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ExtremeBDS [4]
4 years ago
13

The distance between planets varies as they move in their orbits. The minimum distance from the Earth to Mars is about 0.35 AU.

About how many kilometers is that?
Physics
2 answers:
Natalija [7]4 years ago
6 0

Answer:

c on edge

Explanation:

Alina [70]4 years ago
5 0

Answer:

52,360,000km

Explanation:

To solve this problem you use a conversion factor.

By taking into account that 1UA = 1.496*10^{8}km you obtain:

0.35UA*\frac{1.496*10^{8}km}{1UA}=52,360,000km

hence, 0.35UA is about 52,360,000km. This is the least distance between Mars and Earth

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What is the net force exerted by these two charges on a third charge q3 = 49.0nC placed between q1 and q2 at x3 = -1.085m ?Your
densk [106]

Answer:

The net force exerted by these two charges on a third charge is 5.468\times10^{-6}\ N

Explanation:

Given that,

Third charge q_{3}=49.0\ nC

Distancex_{3}=-1.085\ m

Suppose The magnitude of the force F between two particles with charges Q and Q' separated by a distance d. Consider two point charges located on the x axis one charge, q₁ = -12.5 nC , is located at x₁ = -1.650 m, the second charge, q₂ = 31.5 nC , is at the origin.

We need to calculate the total force will be the vector sum of two forces

Using Coulomb's law,

F_{13}=\dfrac{kq_{1}q_{3}}{(x_{1}-x_{3})^2}

Put the value into the formula

F_{13}=\dfrac{9\times10^{9}\times(-12.5\times10^{-9})\times49\times10^{-9}}{(-1.650-(-1.085))^2}

F_{13}=-17268.3\times10^{-9}\ N

We need to calculate the force will be to the negative charge with opposite charges

Using Coulomb's law,

F_{23}=\dfrac{kq_{2}q_{3}}{(x_{2}-x_{3})^2}

Put the value into the formula

F_{23}=\dfrac{9\times10^{9}\times(31.5\times10^{-9})\times49\times10^{-9}}{(-1.085)^2}

F_{23}=11800.2\times10^{-9}\ N

The force also will be to the negative side, charges with same charge sign

We need to calculate the net force exerted by these two charges on a third charge

Using formula of net force

F_{net}=F_{13}+F_{23}

F_{net}=-17268.3\times10^{-9}+11800.2\times10^{-9}

F_{net}=-0.0000054681\ N

F_{net}=-5.468\times10^{-6}\ N

Negative sign shows the negative direction.

Hence, The net force exerted by these two charges on a third charge is 5.468\times10^{-6}\ N

5 0
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Which of these is a good example of Newton's First Law of inertia
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Answer:

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Answer:

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Explanation:

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Given:
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