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2 years ago

14

How did astronomers precisely determine the length of an Astronomical Unit in the 1960s?

1 answer:

zaharov [31]2 years ago

5 0

Answer:

Use of telemetry and radar astronomy

Explanation:

An astronomical Unit (AU) is a unit of measuring distances in outer space, which is based on the approximate distance between the earth and the Sun.

After several years of trying to approximate the distance between the Sun and the Earth using several methods based on geometry and some other calculations, advancements in technology made available the presence of special motoring equipment, which can be placed in outer space to remotely monitor and measure the position of the sun.

The use of direct radar measurements to the sun (radar astronomy) have also made the determination of the AU more accurate.

A standard radar pulse of known speed is sent to the Sun, and the time with which it takes to return is measured, once this is recorded, the distance between the Earth and the Sun can be calculated using

distance = speed X time.

However, most of these means have to be corrected for parallax errors

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2 years ago

Some fish have a density slightly less than that of water and must exert a force (swim) to stay submerged. What force must an 85

vichka [17]

The force require to keep grouper submerged is 8.207N.

According to Archimedes principle buoyant force of any object must equal to weight of fluid it displaced.

The expression for the force exerted to stay submerged in salt water is

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now substitute w(b) for F(b)

→ F = Vρg - w(fish)

where V = volume of sea water

ρ = density of sea water

Now by Archimedes principle V = m(fish) / ρ(fish)

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- (85kg) × 9.8 m/s^2

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brainly.com/question/15076878

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1 year ago

What are some other examples of a force acting on an object without<br> touching it?

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2 years ago

An 67-kg jogger is heading due east at a speed of 2.3 m/s. A 70-kg jogger is heading 61 ° north of east at a speed of 1.3 m/s. F

ololo11 [35]

Answer

given,

mass of jogger = 67 kg

speed in east direction = 2.3 m/s

mass of jogger 2 = 70 Kg

speed = 1.3 m/s in 61 ° north of east.

jogger one

$P_1 = m_1 v_1 \hat{i}$

$P_1 = 67 \times 2.3\hat{i}$

$P_1 = 154.1 \hat{i}$

$P_2 = m_2 v_2 \hat{i} +m_2 v_2 \hat{j}$

$P_2 = 70\times v cos \theta \hat{i} +70\times v sin \theta \hat{j}$

$P_2 = 70\times 1.3 cos 61^0 \hat{i} +70\times 1.3 sin 61^0\hat{j}$

$P_2 = 44.12\hat{i} +79.59\hat{j}$

now

P = P₁ + P₂

$P = 198.22 \hat{i} +79.59 \hat{j}$

magnitude

$P = \sqrt{198.22^2 + 79.59^2}$

$P =213.60 kg.m/s$

$\theta = tan^{-1}\dfrac{79.59}{198.22}$

$\theta = 21.87$

the angle is $\theta = 21.87$ north of east

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3 years ago