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Arrange the distances between Earth and various celestial objects in order from least to greatest. Use the conversion table to h

Kaylis [27]

distance to the star Betelgeuse: 640 ly

As we know that

$1 ly = 63000 AU$

also we know that

$1AU = 1.5 \times 10^8 km$

$1 ly = 63000 (1.5 \times 10^8) = 9.45 \times 10^{12} km$

So the distance of Betelgeuse = 640 ly

$d_1 = 640 \times 9.45 \times 10^{12} = 6.05 \times 10^{15} m$

distance to the star VY Canis Majoris: $3.09 × 10^8 AU$

$d_2 = 3.09\times 10^8 \times 1.5 \times 10^8 km$

$d_2 = 4.64 \times 10^{16} km$

distance to the galaxy Large Magellanic Cloud: 49976 pc

$1 pc = 3.262 ly = 3.262 \times 9.45 \times 10^{12} km$

$1pc = 3.08 \times 10^{13} km$

now we have

$d_3 = 49976 \times 3.08 \times 10^{13}$

$d_3 = 1.54 \times 10^{18} km$

distance to Neptune at the farthest: 4.7 billion km

$d_4 = 4.7 \times 10^9 km$

now the order of distance from least to greatest is as following

1. distance to Neptune at the farthest

2. distance of Betelgeuse

3. distance to the star VY Canis Majoris

4. distance to the galaxy Large Magellanic Cloud

6 0

4 years ago

Course hero N4M.6 A board has one end wedged under a rock having a mass of 380 kg and is supported by another rock that touches

aleksandrvk [35]

Answer:

Therefore it is save to carry a 62kg adult

Explanation:

From the question we are told that:

Mass $m=380kg$

Height of supporting Rock $X=85cm$

Length of Board $L_r=4.5m$

Mass of board $M_b=22kg$

Mass of adult $M_a=62$

Generally the moment of balance about wedge part about is mathematically given by

$N -Q + R = Mg + mg$

$0.85*N - Mg*2.25 - mg*(2.25 + x) = 0$

$0.85*N = + Mg*2.25 + mg*(2.25 + x)$

where

$N+R=4547$

therefore

$N = 570.70588 + 1608.3529 + 714.823 x$

if N=0 at fallen person

$x=3.04m$

Therefore it is save to carry a 62kg adult

8 0

3 years ago

If the radius of an atom is 60 pm and the radius of the Earth is 6000 km, by how many orders of

kherson [118]

Answer:

1 x 10¹⁷

Explanation:

Given data:

Radius of the earth = 6000km

Radius of an atom = 60pm

Now, how many orders is the radius of the earth larger than an atom

Solution:

To solve this problem, let us express both quantity as the same unit;

1000m = 1km

6000km = 6000 x 10³m = 6 x 10⁶m

60pm;

1 x 10⁻¹²m = 1pm

60pm = 60 x 1 x 10⁻¹²m = 6 x 10⁻¹¹m

Now;

The order: $\frac{6 x 10^{6} }{6 x 10^{-11} }$ = 1 x 10¹⁷

6 0

3 years ago

A geologist sees a fault along which blocks of rock in the footwall have moved higher relative to blocks of rock in the hanging

Luda [366]

Answer: Normal fault

Explanation:

The type of fault that is explained above is a normal fault. We should note that normal faults typically takes place in a divergent boundary in a scenario where the crusts may have been pulled apart.

Since the crust is pulled apart in this case, it leads to the downward movement of the hanging wall which leads to the football being above the hanging wall.

6 0

3 years ago

A worker is thinking about two ways to get a box up 1.2 m onto a loading dock. He can use a force of 250 N to lift it straight u

harina [27]

Answer:

<em>The second option has a lower power output. P=30 W</em>

Explanation:

<u>Mechanical Power </u>

It is a physical magnitude that measures the rate a work W is done over time t.

$\displaystyle P=\frac{W}{t}$