3 years ago

Recording the location of a star requires a measurement of:

Physics

1 answer:

Scorpion4ik [409]3 years ago

7 0

Answer:

It requires a measurement of altitude azimuth time.

Hope this helps, if it did, please give it a brainliest.

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3.An object that begins at rest has an acceleration of 2 m/s/s What is its instantaneous speed after 3 seconds?

neonofarm [45]

The speed will be 6m/s i believe

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4 years ago

An applicable analogy for part of this activity: The most direct route between San Diego, California, and Caribou, Maine, is abo

PtichkaEL [24]

Answer:

Total time spend = 48.18 hours (Approx)

Explanation:

Given:

Total distance = 5,300 km

Average speed = 110 km / h

Find:

Total time spend

Computation:

Time = Distance / speed

Total time spend = Total distance / Average speed

Total time spend = 5,300 / 110

Total time spend = 48.18 hours (Approx)

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3 years ago

Why do you think the objects in each collision experience a different change in velocity?

Sphinxa [80]

Answer:

In a collision, there is a collision force which endures for some amount of time to cause an impulse. This impulse acts upon the object to change its velocity and thus its momentum

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3 years ago

Lin suffers from pain in her right wrist. Several doctors check her wrist regularly over a period of time. Doctors' notes and

FromTheMoon [43]

Answer:

Explanation:

They looked at her hand and made a formal conclusion from that

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3 years ago

At the end of the adiabatic expansion, the gas fills a new volume V₁, where V₁ > V₀. Find W, the work done by the gas on the

tino4ka555 [31]

Answer:

$W=\frac{p_0V_0-p_1V_1}{\gamma-1}$

Explanation:

An adiabatic process refers to one where there is no exchange of heat.

The equation of state of an adiabatic process is given by,

$pV^{\gamma}=k$

where,

$p$ = pressure

$V$ = volume

$\gamma=\frac{C_p}{C_V}$

$k$ = constant

Therefore, work done by the gas during expansion is,

$W=\int\limits^{V_1}_{V_0} {p} \, dV$

$=k\int\limits^{V_1}_{V_0} {V^{-\gamma}} \, dV$

$=\frac{k}{\gamma -1} (V_0^{1-\gamma}-V_1^{1-\gamma})\\$

(using $pV^{\gamma}=k$ )

$=\frac{p_0V_0-p_1V_1}{\gamma-1}$

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3 years ago