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Elanso [62]
2 years ago
15

A certain corner of a room is selected as the origin of a rectangular coordinate system. If a fly is crawling on an adjacent wal

l at a point having coordinates (3.1, 0.5), where the units are meters, what is the distance of the fly from the corner of the room? Answer needs to be in appropriate significant figures.
Physics
1 answer:
asambeis [7]2 years ago
5 0

Answer:

The distance is 3.1 m

Explanation:

The position vector of the fly relative to the corner of the wall is

r = (3.1, 0.5).

The distance of the fly from the corner will be calculated as the magnitude of the vector "r"

magnitude of vector  r = \sqrt{(3.1 m)^{2} + (0.5 m)^{2}} = 3.1 m

Since the numbers to be added have only one decimal place 3.<u>1</u> and 0.<u>5</u>, the result of the sum will have to have one decimal place. The result of the square root will also have one decimal place.

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How are energy, work ang power related?
jolli1 [7]

Answer:

Work is the change in energy and power is the rate of doing work.

5 0
3 years ago
Which of the following frequencies falls in the range of RF waves used by commercial radio broadcasting stations?
erma4kov [3.2K]

A).  600,000 Hz  or  600 KHz
Yes.  Commercial broadcasters operate here.
This is the '600' on your AM radio dial.

B).  60 Hz
No.  In principle, this frequency might be used for communication or
commercial broadcasting, but it suffers from two inconvenient truths:
-- An efficient antenna for 60 Hz ... either transmitting or receiving ...
needs to be almost 780 miles long.
-- This is the frequency of the electric power utility in the US and
Canada, so every outlet, wire, cable, lamp cord, and electric line
on a pole RADIATES a little bit of signal at this frequency.  That's
an awful lot of interference.

C).  6,000,000 Hz  or  6 MHz
There's a lot of broadcasting activity here, but it's not commercial
music, news, and sports into local homes and cars. 
It's foreign short-wave broadcast, bringing news, propaganda, and
culture from one country to another.  Pretty interesting to browse.

D).  6,000 Hz  or  6 KHz.
No.  Not used for communication, for an interesting reason:
This frequency is smack in the middle of the human hearing range.
So if it were used for communication ... with high-power transmitters
here and there ... then you wouldn't hear it in the air.  But wherever
wires were being used to carry sound ... your stereo's speaker wires,
wires from your player to your ear-buds, wires to the telephones in
your house etc ... the wires would act as antennas, picking up 
broadcasts at 6 KHz, and the broadcasts would get into everything.
Not a smart plan.

7 0
3 years ago
Read 2 more answers
If a train is going 60 m/s hits the brakes, and it takes the train 1 minute 25 seconds to stop, what is the train’s acceleration
Cloud [144]
Hey there, the answer is ..............................  About 0.7 m/sec^2
<span>
Acceleration is the change in speed / time </span>

<span>Change in speed is 60 m/sec </span>

<span>Time is 1 minute 25 second. Convert that to seconds. </span>

<span>Divide the change in speed by the time in seconds.

About 0.7 m/sec^2

 </span><span>So the acceleration is - 60 / 85 = - 0.71 m/s^2 

HOPE I HELPED!!!!!!!!!!
</span>
8 0
3 years ago
When you drop an object from a certain height, it takes time T to reach the ground with no air resistance. If you dropped it fro
alexandr402 [8]

Answer:

Explanation:

Given

When we drop an object from height , suppose h

it takes time T

using equation of motion

h=ut+\frac{1}{2}at^2

where  

h=displacement

u=initial\ velocity

a=acceleration

t=time

here u=0 because it dropped from a certain height

h=\frac{1}{2}gT^2

T=\sqrt{\frac{2h}{g}}

When height is increases to three times of original height

i.e. h'=3 h

then time period becomes

T'=\sqrt{\frac{2\times 3h}{g}}

T'=\sqrt{3}T

5 0
3 years ago
When the distance between two masses is doubled, the gravitational attraction between
mrs_skeptik [129]

The new gravitational attraction will be 1/4 as much

Explanation:

The magnitude of the gravitational force between two objects is given by

F=G\frac{m_1 m_2}{r^2}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

In this problem, the original force between the two objects is F, when they are separated by a distance r.

Later, the distance between the two objects is doubled, so the new distance is

r'=2r

Therefore, the new force will be

F'=G\frac{m_1 m_2}{(2r)^2}=\frac{1}{4}(\frac{Gm_1 m_2}{r^2})=\frac{F}{4}

Therefore, the new force will be one-fourth as much.

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

5 0
3 years ago
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