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ValentinkaMS [17]
3 years ago
7

Solenoid constructed with 50 turns of wire and has inductance L. if we made another solenoid identical in length and cross secti

onal area but with 200 turns of wire it would have an inductance closest to__________
Physics
1 answer:
Vlad [161]3 years ago
6 0

Answer:

It would have an inductance closest to 16 L.

Explanation:

Inductance for a one solenoid can be calculated with a formula following:

L=μ*N^2*A/l

Then, in this situation we are increasing the number of turns by 4 without any length change. First solenoid with 50 turn has inductance L which is:

L= μ*50^2*A/l=2500*μ*A/l

When we increase the number of turns by four, it will increase to:

L'=μ*200^2*A/l=40000*μ*A/l=16 L

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3 years ago
With what tension must a rope with length 2.90 m and mass 0.125 kg be stretched for transverse waves of frequency 42.0 Hz to hav
Vesnalui [34]

Answer:

41.64 N

Explanation:

Applying,

v = √(T/m')................ Equation 1

Where v = velocity of the wave, T = Tension of the rope, m' = mass per unit length of the rope.

make T the subject of the equation,

T = v²m'................. Equation 2

But,

v = λf............... Equation 3

Where λ = wavelength, f = frequency

And

m' = m/L........... Equation 4

Where m = mass of the rope, L = length of the rope

Substitute equation 3 and equation 4 into equation 2

T = (λf)²(m/L).............. Equation 5

From the question,

Given: λ = 0.740 m, f = 42 Hz, m = 0.125 kg, L = 2.9 m

Substitute these values into equation 5

T = (42×0.74)²(0.125/2.9)

T = 41.64 N

6 0
3 years ago
Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles
Anna007 [38]

Question

Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles  travelled in a straight line and some were deflected at different angles.

Which statement best describes what Rutherford concluded from the motion of the particles?

A) Some particles travelled through empty spaces between atoms and some particles were deflected by electrons.

B) Some particles travelled through empty parts of the atom and some particles were deflected by electrons.

C) Some particles travelled through empty spaces between atoms and some particles were deflected by small areas of high-density positive charge in atoms.

D) Some particles travelled through empty parts of the atom and some particles were deflected by small areas of high-density positive charge in atoms.    

Answer:

 

The right answer is C)    

Explanation:

In the experiment described above, a piece of gold foil was hit with alpha particles, which have a positive charge. Alpha particles <em>α</em> were used because, if the nucleus was positive, then it would deflect the positive particles. The principles of physics posit that electric charges of the same orientation repel.  

So most as expected some of the alpha particles went right through meaning that the gold atoms comprised mostly empty space except the areas that were with a dense population of positive charges. This area became known as the "nucleus".  

Due to the presence of the positive charges in the nucleus, some particles had their paths bent at large angles others were deflected backwards.

Cheers!

6 0
3 years ago
Read 2 more answers
Does the image above correctly illustrate how the coriolis effect impacts global winds?
Sedbober [7]
Yes the winds are moving in a straight line
6 0
3 years ago
In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 78.5 m/s. Th
Andrews [41]

Answer:

FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

Explanation:

First you have to consider that the Ford Thunderbird (FT) follows a rectilinear motion with varying acceleration, while Mercedez Benz (MB) has a constant velocity (no acceleration). So if you finde the time spent by FT in each section, and the distance, then you will find the distance for MB.

1) Vf² = Vi² + 2ad, where Vf: final velocity, Vi: ionitial velocity, a: acceleration and d: distance.

For the first portion  (0 m/s)² = (78.5 m/s)² + 2a(250 m) ⇒

-(78.5 m/s)² / 2(250m) = a ⇒ a = -12.3 m/s².

Now, you can find the corresponding time for this section with the following formule: Vf = Vi + at ⇒ 0 m/s = 78.5 m/s + (-12.3 m/s²) t

⇒ t= (-78.5 m/s)/ (-12.3 m/s²) ⇒ t= 6.4 seconds.

2) Then FT spent 5 seconds in the pit.

3) The the FT accelerates until reach 78.5 m/s again in a distance of 370 m.

Vf² = Vi² + 2ad ⇒ (78.5 m/s)² = (0 m/s)² + 2a(370 m)

⇒ (78.5 m/s)²/ 2(370 m) = a ⇒ a = 8.3 m/s²

Then, Vf = Vi + at ⇒ 78.5 m/s = 0 m/2 + (8.3 m/s²) t

⇒ (78.5 m/s)/(8.3 m/s²) = t ⇒ t = 9.5 seconds.

4) Summarizing, the FT moves 620 meters (250 + 370 mts) in 20.9 seconds ( 6.4 s + 5 s + 9.5 s).

5) During this time, MB moves

Velocity = distance/ time ⇒ Velocity x time = Distance

⇒ Distance = (78.5 m/s) x  (20.9 seconds) ⇒ Distance = 1640.6 meters

6) Finally, the FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

3 0
3 years ago
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