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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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In an elastic collision, a 400-kg bumper car collides directly from behind with a second, identical bumper car that is traveling
kirza4 [7]

Answer:

v₁ = 3.5 m/s

v₂ = 6.4 m/s

Explanation:

We have the following data:

m₁ = mass of trailing car = 400 kg

m₂ = mass of leading car = 400 kg

u₁ = initial speed of trailing car = 6.4 m/s

u₂ = initial speed of leading car = 3.5 m/s

v₁ = final speed of trailing car = ?

v₂ = final speed of leading car = ?

The final speed of the leading car is given by the following formula:

v_2=\frac{2m_1}{m_1+m_2}u_1-\frac{m_1-m_2}{m_1+m_2}u_2\\\\v_2=\frac{(2)(400\ kg)}{400\ kg+400\ kg}(6.4\ m/s)-\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(3.5\ m/s)

<u>v₂ = 6.4 m/s</u>

The final speed of the leading car is given by the following formula:

v_1=\frac{m_1-m_2}{m_1+m_2}u_1+\frac{2m_2}{m_1+m_2}u_2\\\\v_1=\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(6.4\ m/s)+\frac{(2)(400\ kg)}{400\ kg+400\ kg}(3.5\ m/s)

<u>v₁ = 3.5 m/s</u>

4 0
3 years ago
Your friends sit in a sled in the snow. If you apply a force pf 75 N to them, they have an acceleration of 0.9 m/s ^ 2. What is
Elza [17]

Answer:

Mass of the sled in the snow 83.33 kg.

<u>Explanation</u>:

Given that,  

Force applied to move the sled in the snow (F) = 75N

\text { Acceleration }(a)=0.9 \mathrm{m} / \mathrm{s}^{2}

We know that

Newton's second law of motion is  

\text { Force }=\text { mass } \times \text { acceleration }

F = ma (Or "force" is equal to "mass" times "acceleration".)

So if we move this around we can isolate mass and get mass

\text { Mass }=\frac{\text { force }}{\text { accelearation }}

\mathrm{M}=\frac{75}{0.9}

M = 83.33 kg

Mass of the sled in the snow <u>83.33 kg.</u>

3 0
3 years ago
Which part of the wave has the highest frequency?
trapecia [35]

Answer:

The last part on the right side of the diagram

Explanation:

Im on plato and just got it right :)

6 0
3 years ago
Read 2 more answers
One horse is pulling a 755 kg sled straight ahead applying a force of 1988 N. If the acceleration of the sled is 1.36 m/s2, what
Inessa [10]

Answer:

The coefficient of kinetic friction is 0.13

Explanation:

Newton's second law states that the acceleration of an object is proportional to the net force on it, the factor of proportionality is the mass. So, we can express that law mathematically as:

\sum\overrightarrow{F}=m\overrightarrow{a} (1)

With F the net force, m the mass and a the acceleration of the object. In our case we're interested on what's happening to the sled, then we have to analyze the forces on it, those forces are the weight and the normal force on the vertical direction and the pulling force and frictional force in the horizontal direction. So, because (1) is a vector equation we can express that in their vertical (y) and horizontal (x) components:

F_y=ma_y (2)

F_x=ma_x (3)

On y we have that the acceleration is zero because the sled is not moving upward or downward, remember that the net force on y is the weight (W) pointing downward and the normal force pointing upward:

F_y=W+n=0

Following the convention that positive is upward and negative downward, W=mg=(755)(-9.81):

F_y=(755)(-9.81)+n=0

n=7406.55 N (4)

Now on the x direction we have the sum of the forces is the pulling force (T) and friction force (f)

F_x=F+f=ma_x

Choosing the direction where the horse is pulling F=1988N and the acceleration should be positive too, then:

1988+f=m(1.36)

f=(755)(1.36)-1988=-961.2 N

The negative sign means it's in the opposite direction the horse is pulling

The frictional force is related with the coefficient of kinetic friction in the next way:

|f|=\mu_k n

with μk the coefficient of kinetic friction, and n the normal force that we already found on (4), so we simply solve the last equation for μk:

\mu_k=\frac{|f|}{n}=\frac{961.2}{7406.55}=0.13

4 0
3 years ago
We have a 10kg ball and a 15kg ball. We drop them from the top of the science building while objects are at rest and there is no
Mashcka [7]
They will hit the ground at the same time, as mass is negligible when calculating the acceleration of gravity when there is no air resistance
3 0
3 years ago
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