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3 years ago

Choose the list below that properly arranges objects in the Solar System (from closest to farthest from the center). Select one:

Physics

2 answers:

valina [46]3 years ago

4 0

Answer:

Option a.

Explanation:

The correct sequence considering closest to farthest distance from the center, i.e., sun of the solar system is:

Starting from the Sun, the next comes the inner planet or what we call the terrestrial planet of our solar system which are mainly composed of rocks and mainly silicate rocks and metals. Some examples of such planets are: mercury, Earth, Mars and Venus.

After that comes the asteroid belt (a disc full of asteroids) which is in between the Jupiter and Mars separating Terrestrial and Jovian planets.

This is then followed by Jovian planets which are composed oh helium and Hydrogen and are known as gas Giants.

Jupiter, Uranus, Saturn and Neptune comes under the Jovian planets and then Kuiper belt(consists of the remains of the early history of our solar system) follows originating at Neptune's orbit and stretching beyond Pluto.

mr Goodwill [35]3 years ago

3 0

Answer:

Choose the list below that properly arranges objects in the Solar System (from closest to farthest from the center). Select one:

a. Sun, Terrestrial Planets, Asteroid Belt, Jovian Planets, Kuiper Belt ✓

b. Oort Cloud, Asteroid Belt, Terrestrial Planets, Jovian Planets

c. Sun, Kuiper Belt, Terrestrial Planets, Jovian Planets, Asteroid Belt

d. Sun, Terrestrial Planets, Kuiper Belt, Asteroid Belt, Jovian Planets

Hope this helps

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A 25,000 kg traveling east collides with a 2,000 kg truck standing still on the tracks. After the collision the train and truck

Elis [28]

Answer:

24.084 m/s

Explanation:

From the law of conservation of linear momentum

Total momentum before collision equals to the total momentum after collision

Since momentum=mv where m is mass and v is velocity

$M_{truck}V_{truck}=V_{common}*(M_{truck} +M_{standing})$ where $M_{truck}$ is the mass of the truck, $V_{truck}$ is velocity of the truck, $V_{common}$ is the common velocity of moving and standing truck after collision and $M_{standing}$ is the mass of the standing truck

Making $V_{truck}$ the subject we obtain

$V_{truck}=\frac { V_{common}*(M_{truck} +M_{standing})}{M_{truck}}$

Substituting $M_{truck}$ as 25000 Kg, $V_{common}$ as 22.3 m/s, $M_{standing}$ as 2000 Kg we obtain

$V_{truck}=\frac { 22.3 m/s *(25000 Kg +2000 Kg)}{25000}= 24.084 m/s$

Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive

The truck was moving at 24.084 m/s

3 0

3 years ago

An AC adapter for a telephone answering machine uses a transformer to reduce the line voltage of 120 V to a voltage of 5.00 V. T

Maksim231197 [3]

Answer:

Explanation:

We are given that

Initial voltage, $V_1=120 V$

Final voltage, $V_2=5 V$

Number of tuns in primary coil of the transformer, $N_p=840$

Rms current, $I_{rms}=580mA=580\times 10^{-3} A$

$1 mA=10^{-3} A$

We have to find the number of turns are there on the secondary coil.

We know that

$\frac{N_s}{N_p}=\frac{V_2}{V_1}$

Using the formula

$\frac{N_s}{840}=\frac{5}{120}$

$N_s=\frac{5}{120}\times 840=35$

Hence, there are number of turns on the secondary coil=35

8 0

3 years ago

An electron moving parallel to a uniform electric field increases its speed from 2.0 × 10^7 m/s to 4.0 × 10^7 m/s over a distanc

julia-pushkina [17]

Answer:

$E = 2.84 * 10^5 N/C$

Explanation:

The speed increased from 2.0 * 10^7 m/s to 4.0 * 10^7 m/s over a 1.2 cm distance.

Let us find the acceleration:

$v^2 = u^2 + 2as$

$(4.0 * 10^7)^2 = (2.0 * 10^7)^2 + 2 * a * 0.012\\\\(4.0 * 10^7)^2 - (2.0 * 10^7)^2 = 0.024a\\\\1.2 * 10^{15}= 0.024a\\\\a = 1.2 * 10^{15} / 0.024\\\\a = 5 * 10^{16} m/s^2$

Electric force is given as the product of charge and electric field strength:

F = qE

where q = electric charge

E = Electric field strength

Force is generally given as:

F = ma

where m = mass

a = acceleration

Equating both:

ma = qE

E = ma / q

For an electron:

m = 9.11 × 10^{-31} kg

q = 1.602 × 10^{-19} C

Therefore, the electric field strength of the electron is:

$E = \frac{9.11 * 10^{-31} * 5 * 10^{16}}{1.602 * 10^{-19}} \\\\E = 2.84 * 10^5 N/C$

8 0

3 years ago

In what direction does an applied force move an object

mr Goodwill [35]

In the direction the force is directed towards. If the boy kicks the ball to the right, the ball will roll to the right.

4 0

3 years ago

Read 2 more answers

What is the name of the part of the wave that is labeled -

tresset_1 [31]

Answer:

amplitude

Explanation:

sound travel and it amplifies

7 0

3 years ago