Why do we use atomic models?

Every few hundred years most of the planets line up on the same side of the Sun.(Figure 1)Calculate the total force on the Earth

mylen [45]

Answer: $3.7 \times 10^{-4} N$

Explanation:

The gravitational pull between two object is given by:

$F = G\frac{Mm}{r^2}$

Where M and m are the masses of the object, r is the distance between the masses and G = 6.67× 10⁻¹¹ m³kg⁻¹ s⁻² is the gravitational constant.

We have to calculate the net force on Earth due to Venus, Jupiter and Saturn when they are in one line. It means when they are the closest distance.

$F_{net] = G\frac{M_eM_v}{r_v^2}+G\frac{M_eM_j}{r_j^2}+G\frac{M_eM_s}{r_s^2}$

Mass of Earth, Me = 5.98 × 10²⁴ kg

Mass of Venus, Mv = 0.815 Me

Mass of Jupiter, Mj = 318 Me

Mass of Saturn, Ms = 95.1 Me

closest distance between Earth and Venus, rv = 38 × 10⁶ km = 0.25 AU

closest distance between Jupiter and Earth, rj = 588 × 10⁶ km = 3.93 AU

closest distance between Earth and Saturn, rs = 1.2 × 10⁹ km = 8.0 AU

where 1 AU = 1.5 × 10¹¹ m

Inserting the values:

$F_{net} = G\frac{M_e\times 0.815 M_e}{(0.25AU)^2}+G\frac{M_e\times 318 M_e}{(3.93AU)^2}+G\frac{M_e\times 95.1 M_e}{(8.0AU)^2}\\ \Rightarrow F_{net} = \frac{(GM_e^2)}{(1AU)^2}(\frac{0.815}{0.25^2}+\frac{318}{3.93^2}+\frac{95.1}{8.0^2})=\frac{6.67\times 10^{-11} \times (5.98\times 10^{24})^2}{(1.5\times 10^{11})^2}(35.1) = 3.7 \times 10^{-4} N$

4 0

3 years ago

Read 2 more answers

Two ice skaters, each with a mass of 72.0 kg, are skating at 5.45 m/s when they collide and stick together. If the angle between

Rudiy27

Answer:

The skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.

Explanation:

To solve the problem it is necessary to go back to the theory of conservation of momentum, specifically in relation to the collision of bodies. In this case both have different addresses, consideration that will be understood later.

By definition it is known that the conservation of the moment is given by:

$m_1v_1+m_2v_2=(m_1+m_2)v_f$

Our values are given by,

$m_1=m_2=72Kg$

As the skater 1 run in x direction, there is not component in Y direction. Then,

Skate 1:

$v_{x1}=5.45m/s$

$v_{y1}=0$

Skate 2:

$v_{x2} = 5.45*cos105= -1.41m/s$

$v_{y2} = 5.45*sin105 = 5.26m/s$

Then, if we applying the formula in X direction:

m_1v_{x1}+m_2v_{x2}=(m_1+m_2)v_{fx}

75*5.45-75*1.41=(75+75)v_{fx}

Re-arrange and solving for v_{fx}

v_{fx}=\frac{4.04}{2}

v_{fx}=2.02m/s

Now applying the formula in Y direction:

$m_1v_{y1}+m_2v_{y2}=(m_1+m_2)v_{fy}$

$0+75*5.25=(75+75)v_{fy}$

$v_{fy}=\frac{5.25}{2}$

$v_{fy}=2.63m/s$

Therefore the skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.

6 0

3 years ago

If a train is 100 kilometers away, how much sooner would you hear the train coming by listening to the rails (iron) as opposed t

Whitepunk [10]

From tables, the speed of sound at 0°C is approximately
V₁ = 331 m/s (in air)
V₃ = 5130 m/s (in iron)

Distance traveled is
d = 100 km = 10⁵ m

Time required to travel in air is
t₁ = d/V₁ = 10⁵/331 = 302.12 s

Time required to travel in iron is
t₂ = d/V₂ = 10⁵/5130 = 19.49 s

The difference in time is
302.12 - 19.49 = 282.63 s

Answer: 283 s (nearest second)

6 0

3 years ago

How does Mercury's close proximity to the sun and thin atmosphere affect its ability to maintain liquid water

alekssr [168]

Answer

A thin atmosphere does not supply much oxygen, and the heat from the sun would evaporate it, because mercury is close to the sun.

5 0

3 years ago

The radius of the earth is 6378 km. What is the diameter of the earth in meters?

Nesterboy [21]

To determine the diameter of the earth in metres first multiply the original value by 2.

6378 X 2 = 12 756 km.

Then convert km - m

1 km = 1000 m
12 756 km = ? m

12 756 • 1000 = 12 756 000 = 12 756 000 m or 1.2756 X 10 ^ 7 m

The final solution for the diameter is 1.2756 X 10 ^ 7 m.

7 0

3 years ago