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3 years ago

If you hold a gun for the rest of your life do you a cyborg

1 answer:

4 0

No, I think it means you are extremely attached to the weapon. A cyborg, would be a person made of technology or some sort of relative definition.

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Discuss why the article says the fact that we’re in our universe complicates our understanding of the expansion of the universe.

MArishka [77]

Because of our perception of the universe from inside the universe, we are unable to see how and towards what the universe is expanding. Also, our understanding of it is further complicated because we are moving as part of the expansion, thus distorting our perception of it.

4 0

3 years ago

Read 2 more answers

Allison is looking through a telescope at an object in space. The object looks like a very small planet, and it does not have a

Verizon [17]

Answer:

Allison is probably looking at the asteroid.

Explanation:

Asteroids are the giant gas balls of inner solar system that are present in the space.
Asteroids orbits around the sun in very irregular ans strange path unlike the other planets.
This is also the reason why it's called as the minor planets.
It could have been comet but if it had been a comets he must have seen the tail but asteroids do not have any tail like structure.

8 0

3 years ago

We have two solenoids: solenoid 2 has twice the diameter, half the length, and twice as many turns as solenoid 1. The current in

leva [86]

Answer:

the field at the center of solenoid 2 is 12 times the field at the center of solenoid 1.

Explanation:

Recall that the field inside a solenoid of length L, N turns, and a circulating current I, is given by the formula:

$B=\mu_0\, \frac{N}{L} I$

Then, if we assign the subindex "1" to the quantities that define the magnetic field ( $B_1$ ) inside solenoid 1, we have:

$B_1=\mu_0\, \frac{N_1}{L_1} I_1$

notice that there is no dependence on the diameter of the solenoid for this formula.

Now, if we write a similar formula for solenoid 2, given that it has :

1) half the length of solenoid 1 . Then $L_2=L_1/2$

2) twice as many turns as solenoid 1. Then $N_2=2\,N_1$

3) three times the current of solenoid 1. Then $I_2=3\,I_1$

we obtain:

$B_2=\mu_0\, \frac{N_2}{L_2} I_2\\B_2=\mu_0\, \frac{2\,N_1}{L_1/2} 3\,I_1\\B_2=\mu_0\, 12\,\frac{N_1}{L_1} I_1\\B_2=12\,B_1$

5 0

4 years ago

GETTING TIMED PLS HELP! Do the runners in the picture above represent kinetic or potential energy? you need to explain why.

prohojiy [21]

Answer:

They represent kinetic energy

Explanation:

Kinetic Energy

A body can do work due to some of its attributes or states. For example, its mass can do work if used to provide energy, if the object is at a certain height respect to some reference level, it can do work when going downwards (potential energy), if the object moves at a certain speed, it can do work when transferring part of its speed to other objects. It's called kinetic energy and is given by

$\displaystyle K=\frac{mv^2}{2}$

Both runners are moving in a horizontal path, thus they have kinetic energy, given by the above equation. If they could jump below ground level, then they will also have potential energy

8 0

3 years ago

Assume the Earth is a ball of perimeter 40, 000 kilometers. There is a building 20 meters tall at point a. A robot with a camera

torisob [31]

Answer:

Approximately 21 km.

Explanation:

Refer to the not-to-scale diagram attached. The circle is the cross-section of the sphere that goes through the center C. Draw a line that connects the top of the building (point B) and the camera on the robot (point D.) Consider: at how many points might the line intersects the outer rim of this circle? There are three possible cases:

No intersection: There's nothing that blocks the camera's view of the top of the building.
Two intersections: The planet blocks the camera's view of the top of the building.
One intersection: The point at which the top of the building appears or disappears.

There's only one such line that goes through the top of the building and intersects the outer rim of the circle only once. That line is a tangent to this circle. In other words, it is perpendicular to the radius of the circle at the point A where it touches the circle.

The camera needs to be on this tangent line when the building starts to disappear. To find the length of the arc that the robot has travelled, start by finding the angle $\angle \mathrm{B\hat{C}D}$ which corresponds to this minor arc.

This angle comes can be split into two parts:

$\angle \mathrm{B\hat{C}D} = \angle \mathrm{B\hat{C}A} + \angle \mathrm{A\hat{C}D}$ .

Also,

$\angle \mathrm{B\hat{A}C} = \angle \mathrm{D\hat{A}C} = 90^{\circ}$ .

The radius of this circle is:

$\displaystyle r = \frac{c}{2\pi} = \rm \frac{4\times 10^{7}\; m}{2\pi}$ .

The lengths of segment DC, AC, BC can all be found:

$\rm DC = \rm \left(1.75 \displaystyle + \frac{4\times 10^{7}\; m}{2\pi}\right)\; m$ ;
$\rm AC = \rm \displaystyle \frac{4\times 10^{7}}{2\pi}\; m$ ;
$\rm BC = \rm \left(20\; m\displaystyle +\frac{4\times 10^{7}}{2\pi} \right)\; m$ .

In the two right triangles $\triangle\mathrm{DAC}$ and $\triangle \rm BAC$ , the value of $\angle \mathrm{B\hat{C}A}$ and $\angle \mathrm{A\hat{C}D}$ can be found using the inverse cosine function:

$\displaystyle \angle \mathrm{B\hat{C}A} = \cos^{-1}{\rm \frac{AC}{BC}}$

$\displaystyle \angle \mathrm{D\hat{C}A} = \cos^{-1}{\rm \frac{AC}{DC}}$

$\displaystyle \angle \mathrm{B\hat{C}D} = \cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}$ .

The length of the minor arc will be:

$\displaystyle r \theta = \frac{4\times 10^{7}\; \rm m}{2\pi} \cdot (\cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}) \approx 20667 \; m \approx 21 \; km$ .

5 0

3 years ago

If you hold a gun for the rest of your life do you a cyborg​

If you hold a gun for the rest of your life do you a cyborg