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navik [9.2K]
3 years ago
10

A long, straight, vertical wire carries a current upward. due east of this wire, in what direction does the magnetic field point

? a long, straight, vertical wire carries a current upward. due east of this wire, in what direction does the magnetic field point? upward west north south downward east
Physics
1 answer:
Solnce55 [7]3 years ago
8 0
(I assume that the 4 directions north-south-east-west are meant with respect to the wire seen from the top.)

We can use the right-hand rule to understand the direction of the magnetic field generated by the wire. The thumb follows the direction of the current in the wire (upward), while  the other fingers give the direction of the field in every point around the wire. Seen from the top, the field has an anti-clockwise direction. Therefore, if we take a point at east with respect to the wire, in this point the field has direction south.
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A 8.8 cm diameter circular loop of wire is in a 1.04 T magnetic field. The loop is removed from the field in 0.30 s . Assume tha
denis23 [38]

Answer:

0.021 V

Explanation:

The average induced emf (E) can be calculated usgin the Faraday's Law:

E = - \frac{N*\Delta \phi}{\Delta t}  

<u>Where:</u>

<em>N = is the number of turns = 1   </em>

<em>ΔΦ = ΔB*A                                            </em>

<em>Δt = is the time = 0.3 s   </em>

<em>A = is the loop of wire area = πr² = πd²/4 </em>

<em>ΔB: is the magnetic field = (0 - 1.04) T                     </em>

Hence the average induced emf is:

E = - \frac{\Delta B*A}{\Delta t} = - \frac{(0- 1.04 T) \pi (0.088 m)^{2}}{4*0.3 s} = 0.021 V                      

Therefore, the average induced emf is 0.021 V.

I hope it helps you!

8 0
3 years ago
How do galaxies vary?
givi [52]

The universe has trillions of galaxies and counting. Astronomers give names to each galaxy base on its shape (e.i, Sombrero galaxy, Milkyway Galazy, ect,.).

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4 0
3 years ago
*please please help* Suppose you are going on a car trip with your family.
Strike441 [17]

Answer:

1. To determine the average speed for the first day of the trip, the total distance traveled would have to be acquired and then how long it took to arrive at the final destination, only including the time that was actually traveled and not any time that was accumulated by any rest stops. Once you have this information, you have to divide the distance over time and you have the average speed (mph).

2. To determine the instantaneous speed, you would just have to look at the speedometer, which tells you at what speed the car is traveling at that exact moment.

Explanation:

I took physics 121 and got the same question. This is my answer that i used and my teacher said it was right.

3 0
3 years ago
Bill leaves his 60 W desk lamp on every day, including weekends, for eight hours. After one month (30 days), how much total ener
maxonik [38]

' W ' is the symbol for 'Watt' ... the unit of power equal to 1 joule/second.

That's all the physics we need to know to answer this question.
The rest is just arithmetic.

(60 joules/sec) · (30 days) · (8 hours/day) · (3600 sec/hour)

= (60 · 30 · 8 · 3600) (joule · day · hour · sec) / (sec · day · hour)

= 51,840,000 joules
__________________________________

Wait a minute !  Hold up !  Hee haw !  Whoa ! 
Excuse me.  That will never do.
I see they want the answer in units of kilowatt-hours (kWh).
In that case, it's

(60 watts) · (30 days) · (8 hours/day) · (1 kW/1,000 watts)

= (60 · 30 · 8 · 1 / 1,000) (watt · day · hour · kW / day · watt)

= 14.4 kW·hour

Rounded to the nearest whole number:

14 kWh

7 0
3 years ago
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To develop this problem it is necessary to apply the concepts related to the Dopler effect.

The equation is defined by

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Where

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v = Emitter speed

And

f_h = f_0 \frac{c}{c+v}

Therefore using the values given we can find the velocity through,

\frac{f_h}{f_0}=\frac{c-v}{c+v}

v = c(\frac{f_h-f_i}{f_h+f_i})

Assuming the ratio above, we can use any f_h and f_i with the ratio 2.4 to 1

v = 353(\frac{2.4-1}{2.4+1})

v = 145.35m/s

Therefore the cars goes to 145.3m/s

7 0
3 years ago
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