Un objeto que tiene como masa 80 kg acelera a razón de 10 m/s2 ¿que fuerza lo impulzo

What force does a window washer want acting on him when trying to keep the same position

Ksenya-84 [330]

That would be balanced.

8 0

3 years ago

Derive the formula for the moment of inertia of a uniform, flat, rectangular plate of dimensions l and w, about an axis through

Ad libitum [116K]

Answer:

A uniform thin rod with an axis through the center

Consider a uniform (density and shape) thin rod of mass M and length L as shown in (Figure). We want a thin rod so that we can assume the cross-sectional area of the rod is small and the rod can be thought of as a string of masses along a one-dimensional straight line. In this example, the axis of rotation is perpendicular to the rod and passes through the midpoint for simplicity. Our task is to calculate the moment of inertia about this axis. We orient the axes so that the z-axis is the axis of rotation and the x-axis passes through the length of the rod, as shown in the figure. This is a convenient choice because we can then integrate along the x-axis.

We define dm to be a small element of mass making up the rod. The moment of inertia integral is an integral over the mass distribution. However, we know how to integrate over space, not over mass. We therefore need to find a way to relate mass to spatial variables. We do this using the linear mass density of the object, which is the mass per unit length. Since the mass density of this object is uniform, we can write

λ = m/l (orm) = λl

If we take the differential of each side of this equation, we find

d m = d ( λ l ) = λ ( d l )

since

λ

is constant. We chose to orient the rod along the x-axis for convenience—this is where that choice becomes very helpful. Note that a piece of the rod dl lies completely along the x-axis and has a length dx; in fact,

d l = d x

in this situation. We can therefore write

d m = λ ( d x )

, giving us an integration variable that we know how to deal with. The distance of each piece of mass dm from the axis is given by the variable x, as shown in the figure. Putting this all together, we obtain

I=∫r2dm=∫x2dm=∫x2λdx.

The last step is to be careful about our limits of integration. The rod extends from x=−L/2x=−L/2 to x=L/2x=L/2, since the axis is in the middle of the rod at x=0x=0. This gives us

I=L/2∫−L/2x2λdx=λx33|L/2−L/2=λ(13)[(L2)3−(−L2)3]=λ(13)L38(2)=ML(13)L38(2)=112ML2.

4 0

3 years ago

. Inside a laser apparatus, the stimulation and relaxation of electrons in atoms causes many photons with the same to be continu

mrs_skeptik [129]

Inside a laser apparatus, the stimulation and relaxation of electrons in atoms cause many photons with the same wavelength to be continuously emitted.

From the questions given, the main objective is to fill in the gaps and add important information where necessary. The missing information is highlighted in bold and underlined.

Inside a laser apparatus, the stimulation and relaxation of electrons in atoms cause many photons with the same wavelength to be continuously emitted.

2. When these photons are emitted, they travel between two reflective

surfaces to form the wave that is represented in the simulation.

3. This wave is the summation of all the photons being introduced with

every oscillation, and as they continue to travel, the amplitude

increases. 

4. This occurs because the photons are emitted in a coherent fashion;

however, amplitude when the photons overlap in an incoherent

fashion.

5. In a laser device, a small portion of photons are permitted to escape

(for use in an application). This is emulated in the simulation, by

settling the Damping to Lots such that amplitude remains relatively

constant when compared to damping of None. (Damping

represents the Loss of photons.

6. The generation of multiple wavelengths is possible in some laser

producing systems, and the diffraction angle can be varied to allow

the isolation of different wavelengths.

7. Finally, when the power of a laser is described, the wave property

that is being referenced is a function of its frequency and

amplitude.

Therefore, we can conclude that we've fully understood the concept of emission of photons and wavelength in a laser apparatus.

Learn more about wavelength here:

brainly.com/question/23023103?referrer=searchResults

5 0

3 years ago

A sound wave has a frequency of 247 Hz and a wavelength of 1.4 m. What is the speed of the sound wave in air?

zavuch27 [327]

The speed of sound through air at room temperature is almost always 343 m/s. However, since it doesn't tell you that, use the equation wavelength=velocity/frequency. Plug in the numbers: 1.4=v/247, and v=345.8 m/s.

5 0

3 years ago

A film of oil on water will appear dark when it is very thin, because the path length difference becomes small compared with the

VMariaS [17]

Answer:

jhgyhhtuifghhjfgy

Explanation:

jjhuyddfrgzzdssfdffsd fdfdfsdfddsfsdfsd

6 0

3 years ago