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3 years ago

What do astronomers call a system that is composed of more than two stars

Physics

1 answer:

andrew-mc [135]3 years ago

3 0

Astronomers call a system with more than two stars a constellation.

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A car traveling west in a straight line on a highway decreases its speed from 30 meters per second to 23 meters per second in 2

garik1379 [7]

U1= 30m/s
u2= 23m/s
t=2s
a=(u2-u1)/t
a=-7/2=-3.5 m/s^2

6 0

3 years ago

Read 2 more answers

Once a disk forms around a star, the process of planetary formation can begin. Rank the evolutionary stages for the formation of

PolarNik [594]

Answer: See explanation

Explanation:

The evolutionary stages for the formation of planets from earliest to latest will be:

1. Dust keeps matter inside the disk cool enough for planet formation to start

2. Dust grains form condensation nuclei on which surrounding atoms condense to form small clumps of matter.

3. Small clumps of matter stick together via the process of accretion to form planetesimals a few hundred kilometers in diameter.

4. Planetesimals begin to accrete, forming protoplanets.

5. A collection of a few planet-sized protoplanets remain in a fairly cleared out disk around the star

6 0

3 years ago

A piece of wood has a volume of 4.6 cm to the power of 3, and a mass of 3.7 g what is the density of the wood

AysviL [449]

Density is given by:

D = M/V

D = density, M = mass, V = volume

Given values:

M = 3.7g, V = 4.6cm³

Plug in and solve for D:

D = 3.7/4.6

D = 0.80g/cm³

3 0

3 years ago

Consider a container of oxygen gas at a temperature of 23°C that is 1.00 m tall. Compare the gravitational potential energy of a

Sergio039 [100]

Answer:

Yes, it is reasonable to neglect it.

Explanation:

Hello,

In this case, a single molecule of oxygen weights 32 g (diatomic oxygen) thus, the mass of kilograms is (consider Avogadro's number):

$m=1molec*\frac{1mol}{6.022x10^{23}molec} *\frac{32g}{1mol}*\frac{1kg}{1000g}=5.31x10^{-26}kg$

After that, we compute the potential energy 1.00 m above the reference point:

$U=mhg=5.31x10^{-26}kg*1.00m*9.8\frac{m}{s^2}=5.2x10^{-25}J$

Then, we compute the average kinetic energy at the specified temperature:

$K=\frac{3}{2}\frac{R}{Na}T$

Whereas $N_A$ stands for the Avogadro's number for which we have:

$K=\frac{3}{2} \frac{8.314\frac{J}{mol*K}}{6.022x10^{23}/mol}*(23+273)K\\ \\K=6.13x10^{-21}J$

In such a way, since the average kinetic energy energy is about 12000 times higher than the potential energy, it turns out reasonable to neglect the potential energy.

Regards.

8 0

3 years ago

A 38.2 kg wagon is towed up a hill inclined at 17.5 ◦ with respect to the horizontal. The tow rope is parallel to the incline an

Tema [17]

Answer:

v = 8.57 m/s

Explanation:

As we know that the wagon is pulled up by string system

So the net force on the wagon along the inclined is due to tension in the rope and component of weight along the inclined plane

So as per work energy theorem we know that

work done by tension force + work done by force of gravity = change in kinetic energy

$F_t . d - (mgsin\theta)(d) = \frac{1}{2}mv^2 - 0$

so we have

$F_t = 129 N$

$\theta = 17.5^o$

m = 38.2 kg

d = 85.4 m

so now we have

$129(85.4) - (38.2)9.8sin17.5 (85.4) = \frac{1}{2}(38.2) v^2$

$v = 8.57 m/s$

7 0

3 years ago