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Lesechka [4]
3 years ago
5

A 10,000-watt radio station transmits at 535 kHz. Determine the number of joules transmitted per second. 10,000 J/s 10 J/s 535 J

/s 535,000 J/
Physics
1 answer:
REY [17]3 years ago
5 0
1 nanowatt  =  1 nanojoule/sec
1 watt  =  1 joule/sec
10 watts  =  10 joules/sec
100 watts  =  100 joules/sec
742.914 watts  =  742.914 joules/sec
1,000 watts  =  1,000 joules/sec
10,000 watts  =  10,000 joules/sec
100,000 watts  =  100,000 joules/sec
1 megawatt  =  1 megajoule/sec
1 gigawatt  =  1 gigajoule/sec
1 petawatt  =  1 petajoule/sec

We don't care what frequency the transmission is using,
or who their morning DJ is.

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In the diagram, above, marker F is pointing to a __________, which are formed when meanders wear away at a narrow point and a po
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A constant friction force of 25 N acts on a 65-kg skier for 15 s on level snow. What is the skier’s change in velocity
nikdorinn [45]

Answer:

\Delta v=5.77m/s

Explanation:

Newton's 2nd Law relates the net force <em>F</em> on an object of mass <em>m </em>with the acceleration <em>a</em> it experiments by <em>F=ma.</em> In our case the net force is the friction force, since it's the only one the skier is experimenting horizontally and the vertical ones cancel out since he's not moving in that direction. Our acceleration then will be:

a=\frac{F}{m}

Also, acceleration is defined by the change of velocity \Delta v in a given time t, so we have:

a=\frac{\Delta v}{t}

Since we want the change in velocity, <em>mixing both equations</em> we conclude that:

\Delta v=at=\frac{Ft}{m}

Which for our values means:

\Delta v=\frac{Ft}{m}=\frac{(25N)(15s)}{(65Kg)}=5.77m/s

4 0
4 years ago
1.
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Answer:

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8 0
3 years ago
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What is 3/4 of 12 and 24
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Answer:

3/4 of 12 = 16

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Those are the answers based on how your question sounded

Explanation:

6 0
4 years ago
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is to the east. (i) Automobile A travels 50 km due east. (ii) Automobile B travels 50 km due west. (iii) Automobile C travels 60
kenny6666 [7]

Complete Question:

Each of the following trips lasts exactly one hour. The positive direction is to the east.

(i) Automobile A travels 50 km due east. (ii) Automobile B travels 50 km due west. (iii) Automobile C travels 60 km due east, then turns around and travels 10 km due west. (iv) Automobile D travels 70 km due east. (v) Automobile E travels 20 km due west, then turns around and travels 20 km due east. (a) Rank the five trips in order of average x-velocity from most positive to most negative. (b) Which trips, if any, have the same average x-velocity? (c) For which trip, if any, is the average x-velocity equal to zero?

Answer:

a) 1. (iv) 2.) (i) and (iii). 3. (v) 4.(ii)

Explanation:

By definition, the average velocity, can be expressed as the total displacement (the difference between the final and initial position), divided by the time interval during which we define the displacement.

We can put these words in an equation format as follows:

vavg,x = \frac{xf-xo}{t-to}

If we choose x₀ = 0 and t₀ = 0, the average velocity is just the final position over one hour.

The final positions, in the five trips, are as follows:

(i) = +50 Km (2nd)

(ii) = -50 Km (4th)

(iii)  + 60 Km -10 Km = + 50 km (2nd)

(iv) = + 70 km (1st)

(v) = -20 km + 20 Km = 0 (3rd)

b) As stated above, the trips (i) and (iii) have the same average x-velocity.

c) As stated above, in trip (v) as the total displacement is 0, average -velocity is 0 also.

5 0
4 years ago
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