Ask question

Ask question

All categories

3 years ago

12

A spring whose spring constant is 270 lbf/in has an initial force of 100 lbf acting on it. Determine the work, in Btu, required

to compress it another 1 in.

1 answer:

cricket20 [7]3 years ago

6 0

Answer:

0.02585 BTU

Explanation:

Given: Spring constant, k = 270 lbf/in

Initial force, f =100 lbf

Compression, x = 1 in

Work done can be calculated as follows:

$W = \int {(f+kx)} \, dx \\W = fx + \frac{1}{2}kx^2\\W= (100 lbf)(1 in)+ \frac{1}{2}(270 lbf/in)(1 in)^2\\W= 100+135 = 235 lbf in\\W=235 \times 0.00011 BTU = 0.02585 BTU$

You might be interested in

A sinusoidal wave travels along a string. The time for a particular point to move from maximum displacement to zero is 0.13 s. W

vitfil [10]

Answer:

Part a)

T = 0.52 s

Part b)

$f = 1.92 Hz$

Part c)

$speed = 3.65 m/s$

Explanation:

As we know that the particle move from its maximum displacement to its mean position in t = 0.13 s

so total time period of the particle is given as

$T = 4\times 0.13 = 0.52 s$

now we have

Part a)

T = time to complete one oscillation

so here it will move to and fro for one complete oscillation

so T = 0.52 s

Part b)

As we know that frequency and time period related to each other as

$f = \frac{1}{T}$

$f = \frac{1}{0.52}$

$f = 1.92 Hz$

Part c)

As we know that

wavelength = 1.9 m

frequency = 1.92 Hz

so wave speed is given as

$speed = wavelength \times frequency$

$speed = 1.92 \times 1.9$

$speed = 3.65 m/s$

4 0

3 years ago

A father is trying to teach his child to ice skate. As the child stands still, the father pushes him forward with an acceleratio

rjkz [21]

Hi there!

We can use Newton's Second Law:

$\Sigma F = ma$

∑F = net force (N)

m = mass (kg)

a = acceleration (m/s²)

We are given the mass and acceleration, so:

∑F = 20 · 2 = <u>40 N</u>

4 0

2 years ago

How does air resistance affect an object’s speed?

tatuchka [14]

When air resistance<span> acts, acceleration during a fall </span>will<span> be less than g because </span>air resistance affects<span> the motion of the falling </span>objects<span> by slowing it down. </span>Air resistance<span> depends on two important factors - the</span>speed<span> of the </span>object<span> and its surface area. Increasing the surface area of an </span>object<span> decreases its </span>speed<span>.</span>

4 0

3 years ago

One gallon of paint (volume = 3.79 10-3 m3) covers an area of 17.8 m2. What is the thickness of the fresh paint on the wall?

AURORKA [14]

Answer:

0.2129 mm

Explanation:

We have given volume of the paint = 1 gallon $=3.79\times 10^{-3}m^3$

Area that covers the paint $=17.8 m^2$

We have to find the thickness of the fresh paint

So $thickness=\frac{volume}{area}=\frac{3.79\times 10^{-3}}{17.8}=0.2129\times 10^{-3}m=0.2129mm$

So the thickness of fresh paint on the wall is 0.2129 mm

6 0

3 years ago

X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to

lys-0071 [83]

Answer:

a) $\Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

b) $\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

c) $E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

Explanation

Part a

For this case we can use the Compton shift equation given by:

$\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

Part b

For this cas we can calculate the wavelength of the phton with this formula:

$\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

$E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

3 0

3 years ago

Read 2 more answers