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Verdich [7]
3 years ago
12

All scientific theories on the solar system are based on-

Physics
2 answers:
S_A_V [24]3 years ago
6 0
It has to be C, non-observable information. Biased is definitely wrong, current evidences can't be right because scientific theories have been established long before we were born and non-testable was already listed by someone else that they put it and got it wrong.
I hope this helps.
joja [24]3 years ago
5 0
It's not non testable facts..I just took the test and got it wrong.
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A solenoid has a length , a radius , and turns. The solenoid has a net resistance . A circular loop with radius is placed around
geniusboy [140]

This question is incomplete, the complete question is;

A solenoid has a length 11.34 cm , a radius 1.85 cm , and 1627 turns. The solenoid has a net resistance of 144.9 Ω . A circular loop with radius of 3.77 cm is placed around the solenoid, such that it lies in a plane whose normal is aligned with the solenoid axis, and the center of the outer loop lies on the solenoid axis. The outer loop has a resistance of 1651.6 Ω. At a time of 0 s , the solenoid is connected to a battery that supplies a potential 34.95 V. At a time 2.58 μs , what current flows through the outer loop?

Answer:

the current flows through the outer loop is 1.3 × 10⁻⁵ A

Explanation:

Given the data in the question;

Length l = 11.34 cm = 0.1134 m

radius a = 1.85 cm = 0.0185 m

turns N = 1627

Net resistance R_{sol = 144.9 Ω

radius b = 3.77 cm = 0.0377 m

R_o = 1651.6 Ω

ε = 34.95 V

t = 2.58 μs = 2.58 × 10⁻⁶ s

Now, Inductance; L = μ₀N²πa² / l

so

L = [ ( 4π × 10⁻⁷ ) × ( 1627 )² × π( 0.0185 )² ] / 0.1134

L = 0.003576665 / 0.1134

L = 0.03154

Now,

ε = d∅/dt = \frac{d}{dt}( BA ) =  \frac{d}{dt}[ (μ₀In)πa² ]

so

ε = μ₀n \frac{dI}{dt}( πa² )

ε = [ μ₀Nπa² / l ] \frac{dI}{dt}

ε = [ μ₀Nπa² / l ] [ (ε/L)e^( -t/R_{sol) ]

I = ε/R_o = [ μ₀Nπa² / R_ol ] [ (ε/L)e^( -t/R_{sol) ]

so we substitute in our values;

I = [ (( 4π × 10⁻⁷ ) × 1627 × π(0.0185)²) / (1651.6 ×0.1134) ] [ ( 34.95 / 0.03154)e^( -2.58 × 10⁻⁶ / 144.9 ) ]

I = [ 2.198319 × 10⁻⁶ / 187.29144 ] [ 1108.116677 × e^( -1.7805 × 10⁻⁸ )

I = [ 1.17374 × 10⁻⁸ ] × [ 1108.116677 × 0.99999998 ]

I = [ 1.17374 × 10⁻⁸ ] × [ 1108.11665 ]

I = 1.3 × 10⁻⁵ A

Therefore, the current flows through the outer loop is 1.3 × 10⁻⁵ A

7 0
2 years ago
In Hooke's law, what does k represent?
kicyunya [14]

For Plato, K represents the stiffness of the spring. Hope this helps!

3 0
3 years ago
Read 2 more answers
7-10/2<br><br> A.2<br> B.-13<br> C.12
vagabundo [1.1K]

=7-10/2

=7- 5

= 2

A. 2

Hope this helps

3 0
3 years ago
Read 2 more answers
How many significant digits are in the following measurements?<br> a. 1300 m
Fofino [41]

Answer:

For example, 1300 with a bar placed over the first 0 would have three significant figures (with the bar indicating that the number is precise to the nearest ten).

Explanation:

hope it helps :)

5 0
2 years ago
Suppose you take a short piece of wire that is not attached to anything and move it up and down in a magnetic field. Explain whe
Otrada [13]

Answer:

No.

Explanation:

  • According to Faraday's law, the induced emf in the circuit is given by :

         e=\dfrac{d\phi}{dt}, it is proportional to the rate of change of magnetic flux.

  • In this case, a short piece of wire that is not attached to anything and move it up and down in a magnetic field. It means that the circuit is not completed here. It is an open circuit. For the induction of current, a circuit must be completed.
  • Hence, no current will induce.
4 0
3 years ago
Read 2 more answers
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