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Sav [38]
2 years ago
11

A spring with a spring constant of 350 N/m pulls a door closed. How much work is done as the spring pulls the door at a constant

velocity from an 85.0-cam stretch to a 5.0-cm stretch?
Physics
1 answer:
german2 years ago
3 0

The work done to stretch the spring will be 112 J.

<h3>What is spring force?</h3>

The force required to extend or compress a spring by some distance scales linearly with respect to that distance is known as the spring force. Its formula is

F = kx

The given data in the problem is;

F is the spring force =?

K is the spring constant= 8.5 N/m

x is the length by which spring got stretched = 1.2m

The work is done to stretch the spring is;

\rm W= \frac{1}{2} kx^2 \\\\ W=\frac{1}{2} \times 350 \times (0.850-0.050)^2 \\\\ W=0.5 \times 350 \times (0.80)^2 \\\\W=112 \ J

To learn more about the spring force refer to the link;

brainly.com/question/4291098

#SPJ1

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When a new path of lesser resistance is made for an existing circuit a(n) _____________ circuit occurs.
Margaret [11]

When a new path of lesser resistance is made for an existing circuit a(n) short circuit occurs.

<h3>What is short circuit?</h3>

An electrical circuit short circuit is when two nodes that are supposed to be connected at different voltages make an improper connection. This leads to an electric current that can damage circuits, cause overheating, fire, or explosions, and is only constrained by the network's remaining nodes' equivalent Thevenin resistance. While short circuits are typically the result of a failure, they can occasionally be brought on purpose, such as when voltage-sensing crowbar circuit protectors are being installed.

An electrical connection that requires two nodes to have the same voltage is known as a short circuit in circuit analysis. Since there is no resistance and hence no voltage drop across the link in a "perfect" short circuit, there is no short circuit.

To learn more about short circuit, visit:

brainly.com/question/13260673

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5 0
1 year ago
Oscilloscopes have parallel metal plates inside them to deflect the electron beam. These plates are called the deflecting plates
alexdok [17]

Answer:

1.6 pF

Explanation:

The capacitance of a parallel-plate capacitor in air is given by:

C=\frac{\epsilon_0 A}{d}

where

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A is the area of each plate

d is the separation between the plates

In this problem we have:

- Separation between the plates: d = 5.00 mm = 0.005 m

- Area of the plates: A = 3.00 cm \cdot 3.00 cm = 9.00 cm^2 = 9\cdot 10^{-4}m^2

Therefore, the capacitance is

C=\frac{(8.85\cdot 10^{-12}F/m)(9\cdot 10^{-4} m^2)}{0.005 m}=1.6\cdot 10^{-12} F=1.6 pF

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3 years ago
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In a string of length L, the wavelength of the n-th harmonic of the standing wave produced in the string is given by:

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The length of the string in this problem is L=3.5 m, therefore the wavelength of the 1st harmonic of the standing wave is:

\lambda=\frac{2}{1} \cdot 3.5 m=7.0 m


The wavelength of the 2nd harmonic is:

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\lambda=\frac{2}{4} \cdot 3.5 m=1.75 m


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3 years ago
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