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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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In a plate glass factory, sheets of glass move along a conveyor belt at a speed of 15.0 cm/s. An automatic cutting tool descends
Leni [432]

Answer:

angle at which the cutter should be set = 32°

Explanation:

We are given;

Speed of conveyor belt; v1 = 15 cm/s

Width;w = 72 cm

Speed of the cutter that moves across the width; v2 = 24 cm/s

Let's calculate time;

Since the width and the rate at which the cutter works per seconds is given, thus;

Time = w/v2

Time = 72/24 = 3 seconds

Now, we know that, velocity = distance/time

Thus, distance = velocity x time

So, distance covered;d = 15 x 3 = 45 cm

Now, for us to calculate the angle the cutter should be set, we would treat it as a triangle.

When the cutter cuts it, it covers a distance d which is perpendicular to the width.

Thus, to calculate the angle made with the width, we'll use trigonometric ratio.

Thus, d/w = tan θ where θ is the angle at which the cutter should be set.

So, 45/72 = tan θ

tan θ = 0.625

θ = tan^(-1)0.625

θ = 32°

3 0
3 years ago
If an object falling freely were somehow equipped with an odometer to measure the distance it travels, then the amount of distan
ioda

Answer:c

Explanation:

Given

object is falling Freely with an odometer

Suppose it falls with zero initial velocity

so distance fallen in time t is given by

h=ut+\frac{1}{2}gt^2

here u=0 and t=time taken

h=\frac{1}{2}gt^2

for t=1 s

h_1=\frac{1}{2}g

for t=2 s

h_2=\frac{1}{2}g(2)^2=\frac{4}{2}g=2g

distance traveled in 2 nd sec=2g-\frac{1}{2}g=\frac{3}{2}g

for t=3 s

h_3=\frac{1}{2}g(3)^2=\frac{9}{2}g

distance traveled in 3 rd sec=\frac{9}{2}g-2g=\frac{5}{2}g

so we can see that distance traveled in each successive second is increasing

5 0
2 years ago
Which physical phenomena is responsible for the earth’s sky appearing blue? scattering reflection dispersion refraction
klio [65]

Answer:  Scattering reflection

Sunlight reaches earth's atmosphere and is scattered in all directions by all the gasses and particles in the air. Blue light is seen more than others because it travels as shorter, smaller waves. This is why we see a blue sky most of the time.

Explanation:

7 0
3 years ago
Read 2 more answers
A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite
sladkih [1.3K]

Answer:

The  kinetic energy is KE  =  7.59  *10^{10} \  J

Explanation:

From the question we are told that

       The  radius of the orbit is  r =  2.3 *10^{4} \ km  = 2.3  *10^{7} \ m

       The gravitational force is  F_g  = 6600 \ N

The kinetic energy of the satellite is mathematically represented as

       KE  =  \frac{1}{2} * mv^2

where v is the speed of the satellite which is mathematically represented as

     v  = \sqrt{\frac{G  M}{r^2} }

=>  v^2  =  \frac{GM }{r}

substituting this into the equation

      KE  =  \frac{ 1}{2} *\frac{GMm}{r}

Now the gravitational force of the planet is mathematically represented as

      F_g  = \frac{GMm}{r^2}

Where M is the mass of the planet and  m is the mass of the satellite

 Now looking at the formula for KE we see that we can represent it as

     KE  =  \frac{ 1}{2} *[\frac{GMm}{r^2}] * r

=>    KE  =  \frac{ 1}{2} *F_g * r

substituting values

       KE  =  \frac{ 1}{2} *6600 * 2.3*10^{7}

         KE  =  7.59  *10^{10} \  J

 

7 0
3 years ago
A flywheel with radius of 0.400 mm starts from rest and accelerates with a constant angular acceleration of 0.600 rad/s2rad/s2.
charle [14.2K]

Answer: 0.00024\ m/s^2

Explanation:

Given

Radius of flywheel is r=0.4\ mm

Angular acceleration \alpha=0.6\ rad/s^2

For no change in radius, tangential acceleration is  given as

\Rightarrow a_t=a\lpha \times r

Insert the values

\Rightarrow a_t=0.6\times 0.4\times 10^{-3}\ m/s^2\\\Rightarrow a_t=2.4\times 10^{-4}\ m/s^2\ \text{or}\ 0.00024\ m/s^2

5 0
2 years ago
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