What is the name of the National Outreach Project in FCCLA?

1 answer:

4 0

Answer:10. Describe Career Connection. Career Connection is an FCCLA national program that guides students to link their options and skills for success in families, careers and communities.

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How is air pressure related to elevation?

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Air pressure decreases with altitude

hope that helps you!

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Why do you often freeze after a shower? Help mee

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When water evaporates of your body it cools you down, hence why after a shower when the water droplets begin to evaporate you feel cold

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You know your mass is 62 kg but when you stand on a bathroom scale in an elevator it says your mass is 77 kg what is the acceler

lbvjy [14]

In a stationary situation, the weight of person is
$W=mg=(62 kg)(9.81 m/s^2) = 608.2 N$
This is the weight "felt" by the scale, which is basically the normal reaction applied by the scale on the person, and which uses the value of g (9.81) as reference to convert the weight (602.8 N) into a mass (62 kg).

When the person is in the elevator, the scale says 77 kg. The scale is still using the same value of conversion (9.81), so the apparent weight "felt" by the scale is
$W' = m'g=(77 kg)(9.81 m/s^2)=755.4 N$
This is the normal reaction applied by the scale on the person, and which is directed upward. Besides this force, there is still the weight W of the person, acting downward. So, if we use Newton's second law:
$\sum F = ma$
$W-W'=ma$
where a is the acceleration of the elevator. If we solve for a, we find
$a= \frac{W-W'}{m}= \frac{608.2N-755.4N}{62 kg}=-2.37 m/s^2$
The negative sign means the acceleration is in the opposite direction of g (which we take positive), so it means the elevator is going upward.

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3 years ago

Which rf characteristic best determines the range of a 2.4 ghz ism signal?

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The radio frequency characteristic that best determines the range of a 2.4 GHz ism signal is the wavelength.

This frequency can be used in WiFi and can reach up to 46 meters when indoors and about 92 meters when outdoors.

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3 years ago

Two power lines run parallel for a distance of 222 m and are separated by a distance of 40.0 cm. if the current in each of the t

earnstyle [38]

1) Magnitude of the force:

The magnetic field generated by a current-carrying wire is
$B= \frac{\mu_0I}{2 \pi r}$
where
$\mu_0$ is the vacuum permeability
I is the current in the wire
r is the distance at which the field is calculated

Using I=135 A, the current flowing in each wire, we can calculate the magnetic field generated by each wire at distance
$r=40.0 cm=0.40 m$ ,
which is the distance at which the other wire is located:
$B= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} N/A^2)(135 A) }{2 \pi (0.40 m)}=6.75 \cdot 10^{-5} T$

Then we can calculate the magnitude of the force exerted on each wire by this magnetic field, which is given by:
$F=ILB=(135 A)(222 m)(6.75 \cdot 10^{-5}T)=2.03 N$

2) direction of the force:
The two currents run in opposite direction: this means that the force between them is repulsive. This can be determined by using the right hand rule. Let's apply it to one of the two wires, assuming they are in the horizontal plane, and assuming that the current in the wire on the right is directed northwards:
- the magnetic field produced by the wire on the left at the location of the wire on the right is directed upward (the thumb of the right hand is directed as the current, due south, and the other fingers give the direction of the magnetic field, upward)

Now let's apply the right-hand rule to the wire on the right:
- index finger: current --> northward
- middle finger: magnetic field --> upward
- thumb: force --> due east --> so the force is repulsive

A similar procedure can be used on the wire on the left, finding that the force exerted on it is directed westwards, so the force between the two wires is repulsive.

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3 years ago