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vitfil [10]
3 years ago
12

A piece of plastic is uniformly charged with surface charge density eta1. The plastic is then broken into a large piece with sur

face charge density eta2 and a small piece with surface charge density eta3. Rank in order, from largest to smallest, the surface charge densities eta1 to eta3.?
a. eta1>eta2>eta3
b. eta1>eta2=eta3
c. eta1=eta2=eta3
d. eta2=eta3>eta1
Physics
1 answer:
Amiraneli [1.4K]3 years ago
3 0
C is the answer for this problem
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If the Sun subtends a solid angle Ω on the sky, and the flux from the Sun just above the Earth’s atmosphere, integrated over all
Arada [10]

Answer:

A)Ω = 7.8 × 10^−5 steradians.

B) TE = 5800K

C) fλ(λ1) = (π ^2 ) /ΩBλ(T)

Explanation:

A) First of all, if we assume that the Sun emits isotropically at a luminosity (L⊙) , the flux at a given distance R from the sun would be f(d) = L⊙/ (4πd^2)

The ratio of flux at the solar photosphere to the flux at the Earth’s atmosphere would be: F⊙/{f(d⊙)} = (R⊙)^2 / (d⊙)^2

Now if we think of this relationship of the flux and the earth as a conical pattern, we'll deduce that the solid angle subtended by the sun at Earth’s surface to be;

Ω = π[(R⊙)^2 / (d⊙)^2]

Combining this with the ratio earlier gotten, well arrive at;

F⊙ = {f(d⊙ )π} /Ω

Now let's express The radius of the sun (R) in terms of its angular diameter (2α) and this gives;

R⊙ ≈ αd⊙

Now combining this with the equation for Ω earlier, we get;

Ω ≈ πα^2

So, = π((0.57/2π) /180)^2 = 7.8 × 10^−5 steradians.

B) from Stefan-Boltzmann Law,

F⊙ = σ(TE)^4

From the beginning, we know that;

F⊙ = {f(d⊙ )π} /Ω

And so replacing that in the stephan boltzmann law, we get ;

{f(d⊙ )π} /Ωσ = (TE)^4

So, (TE)^4 = {π (1.4 kWm^(−2))} / [(7.8 × 10^(−5 ) steradians x (5.66961 × 10^(−8))]

In stephan boltzmann law, σ = 5.66961 × 10^(−8)

And so, TE is approximately 5800K.

C) In order to relate fλ(λ1) with T, let's assume the sun’s surface to be an isotropically emitting blackbody, i.e its specific intensity is Iλ = Bλ(T). Hence, the flux at Sun’s surface for a given wavelength would be;

Fλ(λ1) = πBλ(T)

Now, if we combine this with the expression of F⊙ gotten earlier, well get the relation;

fλ(λ1) = (π ^2 ) /ΩBλ(T)

7 0
3 years ago
Use the work—energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resist
andreyandreev [35.5K]

Answer:

a) It is moving at 43.15\frac{m}{s^{2}} when reaches the ground.

b) It is moving at 101.44\frac{m}{s^{2}} when reaches the ground.

Explanation:

Work energy theorem states that the total work on a body is equal its change in kinetic energy, this is:

W=K_f-K_i (1)

with W the total work, Ki the initial kinetic energy and Kf the final kinetic energy. Kinetic energy is defined as:

K=\frac{mv^2}{2} (2)

with m the mass and v the velocity.

Using (2) on (1):

W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2} (3)

In both cases the total work while the objects are in the air is the work gravity field does on them. Work is force times the displacement, so in our case is weight (w=mg) of the object times displacement (d):

W=Fd=wd=mgd (4)

Using (4) on (3):

mgd=\frac{mv_f^2}{2}-\frac{mv_i^2}{2} (5)

That's the equation we're going to use on a) and b).

a) Because the branch started form rest initial velocity (vi) is equal zero, using this and solving (5) for final velocity:

v_f=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*95}

v_f=43.15\frac{m}{s^{2}}

b) In this case the final velocity of the boulder is instantly zero when it reaches its maximum height, another important thing to note is that in this case work is negative because weight is opposing boulder movement, so we should use -mgd:

-mgd=-\frac{mv_i^2}{2}

Solving for initial velocity (when the boulder left the volcano):

v_i=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*525}

v_i=101.44 \frac{m}{s^{2}}

3 0
3 years ago
Two friends leave a movie theater and take different busses to the same ice cream shop. One bus takes a longer route driving on
lisov135 [29]

Answer: short displacement has shorter road and long displacement has longer displacement . now think your self Which one is right

6 0
3 years ago
A 2.5 g bullet traveling at 350 m/s hits a tree and slows uniformly to a stop while penetrating a distance of 12 cm into the tre
Cloud [144]

Answer: Work done = 153.125Joules, Work done = 0.003Nm

Explanation:

Kinetic energy of a body is the energy possessed by a body by virtue of its motion.

Mathematically,

K.E = 1/2MV²

Where;

M = mass of the body = 2.5g = 0.0025kg

V = velocity of the body = 350m/s

Substituting this values in the formula, we have;

K.E = 1/2× 0.0025×350²

K.E = 153.125Joules

Work done is the force applied to body to cause it to move through a distance.

Work = Force × distance

Force = ma = 0.0025 × 10

Force = 0.025N

Distance = 12cm = 0.12m

Work = 0.025×0.12

Work = 0.003Nm

work done by the tree in stopping the bullet is 0.003N

4 0
3 years ago
Read 2 more answers
PLS HELP ASAP (no links)
nikklg [1K]

Answer:

The mass goes down

Explanation:

Because mass is the quantity of matter contained in a substance And Volume is the space occupied by a substance. So when the volume is less the mass decreases

6 0
3 years ago
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