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d1i1m1o1n [39]
3 years ago
12

A spaceship is traveling through deep space to a space station and needs to make a course correction to go around a nebula. The

captain orders the ship to travel 2.3 106 kilometers before turning 70° and traveling 1.5 106 kilometers to reach the space station. If the captain had not ordered a course correction, what would have been the magnitude and direction of the path of the spaceship if it had traveled to the space station through the nebula?
Physics
1 answer:
zlopas [31]3 years ago
8 0

Answer:

Explanation:

Using the given values

Let

Px= 2.3 x 10^6, Py=0, Qx=1.5x10^6cos(70),

Qy= 1.5x10^6sin(70)

So let

Hx= Px+Qx=2.813 x 10^6, Hy=Py+Qy=1.401 x 10^6

Magnitude = √ ((2.813 x 10^6)^2+(1.401 x 10^6)^2)= 3.14 x 10^6

To find direction

tan စ = Hy/Hx = 1.401/2.813= 0.4980

စ = 26.5°

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Three equal charges of magnitude 'e' are located at the vertices of an equilateral triangle of side 1m. Where should you place a
antoniya [11.8K]

Answer:

We would have to locate the charge at the center of the triangle.

Explanation:

I tried my best to make geometric arguments, you can see the attached picture if it helps.

Let's say the triangle has three corners names 1, 2 and 3 as seen in the image. Let's determine the electric field at the center of the triangle. We can see that the x-component of electric field produced by the charge at 1 and the one produce by the charge at 2 will cancel each other out because they have the same but opposite values. The y components on the other hand are equal and are added. The magnitude of all three electric fields are the same (because each corner is at equal distance from the center and all three charges are equal, see the formula for the electric field: |E|=\frac{1}{4.\pi.\epsilon_0}\frac{q}{d^2} where d is the distance between the point of observation and the charge). Thus lets call E the magnitude of the electric field produced by any of the charges.

We can clearly see that the y-component of the electric filed produced by 1 is E\sin(30), the same goes for the one produced by the charge at 2, knowing that \sin(30)=\frac{1}{2}, we have that the total field due to the charges at 1 and 2 are equal to E in the positive y-direction. The field produced by the third charge is clearly -E in the negative y-direction thus canceling the contribution from 1 and 2. The total electric field at the center of the triangle is zero, this also mean that any charge placed at that point will not feel any force.

4 0
3 years ago
two charges attract each other with a force of 1.2 * 10⁻⁶N. Calculate the force that would act between the two charges when dist
Likurg_2 [28]

1. When the distance is doubled: 3\cdot 10^{-7}N

The electrostatic force between two charges is given by:

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the distance between the two charges

The initial force between the two charges is F=1.2\cdot 10^{-6}N. In this part of the problem, the distance between the two charges is doubled, so we can write

r'=2r

And substituting into the formula, we find the new force:

F'=k\frac{q_1 q_2}{r'^2}=k\frac{q_1 q_2}{(2r)^2}=\frac{1}{4}k\frac{q_1 q_2}{r^2}=\frac{F}{4}

So, the force is reduced to 1/4 of its original value. Therefore, it is

F'=\frac{1.2\cdot 10^{-6} N}{4}=3\cdot 10^{-7}N

2. When the distance is halved: 4.8\cdot 10^{-6}N

The initial force between the two charges is F=1.2\cdot 10^{-6}N. In this part of the problem, the distance between the two charges is halved, so we can write

r'=r/2

And substituting into the formula, we find the new force:

F'=k\frac{q_1 q_2}{r'^2}=k\frac{q_1 q_2}{(r/2)^2}=4k\frac{q_1 q_2}{r^2}=4F

So, the force is quadrupled. Therefore, it is

F'=4(1.2\cdot 10^{-6} N)=4.8\cdot 10^{-6}N

3. When the distance is tripled: 1.33\cdot 10^{-7}N

The initial force between the two charges is F=1.2\cdot 10^{-6}N. In this part of the problem, the distance between the two charges is tripled, so we can write

r'=3r

And substituting into the formula, we find the new force:

F'=k\frac{q_1 q_2}{r'^2}=k\frac{q_1 q_2}{(3r)^2}=\frac{1}{9}k\frac{q_1 q_2}{r^2}=\frac{F}{9}

So, the force is reduced to 1/9 of its original value. Therefore, it is

F'=\frac{1.2\cdot 10^{-6} N}{9}=1.33\cdot 10^{-7}N

4 0
4 years ago
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Solnce55 [7]

Answer:

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the energy of the photons is not enough to carry out an electronic transition between two states of the material, when we decrease the wavelength (the energy of the photons increases), the point is reached where the energy of the beam is equal to some energy of a transition, by which the electrons are promoted and since we can see a certain charge, as the atoms are neutral, some electrons must be removed from the material, this is represented in the macroscopic case as the work function of the material, consequently a unbalanced load that is what we can measure.

When we increase the lightning intensity, what we do is that we increase the number of photons and if each photon can remove an electron, by removing the electrons the difference between it and the positive charge (fixed in the nuclei) increases.

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ivanzaharov [21]
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4 0
3 years ago
Two cars are sitting still. Car A has a mass of 2000 kilograms. Car B has a mass of 4000 kg. If the goal is to push both cars to
astra-53 [7]
Since car B has twice as much mass as car A has, it would require 2x as much force that it would need to push car A down.

If that was kinda confusing lemme say it again,

It would take twice as much that would be needed to push car B down than car A

Since everything is the same, basically two people would need to push down car B and one person for car A

Hope this helps
7 0
3 years ago
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