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TEA [102]
3 years ago
12

Consider the magnetic field (B) of a wire with a constant current (I). A Magnetic Field sensor is placed at a radius (r). Will t

he magnitude of the magnetic field change if the sensor was moved around the wire while holding the current and radius constant?
Physics
1 answer:
iren [92.7K]3 years ago
8 0

Answer:

No, the magnitude of the magnetic field won't change.

Explanation:

The magnetic field produced by a wire with a constant current is circular and its flow is given by the right-hand rule. Since this field is circular with center on the wire the magnitude of the magnetic field around the wire will be given by B = [(\mi_0)*I]/(2\pi*r) where (\mi_0) is a constant, I is the current that goes through the conductor and r is the distance from the wire. If the field sensor will move around the wire with a fixed radius the distance from the wire won't change so the magnitude of the field won't change.

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HELP!! AM I CORRECT?? PLS TELL ME, IF U ANSWER PROPERLY I'LL GIVE U BRAINLIEST!!
S_A_V [24]

Answer:ur right I think

Explanation:

3 0
3 years ago
Read 2 more answers
A small town has decided to forego the use of electrical power and send energy through town via mechanical waves on ropes. They
mojhsa [17]

Answer:

the required frequency of waves is 2.066 Hz

Explanation:

Given the data in the question;

μ = 1.50 kg/m

T = 6000 N

Amplitude A = 0.500 m

P = 2.00 kW = 2000 W

we know that, the average power transmit through the rope can be expressed as;

p = \frac{1}{2}vμω²A²

p = \frac{1}{2}√(T/μ)μω²A²

so we solve for ω

ω² = 2P / √(T/μ)μA²

we substitute

ω² = 2(2000) / √(6000/1.5)(1.5)(0.500)²

ω² = 4000 / 23.71708

ω² = 168.65

(2πf)² = ω²

so

(2πf)² = 168.65

4π²f² = 168.65

f² = 168.65 / 4π²

f² = 4.27195

f = √4.27195

f = 2.066 Hz

Therefore, the required frequency of waves is 2.066 Hz

3 0
3 years ago
A computer is purchased for $2816 and depreciates at a constant rate to $0 in 8 years. Find a formula for the value, V , of the
marusya05 [52]

Answer:

  • The formula its f(t) \ = \ - \ 352 \ \frac{\$ }{years} \ t \ + \ \$ \ 2816
  • After 5 years, the computer value its $ 1056

Explanation:

<h3>Obtaining the formula</h3>

We wish to find a formula that

  • Starts at 2816. f(0 \ years) \ = \ \$ \ 2816
  • Reach 0 at 8 years. f( 8 \ years) \ = \ \$ \ 0
  • Depreciates at a constant rate. m

We can cover all this requisites with a straight-line equation. (an straigh-line its the only curve that has a constant rate of change) :

f(t) \ = \ m\ t \ + \ b,

where m its the slope of the line and b give the place where the line intercepts the <em>y</em> axis.

So, we can use this formula with the data from our problem. For the first condition:

f ( 0 \ years ) = m \ (0 \ years) + b = \$ \ 2816

b = \$ \ 2816

So, b = $ 2816.

Now, for the second condition:

f ( 8 \ years ) = m \ (8 \ years) + \$ \ 2816 = \$ \ 0

m \ (8 \ years) = \ - \$ \ 2816

m = \frac{\ - \$ \ 2816}{8 \ years}

m = \frac{\ - \$ \ 2816}{8 \ years}

m = \ - \ 352 \frac{\$ }{years}

So, our formula, finally, its:

f(t) \ = \ - \ 352 \ \frac{\$ }{years} \ t \ + \ \$ \ 2816

<h3>After 5 years</h3>

Now, we just use <em>t = 5 years</em> in our formula

f(5 \ years) \ = \ - \ 352 \ \frac{\$ }{years} \ 5 \ years \ + \ \$ \ 2816

f(5 \ years) \ = \ - \$ \ 1760 + \ \$ \ 2816

f(5 \ years) \ = $ \ 1056

4 0
4 years ago
If a car used 260,000 W of power to complete a race in 15 s, how much work did the car do?
suter [353]

Answer: 3.9 MW

Explanation:

1 W = 1 J/s

260000 J/s (15 s) = 3,900,000 = 3.9 MW

3 0
3 years ago
The boiling point of water is 91.30 °C on the
Taya2010 [7]

Answer:

196.34 °F

Explanation:

To convert from degrees celsius to degrees fahrenheit, use this equation:

(°C * 9/5) + 32 = °F

So, using this equation:

(91.30 * 9/5) + 32 = °F

196.34 + 32 = °F

°F = 196.34

Hope this helps!

3 0
3 years ago
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