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Troyanec [42]
3 years ago
12

Which of the following is accurate in describing converging and diverging lenses?

Physics
2 answers:
kenny6666 [7]3 years ago
3 0

B. Both of these types of lenses have the ability to produce real images.

koban [17]3 years ago
3 0

Answer:

A. Both of these types of lenses have the ability to produce upright images.

Explanation:

Let's analyze each statement proposed:

A. Both of these types of lenses have the ability to produce upright images.  --> TRUE. In fact, depending on the position of the object relative to the lens, both converging and diverging lenses can produce an image which is upright.

B. Both of these types of lenses have the ability to produce real images.  --> FALSE. In fact, diverging lenses can produce virtual images only, because the rays of light are not focused into a single point.

C. Parallel rays converge into a single point at the focal point of a diverging lens, whereas parallel rays diverge at the focal point of a converging lens and spread out.  --> FALSE. It is actually the opposite: in a diverging lens, parallel rays diverge and spread out, while in a converging lens, parallel rays converge into a single point (the focal point)

D. Diverging lenses will produce images behind the lens, and converging lenses will produce images in front of the lens.  --> FALSE. It's actually the opposite: the image produced by a diverging lens is in front of the lens, while the image produced by a converging lens is behind the lens.


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H2SO4 + NaOH —> 2H2O (aq.) + Na2SO4 (salt)

The resulted salt/compound (Na2SO4) when reacting with Methyl Orange (MO) is called ”Removal of methyl orange dye and Na2SO4 salt from synthetic wastewater using reverse osmosis (RO)”

The efficiency of reverse osmosis (RO) membranes used for treatment of colored water effluents can be affected by the presence of both salt and dyes.

Concentration polarization of each of the dye and the salt and the possibility of a dynamic membrane formed by the concentrated dye can affect the performance of the RO membrane.

The objective of the current work was to study the effect of varying the Na2SO4 salt and methyl orange (MO) dye concentrations on the performance of a spiral wound polyamide membrane.

The work also involved the development of a theoretical model based on the solution diffusion (SD) mass transport theory that takes into consideration a pressure dependent dynamic membrane resistance as well as both salt and dye concentration polarizations.

Control tests were performed using distilled water, dye/water and salt/water feeds to determine the parameters for the model.

The experimental results showed that increasing the dye concentration from 500 to 1000 ppm resulted in a decrease in the salt rejection at all of the operating pressures and for both feed salt concentrations of 5000 and 10,000 ppm.

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7 0
3 years ago
How many hydrogen and carbon atoms in a diamond
Wewaii [24]

Answer:

Explanation:

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4 0
3 years ago
To apply Problem-Solving Strategy 12.2 Sound intensity. You are trying to overhear a most interesting conversation, but from you
Ivenika [448]

Answer:

r₂ = 0.316 m

Explanation:

The sound level is expressed in decibels, therefore let's find the intensity for the new location

            β = 10 log \frac{I}{I_o}

let's write this expression for our case

           β₁ = 10 log \frac{I_1}{I_o}

           β₂ = 10 log \frac{I_2}{I_o}

           

          β₂ -β₁ = 10 ( log \frac{I_2}{I_o} - log \frac{I_1}{I_o})

          β₂ - β₁ = 10 log \frac{I_2}{I_1}

          log \frac{I_2}{I_1} = \frac{60 - 20}{10} = 3

           \frac{I_2}{I_1} = 10³

           I₂ = 10³ I₁

having the relationship between the intensities, we can use the definition of intensity which is the power per unit area

           I = P / A

           P = I A

the area is of a sphere

          A = 4π r²

           

the power of the sound does not change, so we can write it for the two points

          P =  I₁ A₁ =  I₂ A₂

          I₁ r₁² = I₂ r₂²

we substitute the ratio of intensities

          I₁ r₁² = (10³ I₁ ) r₂²

         r₁² = 10³ r₂²

         

         r₂ = r₁ / √10³

         

we calculate

          r₂ = \frac{10.0}{\sqrt{10^3} }

          r₂ = 0.316 m

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allochka39001 [22]

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2 years ago
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elena-s [515]

Answer:

The speed after being pulled is 2.4123m/s

Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

W_T+W_F=K_f-K_i (1)

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W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to \frac{1}{2}mv^{2}, so if the initial velocity v_i is equal to zero, the initial kinetic energy K_i is equal to zero.

Then, replacing the values on the equation and solving for v_f, we get:

W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}

\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f

So, the speed after being pulled 3.2m is 2.4123 m/s

8 0
3 years ago
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