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neonofarm [45]
2 years ago
10

The aluminum alloy (2024-t6) absorber plate is 6 mm thick and well insulated on its bottom. the top surface of the plate is sepa

rated from a transparent cover plate by an evacuated space. the tubes are spaced a distance / of 0.20 m from each other, and water is circulated through the tubes to remove the collected energy. the water may be assumed to be at a uniform temperature of 7z 60c. under steady-state operating conditions for which the qhw radiation heat flux to the surface is trad 800 w/m2 , what is the maximum temperature on the plate and the heat transfer rate per unit length of tube? note that trad represents the net effect of solar radiation absorption by the absorber plate and radiation exchange between the absorber and cover plates. you may assume the temperature of the absorber plate directly above a tube to be equal to that of the water.
Physics
1 answer:
Xelga [282]2 years ago
5 0
The aluminum alloy(2024-T6; k=180 W/m<span>!K,) </span>absorber plate<span> is </span>6 mm thick<span> and </span>well insulated<span> on </span>its bottom<span>. The </span>top surface<span> of the </span>plate<span> is </span>separated<span> from a </span>transparent cover plate<span> by an </span>evacuated space<span>. The </span>tubes<span> are</span>spaced<span> a </span>distance<span> L of ...</span>
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Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s
frosja888 [35]

Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀)  we see that for the same t v₁> v₂

Explanation:

You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

    y₁ = v₀₁ t + ½ g t²

    y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

     t ’= (t-t₀)

    y₂ = v₀₂ t ’+ ½ g t’²

    y₂ = 0 + ½ g (t-t₀)²

    y₂ = + ½ g (t-t₀)²

Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

    S = y₁ -y₂

    S = ½ g t²– ½ g (t-t₀)²

    S = ½ g [t² - (t²- 2 t to + to²)]  

    S = ½ g (2 t t₀ - t₀²)

    S = ½ g t₀ (2 t -t₀)

This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to.  The rock y has not left and the distance increases

For t> = to.  the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t   v₁> v₂

3 0
2 years ago
A proton moves along the x-axis with vx=1.0×107m/s. As it passes the origin, what are the strength and direction of the magnetic
Sunny_sXe [5.5K]

Answer:

Magnetic field will be ZERO at the given position

Explanation:

As we know that the magnetic field due to moving charge is given as

B = \frac{\mu_0 qv sin\theta}{4\pi r^2}

so here we know that for the direction of magnetic field we will use

\hat B = \hat v \times \hat r

so we have

\hat B = \hat i \times (\hat i + 0\hat j + 0\hat k)

so magnetic field must be ZERO

So whenever charge is moving along the same direction where the position vector is given then magnetic field will be zero

3 0
2 years ago
A square loop of wire with initial side length 10 cm is placed in a magnetic field of strength 1 T. The field is parallel to the
Fofino [41]

Answer:

2 x 10⁻³ volts

Explanation:

B = magnetic of magnetic field parallel to the axis of loop = 1 T

\frac{dA}{dt} = rate of change of area of the loop = 20 cm²/s = 20 x 10⁻⁴ m²

θ = Angle of the magnetic field with the area vector = 0

E = emf induced in the loop

Induced emf is given as

E = B \frac{dA}{dt}

E = (1) (20 x 10⁻⁴ )

E = 2 x 10⁻³ volts

E = 2 mV

7 0
2 years ago
You throw a tennis ball (mass 0.0570 kg) vertically upward. It leaves your hand moving at 15.0 m/s. Air resistance cannot be neg
Deffense [45]

Answer:195 J

Explanation:

Given

mass of ball m=0.0570\ kg

ball leaves the hand with u=15\ m/s

maximum height reached by ball h=8\ m

Initial Mechanical energy when ball just leaves the hand

M.E._1=(P.E.+K.E.)_1

M.E._1=(mgh)_1+(\frac{1}{2}mv^2)_1

considering hand to be datum so h_1=0[/tex]

so Potential energy at ground is zero

M.E._1=\frac{1}{2}\times m\times (15)^2

M.E._1=6.41\ J

Mechanical Energy at highest point

(M.E.)_2=(P.E.+K.E.)_2

at highest Point velocity is zero

(M.E.)_2=mgh_2+0

(M.E.)_2=0.0570\times 9.8\times 8

(M.E.)_2=4.46\ J

Decrease in Mechanical energy

(M.E.)_1-(M.E.)_2=6.41-4.46

(M.E.)_1-(M.E.)_2=1.95\ J

3 0
2 years ago
A star orbiting a black hole in a clockwise direction at a radial distance of 1.0 × 106 km is acted upon by a counterclockwise f
snow_tiger [21]

Answer:

2.5 \times 10^{11} N-m upwards        

Explanation:

Torque is the vector cross product of the force and radial distance.

\tau = rF sin \theta

\tau = (1.0\times 10^6 \times 10^3) m \times 250 N\times sin 90^o \\\Rightarrow \tau= 2.5 \times 10^{11} N-m

The direction of the torque would be perpendicular to the direction of the force and radial distance. The direction of the force is counter-clockwise. The direction of the torque would be upwards.

4 0
2 years ago
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