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adell [148]
3 years ago
9

In an electric motor, electrical energy is converted into

Physics
1 answer:
N76 [4]3 years ago
5 0

Answer:

idrk im jus here for the points\lim_{n \to \infty} a_n  \lim_{n \to \infty} a_n  \lim_{n \to \infty} a_n  \lim_{n \to \infty} a_n  \lim_{n \to \infty} a_n  \lim_{n \to \infty} a_n∈ω≠≠

Explanation:

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Suppose you are at the center of a large freely-rotating horizontal turntable in a carnival funhouse. As you crawl toward the ed
I am Lyosha [343]

Answer:

c. remains the same, but the RPMs decrease.

Explanation:

Because there aren't external torques on the system composed by the person and the turntable it follows that total angular momentum (I) is conserved, that means the total angular momentum is a constant:

\overrightarrow{L}=constant

The total angular momentum is the sum of the individual angular momenta, in our case we should sum the angular momentum of the turntable and the angular momentum of a point mass respect the center of the turntable (the person)

\overrightarrow{L_{turnatble}}+\overrightarrow{L_{person}}=constant (1)

The angular momentum of the turntable is:

\overrightarrow{L_{turnatble}}=I\overrightarrow{\omega} (2)

with I the moment of inertia and ω the angular velocity.

The angular momentum of the person respects the center of the turntable is:

\overrightarrow{L_{person}}=\overrightarrow{r}\times m\overrightarrow{v} (3)

with r the position of the person respects the center of the turntable, m the mass of the person and v the linear velocity

Using the fact v=\omega r:

\overrightarrow{L_{person}}=\overrightarrow{r}\times rm\overrightarrow{\omega}(3)

By (3) and (2) on (1) and working only the magnitudes (it's all that we need for this problem):

I\omega+r^{2}m\omega=constant

\omega(I+r^{2}m)=constant

Because the equality should be maintained, if we increase the distance between the person and the center of the turntable (r), the angular velocity should decrease to maintain the same constant value because I and m are constants, so the RPM's (unit of angular velocity) are going to decrease.

4 0
3 years ago
Which of the following graphs shows the relationship between two variables that obey the inverse square law?​
blondinia [14]

Well, if we've been paying attention in class, we already KNOW that the electrostatic force changes as the inverse square of the distance, and the top graph is conveniently labeled "Electrostatic Force".

But if we didn't already know that, we'd have to examine the graphs, and find the one where 'y' changes like 1/x² .

The top graph does that.  After 1 unit of time, the force is 350.  Double the time to 2 units, and the force should drop to 1/4 of 350 ... sure enough, it's a little less than 90.  Double the time again, to 4 units, and it should drop to 1/4 of a little less than 90 ... by golly, it's down below 30.

The first graph is what an inverse square looks like.  Now that you've worked out this graph, you'll know an inverse square relationship whenever you see it.  

8 0
3 years ago
The vacuum tube was replaced with what electrical component?
Rufina [12.5K]

Answer:

The vacuum tube was replaced with transistor.

Explanation:

  • The invention of semiconductor was very useful in making solid state transistor that allowed the production of small yet faster, cheaper, and more trusted and reliable computers.
  • These solid state transistor is so often used that, it nearly replaced all the use of transistor.
  • This replacement took place after the invention of semiconductor in the year around 1940.
  • Vacuum tubes also known as thermionic tubes are not used anymore in computers and electronics.  
7 0
3 years ago
Why is hip-hop & Zumba dance exercises so popular? does dancing make you a better person?
Studentka2010 [4]

Answer:

not realy

Explanation:

dancing is a way of exercise and liking a song or you're just happy etc.

7 0
2 years ago
How long must a 0.70-mm-diameter aluminum wire be to have a 0.40 a current when connected to the terminals of a 1.5 v flashlight
Natalka [10]
By using Ohm's law, we can find what should be the resistance of the wire, R:
R= \frac{V}{I}= \frac{1.5 V}{0.40 A} =3.75 \Omega

Then, let's find the cross-sectional area of the wire. Its radius is half the diameter,
r=35 mm=0.35 \cdot 10^{-3} m
So the area is
A=\pi r^2 = \pi (0.35 \cdot 10^{-3} m)^2=3.85 \cdot 10^{-7} m^2

And by using the resistivity  of the Aluminum, \rho=2.65 \cdot 10^{-8} \Omega m, we can use the relationship between resistance R and resistivity:
R= \frac{\rho L}{A}
to find L, the length of the wire:
L= \frac{RA}{\rho}= \frac{(3.75 \Omega)(3,85 \cdot 10^{-7} m^2)}{2.65 \cdot 10^{-8} \Omega m}=54.48 m
4 0
3 years ago
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