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3 years ago

What is the definition of zenith, altitude, horizon, circompolar stars, astrolabe, and doppler effect in astronomy?

Physics

1 answer:

Nikitich [7]3 years ago

3 0

zenith- the point in the sky or celestial sphere directly above an observer

altitude- the apparent height of a celestial object above the horizon, measured as an angle

horizon- a great circle of the celestial sphere, the plane of which passes through the center of the earth and is parallel to that of the apparent horizon of a place.

circompolar stars- (of a star or motion) above the horizon at all times in a given latitude.

astrolabe- One of the fundamental calculations in astrology is the position of the sun relative to the signs of the zodiac. The path of the sun as seen from Earth is called the ecliptic and the astrolabe calculates the position of the sun on this path.

doppler effect- The Doppler effect is the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source. It is named after the Austrian physicist Christian Doppler, who described the phenomenon in 1842.

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750 kg car zooms away from a red light with an acceleration of 7.8 m/s squared . What is the average net force in Newtons that t

natta225 [31]

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 750 × 7.8

We have the final answer as

Hope this helps you

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3 years ago

Explain why ventilation is very important if there is risk of exposure to random gas in your home school

bonufazy [111]

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3 years ago

A 2.98-kg object oscillates on a spring with an amplitude of 8.05 cm. Its maximum acceleration is 3.55 m/s2. Calculate the total

Aloiza [94]

Answer:

a = ω^2 A formula for max acceleration (ignoring sign)

V = ω A formula for max velocity

V^2 = ω^2 A^2 = a A from first equation

E = 1/2 M V^2 = 1/2 * 2.98 * 3.55 * .0805 = .426 J

(kg * m/sec^2 * m = kg m^2 / sec^2 = Joule

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2 years ago

An artillery shell is fired with an initial velocity of 300 m/s at 52.0° above the horizontal. To clear an avalanche, it explode

sasho [114]

The x- and y-coordinates are 9142.57 m and -304.425 m

<u>Explanation:</u>

As the motion of the shell is in a plane (two dimensional space) and the acceleration is that due to gravity which is vertically downward, we resolve initial velocity of the shell $v_{0}$ in horizontal and vertical directions. If the initial velocity of the shell is making angle with the horizontal, the horizontal component of initial velocity will be

$v_{x}=v_{0} \times \cos \theta$

As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of $v_{x}$ and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by

$v_{x}=v_{o} \times \cos \theta=300 \times \cos \left(52^{\circ}\right)=184.69 \mathrm{m} / \mathrm{s}$

$v_{y}=v_{o} \times \sin \theta==300 \times \sin \left(52^{\circ}\right)=236.4 \mathrm{m} / \mathrm{s}$

For motion with constant acceleration, we know

$s=s_{0}+v_{0} t+\left(\frac{(1)}{2}\right) a t^{2}$

Along the horizontal, x-axis, we might write this as

$x=x_{0}+v_{x 0} t+\left(\frac{1}{2}\right) a_{x} t^{2}$

Measuring distances relative to the firing point means

$x_{0}=0$

we know that,

$a_{x}=0$

or,

$v_{x}=v_{x 0}=\text { constant }$

By applying the values, we get,

$x=0+(184.69 \times 49.5)+\left(\left(\frac{1}{2}\right) \times 0 \times(49.5)^{2}\right)=9142.57 \mathrm{m}$

The acceleration of gravity is vertically downward and is $g=-9.8 \mathrm{m} / \mathrm{s}^{2}$ , hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion