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Svetlanka [38]
3 years ago
13

The force F required to compress a spring a distance x is given by F 2 F0 5 kx where k is the spring constant and F0 is the prel

oad. Determine the work required to compress a spring whose spring constant is k 5 200 lbf/in a distance of one inch starting from its free length where F0 5 0 lbf. Express your answer in both lbf·ft and Btu.
Physics
1 answer:
IrinaVladis [17]3 years ago
6 0

Answer:

a)W=8.333lbf.ft

b)W=0.0107 Btu.

Explanation:

<u>Complete question</u>

The force F required to compress a spring a distance x is given by F– F0 = kx where k is the spring constant and F0 is the preload. Determine the work required to compress a spring whose spring constant is k= 200 lbf/in a distance of one inch starting from its free length where F0 = 0 lbf. Express your answer in both lbf-ft and Btu.

Solution

Preload = F₀=0 lbf

Spring constant k= 200 lbf/in

Initial length of spring x₁=0

Final length of spring x₂= 1 in

At any point, the force during deflection of a spring is given by;

F= F₀× kx  where F₀ initial force, k is spring constant and x is the deflection from original point of the spring.

W=\int\limits^2_1 {} \, Fds \\\\\\W=\int\limits^2_1( {F_0+kx} \,) dx \\\\\\W=\int\limits^a_b {kx} \, dx ; F_0=0\\\\\\W=k\int\limits^2_1 {x} \, dx \\\\\\W=k*\frac{1}{2} (x_2^{2}-x_1^{2}  )\\\\\\W=200*\frac{1}{2} (1^2-0)\\\\\\W=100.lbf.in\\\\

Change to lbf.ft by dividing the value by 12 because 1ft=12 in

100/12 = 8.333 lbf.ft

work required to compress the spring, W=8.333lbf.ft

The work required to compress the spring in Btu will be;

1 Btu= 778 lbf.ft

?= 8.333 lbf.ft----------------cross multiply

(8.333*1)/ 778 =0.0107 Btu.

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A magnetic field, a vector field that describes the magnetic influence on moving electric charges, electric currents, and magnetic materials. When a charge moves through a magnetic field, a force that is perpendicular to both its own velocity and the magnetic field operates on it.

Electrons go downward and positive ions move upward in a long, straight, vertical lightning stroke, creating a current of magnitude I = 20.0 kA.

A free electron travels through the air at a speed of v = 300 m/s at a place r = 50.0 m east of the stroke's center.

Let the magnetic field be B, and F be the magnetic force.

Counterclockwise horizontal arcs of field lines are produced by the upward lightning current.

We have, B = 8 × 10⁻⁵ T and;

The mass of an electron is, m = 9.11 × 10⁻³¹ kg

The time interval is Δt = 60 μs = 60 × 10⁻⁶

The angular frequency is given as:

ω = qB /m = 2πN / Δt

Where the number of revolutions is N.

So,

N = qBΔt /2πm

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N = 134 revolutions

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Why does the moon turn red during a lunar eclipse?.
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2 years ago
A mass-spring system has k = 56.8 N/m and m = 0.46 kg.
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Answer:

A. \omega=11.1121\ rad.s^{-1}

B. f=1.7685\ Hz

C. T=0.5654\ s

Explanation:

Given:

  • spring constant, k=56.8\ N.m^{-1}
  • mass attached, m=0.46\ kg

A)

for a spring-mass system the frequency is given as:

\omega=\sqrt{\frac{k}{m} }

\omega=\sqrt{\frac{56.8}{0.46}}

\omega=11.1121\ rad.s^{-1}

B)

frequency is given as:

f=\frac{\omega}{2\pi}

f=\frac{11.1121}{2\pi}

f=1.7685\ Hz

C)

Time period of a simple harmonic motion is given as:

T=\frac{1}{f}

T=0.5654\ s

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3 years ago
Points A, B, and C form the vertices of a triangle in a nonuniform electrostatic field. The electrostatic work done on a particl
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Answer:

Explanation:

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Work done = charge x potential difference

Wab = q ( Va - Vb )

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Given

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Answer:

A). A few of the positive particles aimed at a gold foil seemed to bounce back off of the thin metallic foil.

Explanation:

Scientists decided to change the model of the atom when they discovered new evidence that showed 'few of the positive particles aimed at a gold foil seemed to bounce back off of the thin metallic foil.' On this ground, <u>Rutherford concluded that atom is mostly made up of empty space and thus, he proposed a nucleus model of atom in which the atom comprises of the tiny and positively charged nucleus is surrounded by electrons with a negative charge</u>. Thus, <u>option A</u> is the correct answer.

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