ago

A spaceship whose rest length is 350m has a speed of .82c

igomit [66]

Answer:

$t'=1.1897*10^{-6} s$

t'=1.1897 μs

Explanation:

First we will calculate the velocity of micrometeorite relative to spaceship.

Formula:

$u=\frac{u'+v}{1+\frac{u'*v}{c^{2}}}$

where:

v is the velocity of spaceship relative to certain frame of reference = -0.82c (Negative sign is due to antiparallel track).

u is the velocity of micrometeorite relative to same frame of reference as spaceship = .82c (Negative sign is due to antiparallel track)

u' is the relative velocity of micrometeorite with respect to spaceship.

In order to find u' , we can rewrite the above expression as:

$u'=\frac{v-u}{\frac{u*v}{c^{2} }-1 }$

$u'=\frac{-0.82c-0.82c}{\frac{0.82c*(-0.82c)}{c^{2} }-1 }$

u'=0.9806c

Time for micrometeorite to pass spaceship can be calculated as:

$t'=\frac{length}{Relatie seed (u')}$

$t'=\frac{350}{0.9806c}$ (c = 3*10^8 m/s)

$t'=\frac{350}{0.9806* 3.0*10^{8} }$

$t'=1.1897*10^{-6} s$

t'=1.1897 μs

4 0

3 years ago