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Morgarella [4.7K]
3 years ago
14

You wrap a wire around a piece of iron. If you slowly increase the strength of an electric current flowing through the wire, you

would expect _____.
A. the magnet to become stronger
B. the magnet to stop producing a field
C. the wire to move
D. the wire to unwrap
Physics
2 answers:
Maksim231197 [3]3 years ago
4 0
The correct answer is A. the magnet to become stronger

The stronger the electric current in the piece of metal, the stronger the magnetic field will be.
GarryVolchara [31]3 years ago
3 0
If we wrap a wire around a piece of iron. If we slowly increase the strength of an electric current flowing through the wire, as a result  the magnetic field become stronger so the most appropriate answer will be A. the magnet to become stronger. 
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Why an Aeroplane flying has kinetic  

A flying aeroplane has potential energy has it flies above the ground level. And since the aeroplane is flying motion is associated with it and thus possesses kinetic energy. Hence a flying aeroplane has both potential and kinetic energ

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3 years ago
The graph shows the amplitude of a passing wave over time in seconds (s). What is the approximate frequency of the wave shown?
belka [17]

Answer:

B) 0.3Hz

Explanation:

I just took the test i hope i helped and i hope you pass the test

7 0
3 years ago
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A 225-g object is attached to a spring that has a force constant of 74.5 N/m. The object is pulled 6.25 cm to the right of equil
snow_lady [41]

Answer:

1.137278672 m/s

+5.9 cm or -5.9 cm

Explanation:

A = Amplitude = 6.25 cm

m = Mass of object = 225 g

k = Spring constant = 74.5 N/m

Maximum speed is given by

v_m=A\omega\\\Rightarrow v_m=A\sqrt{\dfrac{k}{m}}\\\Rightarrow v_m=6.25\times 10^{-2}\times \sqrt{\dfrac{74.5}{0.225}}\\\Rightarrow v_m=1.137278672\ m/s

The maximum speed of the object is 1.137278672 m/s

Velocity is at any instant is given by

\dfrac{v_m}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A\omega}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A}{3}=\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A^2}{9}=A^2-x^2\\\Rightarrow A^2-\dfrac{A^2}{9}=x^2\\\Rightarrow x^2=\dfrac{8}{9}A^2\\\Rightarrow x=A\sqrt{\dfrac{8}{9}}\\\Rightarrow x=6.25\times 10^{-2}\sqrt{\dfrac{8}{9}}\\\Rightarrow x=\pm 0.0589255650989\ m

The locations are +5.9 cm or -5.9 cm

4 0
3 years ago
HELP ASAP!!! 100 points!!!!!!!!!
Zepler [3.9K]

Answer:

Shown from explanation.

Explanation:

The resonant frequency is related by the formula;

Fo= 1/2π√(K/M)

Where Fo is resonant frequency

M is mass

K is the spring constant

From the expression above;

When K is increased frequency would increases since frequency is directly proportional to spring constant.

Similarly when the mass is increased the frequency decreases since frequency is inversely proportional to the mass.

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6 0
3 years ago
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Juri is tugging her wagon behind her on the way to... wherever her wagon needs to go. The wagon repair shop. She has a trek ahea
vitfil [10]

Answer:

W = 819152 J = 819.15 KJ

Explanation:

The work done by Juri can be given by the following formula:

W = FdCos\theta

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W = Work done = ?

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θ = angle to horizontal = 35°

Therefore,

W = (200 N)(5000 m)Cos 35°

<u>W = 819152 J = 819.15 KJ</u>

5 0
3 years ago
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