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2 years ago

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Consider a 2 m wide and 5 m long slab resting on flat, fixed, earthen bed. The slab is 20 cm thick with uniform properties. You

may assume the slab is much longer and wider than it is thick. Write the reduced form of the governing equation for the slab. Clearly identify your coordinate system on a sketch and state your assumptions identifying which terms you are eliminating and why.

1 answer:

4vir4ik [10]2 years ago

8 0

Answer:

Explanation:

The coordinate sketch for the system is shown in the attached file below. Also, in the cartesian coordinate system, since the height is less than the length and width, we did neglect the height. Thus, we eliminate the height and converted it to a two-dimension.

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This diagram shows particles that make up an atom. Which label BEST completes the diagram ?

worty [1.4K]

Answer: option A. strong nuclear force.

Explanation:

The diagram shows the subatomic particles inside the nucelous: protons and neutrons.

As you know, the protons are positively charged partilces inside the nucleous.

Being those particles charged with the same kind of charge they experiment electrostatic repulsion. So, how do you explain that they can stand together in such small space as it is the nucleous?

The responsible of keeping the subatomic particles together is the so called strong nuclear force.

Strong nuclear force or simply strong force is one of the four fundamental interactions or forces: i) gravitational, ii) electromagnetic, iii) weak nuclear force, and iv) strong nuclear force.

Strong nuclear force is the strongest force of nature and acts only in short distances as those inside the nucleous and is responsible for both the atraction among quarks and the atraction among protons to bind them together inside the atomic nucleous.

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3 years ago

A 62 kg box is lifted 12 meters off the ground. How much work was done?

OleMash [197]

7291.2 I hope this is right

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3 years ago

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A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o

Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field $dE$ at the center of curvature of the arc is

$dE = k\dfrac{dQ}{r^2}cos(\theta )$ <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where $r$ is the radius of curvature.

Now

$dQ = \lambda rd\theta$ ,

where $\lambda$ is the charge per unit length, and it has the value

$\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.$

Thus, the electric field at the center of the curvature of the arc is:

$E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta }{r^2} cos(\theta)$

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.$

Now, we find $\theta_1$ and $\theta_2$ . To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

$fraction = \dfrac{3.0m}{2\pi (2.3m)} = 0.2076$

and this is

$0.2076*2\pi =1.304$ radians.

Therefore,

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.$

evaluating the integral, and putting in the numerical values we get:

$E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\$

$\boxed{ E = 31.329N/C.}$

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2 years ago

Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m ea

Kipish [7]

Answer:

6.5 m/s

Explanation:

We are given that

Distance, s=100 m

Initial speed, u=1.4 m/s

Acceleration, $a=0.20 m/s^2$

We have to find the final velocity at the end of the 100.0 m.

We know that

$v^2-u^2=2as$

Using the formula

$v^2-(1.4)^2=2\times 0.20\times 100$

$v^2-1.96=40$

$v^2=40+1.96$

$v^2=41.96$

$v=\sqrt{41.96}$

$v=6.5 m/s$

Hence, her final velocity at the end of the 100.0 m=6.5 m/s

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2 years ago

When a 0.350-kg package is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The package is now di

AleksandrR [38]

To solve the problem it is necessary to use the concepts related to the calculation of periods by means of a spring constant.

We know that by Hooke's law

$F=kx$

Where,

k = Spring constant

x = Displacement

Re-arrange to find k,

$k= \frac{F}{x}$

$k= \frac{mg}{x}$

$k= \frac{(0.35)(9.8)}{12*10^{-2}}$

$k = 28.58N/m$

Perioricity in an elastic body is defined by

$T = 2\pi \sqrt{\frac{m}{k}}$

Where,

m = Mass

k = Spring constant

$T = 2\pi \sqrt{\frac{0.35}{28.58}}$

$T = 0.685s$

Therefore the period of the oscillations is 0.685s

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2 years ago