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slava [35]
2 years ago
6

E-mail is usually a poor choice for the distribution of sensitive electronic files because __________. A. Paper copies seem more

official than e-mail B. Not all computers can read e-mails correctly C. It is easy to send e-mail to the wrong person D. E-mail is saved on the server that transfers it Please select the best answer from the choices provided A B C D.
Physics
1 answer:
iragen [17]2 years ago
3 0

Answer:  D.) e-mail is saved on the server that transfers it

Explanation:

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A car is traveling at 40 m/s for 20 seconds. How far did it travel in this time?
barxatty [35]
<h3>Answer:</h3>

800 meters

<h3>Explanation;</h3>

<u>We are given;</u>

  • Speed as 40 m/s
  • Time as 20 seconds

We are required to determine the distance traveled

  • Speed refers to the rate of change in distance.
  • It is given by;

Speed = Distance ÷ time

Rearranging the formula;

Distance = speed × time

In this case;

Distance = 40 m/s × 20 sec

              = 800 meters

Thus, the distance traveled by the car is 800 m

7 0
2 years ago
Andre the Giant and Thumbelina are standing in the middle of an ice-skating rink facing each other. Andre the Giant pushes Thumb
IceJOKER [234]
I don't know for sure but i think that it is because the force him pushing her pushed him back as well

Hope i helped

3 0
3 years ago
Read 2 more answers
Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and ex
Ahat [919]

The question is incomplete. The complete question is :

Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s.

What is the frequency of the sound?

Solution :

Given :

The distance between the two loud speakers, d = 1.8 \ m

The speaker are in phase and so the path difference is zero constructive interference occurs.

At the point D, the speakers are out of phase and so the path difference is $=\frac{\lambda}{2}$

Therefore,

$AD-BD = \frac{\lambda}{2}

$\sqrt{(1.8)^2+(3)^2-3} =\frac{\lambda}{2}$

$\lambda = 2 \times 0.4985$

$\lambda = 0.99714 \ m$

Thus the frequency is :

$f=\frac{v}{\lambda}$

$f=\frac{340}{0.99714}$

f=340.9744 Hz

3 0
2 years ago
Ls are used to analyze large unstructured data sets, such as e-mail, memos, survey responses, etc., to discover patterns and rel
Kruka [31]
The word that could fit to this is Text Mining. This word also referred to as text data mining, roughly equivalent to text analytics, a process of deriving high-quality information from text. This high quality information derived through the devising of patterns and trends through means such as statistical patter learning. 
6 0
3 years ago
Why is it not necessary for radio telescope surfaces to be as smooth as a mirror?
Law Incorporation [45]
It doesn't on account of radio waves are longer than optical waves. Radio waves are a sort of electromagnetic radiation with wavelengths in the electromagnetic range longer than infrared light. These long waves are in the radio locale of the electromagnetic range.
8 0
3 years ago
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