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allochka39001 [22]
3 years ago
13

According to the nebular theory of solar system formation, what key difference in their early formation explains why the jovian

planets ended up so different from the terrestrial planets
Physics
1 answer:
svp [43]3 years ago
3 0

Answer:

The Jovian planets formed beyond the Frostline while the terrestrial planets formed in the Frostline in the solar nebular

Explanation:

The Jovian planets are the large planets namely Saturn, Jupiter, Uranus, and Neptune. The terrestrial planets include the Earth, Mercury, Mars, and Venus. According to the nebular theory of solar system formation, the terrestrial planets were formed from silicates and metals. They also had high boiling points which made it possible for them to be located very close to the sun.

The Jovian planets formed beyond the Frostline. This is an area that can support the planets that were made up of icy elements. The large size of the Jovian planets is as a result of the fact that the icy elements were more in number than the metal components of the terrestrial planets.

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When a container decreases in size, what will happen to an amount of gas in the container??
alexgriva [62]
The gas would also decrease in size since the container lost gas to decrease the size of the container.
7 0
3 years ago
When attempting to determine the coefficient of kinetic friction, why is it necessary to move the block with constant velocity
Readme [11.4K]

Answer:

Explanation:

In order to measure the coefficient of friction , we apply external force to move the body . When external force comes in motion , we adjust the external force so that it moves with zero acceleration or uniform velocity . In this case external force becomes equal to kinetic frictional force and then net force becomes zero because

net force = mass x acceleration = m x 0 = 0

Now frictional force = μ mg where μ is coefficient of kinetic friction

so F = μ mg where F is external force applied

μ = F / mg

Hence , to make external force equal to frictional force , it is necessary to make acceleration of body zero .

4 0
3 years ago
a red ball moves horizontally in a 30 m long tube what is the displacement of the red ball between 0s and 24s?
gayaneshka [121]

Answer:

30 metres.

Explanation:

Given that a red ball moves horizontally in a 30 m long tube.

Displacement is the distance travelled in a specific direction. It has both magnitude and direction.

Since the motion is horizontal, it moves is a certain direction.

Within the stipulation of time, the displacement will be the distance covered in the horizontal direction which is 30 metres.

Therefore, the displacement of the motion of the red ball is 30 metres.

7 0
3 years ago
M=3000km v=25m/s what’s the momentum
valina [46]
<h3>Answer:</h3>

Momentum of the given body will be : 75000 Kg m/s

<h3>Explanation:</h3>

According to Newton's first law of motion, all bodies continue to be in the state of rest or motion unless an external force is applied on the body. We can use this in the case of momentum also

The formula of momentum is given by :

:\implies \sf\quad \sf \:  P = mv

Here, we are given the mass of the body ( m ) as 3000kg and the velocity of the body ( v ) as 25 m/s. On putting the values in the formula:

\begin{aligned}&:\implies \sf\quad \sf \:  P = mv \\& :\implies \sf\quad \sf \:  P = 3000 \times 25 \\ & :\implies \sf\quad \sf \:   \boxed{ \sf \: P = 75000kgm {s}^{ - 1} } \end{aligned}

Momentum is associated with the mass of the moving body and can be defined as the quantity of motion measured as a product of mass and velocity.

8 0
2 years ago
A boy whirls a stone in a horizontal circle of radius 1.4 m and at height 1.5 m above ground level. The string breaks, and the s
balandron [24]

Answer:233.23 m/s^2

Explanation:

Given

radius of circle=1.4 m

Height of stone above ground=1.5 m

Horizontal distance(R)=10 m

It is given at the time of break stone flies horizontally thus stone to cover a height of 1.5 m in time t before reaching ground

1.5=0+\frac{gt^2}{2}

t=0.55 s

Initial horizontal velocity at the time of break is given by u

R=u\times t

10=u\times 0.55

u=18.07 m/s

Therefore magnitude of centripetal acceleration is given by

a_c=\frac{u^2}{r}=\frac{18.07^2}{1.4}=233.23 m/s^2

6 0
3 years ago
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