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Lady_Fox [76]
3 years ago
12

Which of the following criteria do astronomers use to classify an object as a planet? orbits a star gravity makes spherical at l

east as large as mercury clears orbits of other objects
Physics
2 answers:
scoray [572]3 years ago
4 0
The correct answer is Orbits a star, gravity makes it spherical, and clears orbits of other objects. Planets can be smaller than mercury so it's not a criterium.
Flauer [41]3 years ago
3 0

The correct answer of the question is : A) Orbits a star, B) Gravity makes spherical and D) Clear orbits of other objects.

EXPLANATION :

The classification of a celestial object to be a planet depends on some parameters.

The most important thing about the planet is that it must clear its orbit. It means it won't share its orbit with other planets. Otherwise, it will be considered as a dwarf planet.

The second thing is that it must move around any stars. The star becomes the centre of that solar system around which planets start revolving in different orbits.

The third thing is that planets are approximately spherical in shape.

The size of the planets does not play a crucial role in the classification of planets. Planets may be smaller than mercury also.

Hence, all the options except the third one are suitable.



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Find the value of F1 + F2 + F3.<br>​
Dovator [93]

Answer:

F = 0.78[N]

Explanation:

The given values correspond to forces, we must remember or take into account that the forces are vector quantities, that is, they have magnitude and direction. Since we have two X-Y coordinate axes (two-dimensional), we are going to decompose each of the forces into the X & y components.

<u>For F₁</u>

<u />F_{y}=2[N]<u />

<u>For F₂</u>

F_{x}=2*cos(60)\\F_{x}=1[N]\\F_{y}=-2*sin(60)\\F_{y}=-1.73[N]

<u>For F₃</u>

<u />F_{x}=-1*sin(60)\\F_{x}=-0.866[N]\\F_{y}=1*cos(60)\\F_{y}=0.5 [N]<u />

Now we can sum each one of the forces in the given axes:

F_{x}=1-0.866=0.134[N]\\F_{y}=2-1.73+0.5\\F_{y}=0.77[N]

Now using the Pythagorean theorem we can find the total force.

F=\sqrt{(0.134)^{2} +(0.77)^{2}}\\F= 0.78[N]

8 0
3 years ago
An ice-making machine inside a refrigerator operates in a Carnot cycle. It takes heat from liquid water at 0.0 ∘C and rejects he
Agata [3.3K]

Answer:

2.36 x 10^6 J

Explanation:

Tc = 0°C = 273 K

TH = 22.5°C = 295.5 K

Qc = heat used to melt the ice

mass of ice, m = 85.7 Kg

Latent heat of fusion, L = 3.34 x 10^5 J/kg

Let Energy supplied is E which is equal to the work done

Qc = m x L = 85.7 x 3.34 x 10^5 =  286.24 x 10^5 J

Use the Carnot's equation

\frac{Q_{H}}{Q_{c}}=\frac{T_{H}}{T_{c}}

Q_{H}=286.24\times 10^{5}\times \frac{295.5}{273}

QH = 309.8 x 10^5 J

W = QH - Qc

W = (309.8 - 286.24) x 10^5

W = 23.56 x 10^5 J

W = 2.36 x 10^6 J

Thus, the energy supplied is 2.36 x 10^6 J.

8 0
3 years ago
A hippo drives 42 km due East. He then turns and drives 28 km at 25° East of South. He turns again and drives 32 km at 40° North
ch4aika [34]

Answer:

a) Please, see the attched figure

b) Total displacement R = (78.3 km; -4.8 km)

c) R = (78.4 km * cos (-3.5°); 78.4 km * sin (-3.5°))

d) The hippo is 78.4 km from his starting point.

The total distance traveled is 102 km

Explanation:

a)Please, see the attached figure.

b) The vector A can be expressed as:

A = (magnitude * cos α; magnitude * sin α)

Where

magnitude = 42 km

α= 0

Then,

A = (42 km ; 0) or 42 km i

In the same way, we can proceed with the other vectors:

B = ( Bx ; By)

where

(apply trigonometry of right triangles: sen α = opposite / hypotenuse and

cos α = adjacent / hypotenuse. See the figure to determine which component of vector B is the opposite and adjacent side to α)

Bx = 28 km * sin 25 = 11.8 km

By = 28 km * cos 25 = -25.4 km (it has to be negative since it is directed towards the negative vertical region according to our reference system)

B = (11.8 km; -25.4 km) or 11.8 km i - 25.4 km j

C = (Cx; Cy)

where

Cx = 32 km * cos 40° = 24.5 km

Cy = 32 km * sin 40 = 20.6 km

C = (24.5 km; 20.6 km)

Then:

R = A+B+C = (42 km + 11.8 km + 24.5 km; 0 - 25.4 km + 20.6 km)

= (78.3 km; -4.8 km) or 78.3 km i -4.8 km j

c) R = (78.3 km; -4.8 km)

The magnitude of R is:

magnitude = \sqrt{(78.3)^{2 }+ (-4.8)^{2}}= 78.4 km

Using trigonometry, we can calculate the angle:

Knowing that

tan α = opposite / adjacent

and that

opposite = Ry = -4.8 km

adjacent = Rx = 78.3 km

Then:

tan α = -4.8 km / 78.4 km

α = -3.5°

We can now write the vector R in magnitude and direction form:

R = (78.4 km * cos (-3.5°); 78.4 km * sin (-3.5°))

d) The displacement of the hipo relative to the starting point is the magnitude of vector R calculated in c):

magnitude R = 78. 4 km

The total distance traveled is the sum of the magnitudes of each vector:

Total distance = 42 km + 28 km + 32 km = 102 km  

3 0
3 years ago
Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th
aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

-625 = 100A

Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

-225 = -12.5s

Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

3 0
3 years ago
The law of reflection states that the angle of reflection
mote1985 [20]
If the object, ends up with a positive charge, then it is missing electrons. if it is missing electrons, then it must have been removed form the object during the rubbing process.
5 0
3 years ago
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