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2 years ago

10

Suppose that the mirror described in Part A is initially at rest a distance R away from the sun. What is the critical value of a

rea density for the mirror at which the radiation pressure exactly cancels out the gravitational attraction from the sun?

1 answer:

Sati [7]2 years ago

5 0

Ans;

6.25m²/Kd

Explanation: see attached file

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Un cable está tendido sobre dos postes colocados con una separación de 10 m. A la mitad del cable se cuelga un letrero que provo

lisabon 2012 [21]

Answer:

El peso del cartel es 397,97 N

Explanation:

La tensión dada en cada segmento del cable = 2000 N

El desplazamiento vertical del cable = 50 cm = 0,5 m

La distancia entre los polos = 10 m

La posición del letrero en el cable = En el medio = 5

El ángulo de inclinación del cable a la vertical = tan⁻¹ (0.5 / 5) = 5.71 °

El peso del letrero = La suma del componente vertical de la tensión en cada lado del letrero

El peso del signo = 2000 × sin (5.71 grados) + 2000 × sin (5.71 grados) = 397.97 N

El peso del signo = 397,97 N.

8 0

2 years ago

Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. Part A If the surfac

alukav5142 [94]

Answer:

5308.34 N/C

Explanation:

Given:

Surface density of each plate (σ) = 47.0 nC/m² = $47\times 10^{-9}\ C/m^2$

Separation between the plates (d) = 2.20 cm

We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:

$E=\dfrac{\sigma}{2\epsilon_0}$

Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,

$E_{between}=E+E=2E=\frac{2\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}$

Now, plug in $47\times 10^{-9}\ C/m^2$ for 'σ' and $8.85\times 10^{-12}\ F/m$ for $\epsilon_0$ and solve for the electric field. This gives,

$E_{between}=\frac{47\times 10^{-9}\ C/m^2}{8.854\times 10^{-12}\ F/m}\\\\E_{between}= 5308.34\ N/C$

Therefore, the electric field between the plates has a magnitude of 5308.34 N/C

5 0

2 years ago

Sort the phrases based on whether they represent kinetic energy or potential energy.

alexandr1967 [171]

Kinetic energy:
*the energy of a moving body*
rising water vapor.
the wings of flying hummingbird.
rolling marble.

Potential energy:
*the energy that is stored in a body so that any small change in position or state of the body, will result in this body movement by transforming all this potential energy into kinetic energy*
stone resting.
disconnected battery (the potential energy is what called voltage here)
stretched rubber band.

Hope this helps.

8 0

2 years ago

Read 2 more answers

A long coaxial cable (Fig. 2.26) carries a uniform volume charge density rho on the inner cylinder (radius a), and a uniform sur

Yuki888 [10]

Answer:

a) E = ρ / e0

b) E = ρ*a / (e0 * r)

c) E = 0

Explanation:

Because of the geometry, the electric field lines will all have a radial direction.

Using Gauss law

$Q/e0 = \int \int E * dA$

Using a Gaussian surface that is cylinder concentric to the cable, the side walls will have a flux of zero, because the electric field lines will be perpendicular. The round wall of the cylinder will have the electric field lines normal to it.

We can make this cylinder of different radii to evaluate the electric field at different points.

Then:

A = 2*π*r (area of cylinder per unit of length)

Q/e0 = 2*π*r*E

E = Q / (2*π*e0*r)

Where Q is the charge contained inside the cylinder.

Inside the cable core:

There is a uniform charge density ρ

Q(r) = ρ * 2*π*r

Then

E = ρ * 2*π*r / (2*π*e0*r)

E = ρ / e0 (electric field is constant inside the charged cylinder.

Between ther inner cilinder and the tube:

Q = ρ * 2*π*a

E = ρ * 2*π*a / (2*π*e0*r)

E = ρ*a / (e0 * r)

Outside the tube, the charges of the core cancel each other.

E=0

4 0

2 years ago

When a gas is rapidly compressed (say, by pushing down a piston) its temperature increases. When a gas expands against a piston,

shusha [124]

Answer:

Explained in explanation

Explanation:

The first law of thermodynamics states that the change in internal energy of a system(ΔU) is equal to the sum of the net heat transfer into the system(Q) and the net work done on the system(W). In equation, this law is;

ΔU = Q + W

Now, when there's gas inside a container with a movable piston that's tightly fitting, we will assume that the piston can move up and down thereby compressing the gas or allowing the gas to expand against it.

Now these gas molecules inside the container possess kinetic energy. Thus, the internal energy(U) of the system is simply the sum of all the kinetic energies of the individual gas molecules present in the container.

Therefore, if the temperature(T) of the gas increases, then the speed and internal energy(U) of the gas molecules will also increase. In the same way, if the temperature of the gas decreases, the speed and internal energy of the gas molecules would also decrease.

Now, back to the question, when the piston is pushed down, it does work on the gas and the gas does negative work on the piston. Thus, the gas will be get compressed to a smaller space, and thereby making the gas molecules to hit the piston at a faster rate. Thus, there is a decrease in speed and as we saw earlier that when there is a decrease in speed, it means temperature has decreased.

Whereas, when the piston is moved up, the gas does positive work on the piston and the speed of the gas molecules will increase. Like I said earlier that increase in speed means increase in temperature.

4 0

2 years ago