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Which shows the correct order of forces, from weakest to strongest?

Norma-Jean [14]

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7 0

3 years ago

What force acts between Earth and the moon? gravity mass motion newtons

Serggg [28]

Answer:

Gravity

Explanation:

Mass and motion are not forces and newton is unit of force. Gravity is defined as force of attraction or repulsion between two bodies and hence is the answer.

4 0

3 years ago

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Imagine another solar system, with a star of the same mass as the Sun. Suppose a planet with a mass twice that of Earth (2 Earth

sashaice [31]

Answer:

Time period, T = 403.78 years

Explanation:

It is given that,

Orbital distance, $a=1\ AU=1.496\times 10^{11}\ m$

Mass of the Earth, $m_e=5.972\times 10^{24}\ kg$

Mass of the planet, $m_p=2m_e=11.944\times 10^{24}\ kg$

Let T is the orbital period of this planet. The Kepler's third law of motion gives the relation between the orbital period and the orbital distance.

$T^2=\dfrac{4\pi^2}{Gm_p}a^3$

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 11.944\times 10^{24}}\times (1.496\times 10^{11})^3$

$T=1.28\times 10^{10}\ s$

or

T = 403.78 years

So, the orbital period of this planet is 404 years. Hence, this is the required solution.

4 0

3 years ago

ASK YOUR TEACHER As an astronaut, you observe a small planet to be spherical. After landing on the planet, you set off, walking

Vsevolod [243]

Given Information:

circumference of planet = C = 25.7 km

height of release = h = 1.43 m

time = t = 30 s

Required Information:

Mass of planet = M = ?

Answer:

Mass of planet = M = 7.963×10¹⁴ kg

Explanation:

The mass of the planet can be found using Newton's law of gravitation,

g = GM/R²

M = gR²/G

Where g is the gravitational acceleration at the planet, R is the radius of the planet and G is the gravitational constant.

G = 6.6743×10⁻¹¹ m³/kg⋅s²

First we need to find R and g

The relation between circumference and radius is given by

C = 2πR

Where R is the radius of planet

25.7 = 2πR

R = 25.7/2π

R = 4.09 km

From the equations of motion we have,

h = ut + ½gt²

Where u is the initial speed which is zero, t is the time and h is the height of release.

Re-arrange the above equation for g

h = ½gt²

g = 2h/t²

g = (2*1.43)/(30)²

g = 0.003177 m/s²

Finally, the mass of the planet is

M = gR²/G

M = 0.003177*(4.09×10³)²/6.6743×10⁻¹¹

M = 7.963×10¹⁴ kg

Therefore, the mass of the given planet is 7.963×10¹⁴ kg

3 0

4 years ago

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An uncharged capacitor is connected to the terminals of a 4.0 V battery, and 9.0 μC flows to the positive plate. The 4.0 V batte

Lelechka [254]

Answer:

$2.25\mu C$

Explanation:

At the beginning, we have:

V = 4.0 V potential difference across the capacitor

$Q=9.0 \mu C=9.0\cdot 10^{-6}C$ charge stored on the capacitor

Therefore, we can calculate the capacitance of the capacitor:

$C=\frac{Q}{V}=\frac{9.0 \cdot 10^{-6} C}{4.0 V}=2.25\cdot 10^{-6} F$

Later, the battery is replaced with another battery whose voltage is

V = 5.0 V

Since the capacitance of the capacitor does not change, we can calculate the new charge stored:

$Q=CV=(2.25\cdot 10^{-6} F)(5.0 V)=11.25 \cdot 10^{-6} C=11.25 \mu C$

Since the capacitor has been connected exactly as before, we have that the charge on the positive plate has increased from $9.0 \mu C$ to $11.25 \mu C$ . Therefore, the additional charge that moved to the positive plate is

$\Delta Q = 11.25 \mu C-9.0 \mu C=2.25 \mu C$

5 0

3 years ago

PLEASE HELP, PLEASE A CORRECT ANSWER!​

PLEASE HELP, PLEASE A CORRECT ANSWER!