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Wewaii [24]
1 year ago
14

an object is placed at 19.3 cm in front of a diverging lens with a focal length of -18.7 cm. what is the magnification? enter a

number with 2 digits behind the decimal point.
Physics
1 answer:
Bezzdna [24]1 year ago
8 0

The magnification <u>is 31.16.</u>

Magnification is the process of increasing the apparent size of something rather than its physical size. This increase is quantified by a calculated number, also called the "factor". If this number is less than 1, it means size reduction, sometimes called size reduction or reduction.

u =  -19.3

f = -18.7 cm.

m = f/f-u

    = -18.7/-18.7 +19.3

   <u>= 31.16</u>

The term magnification refers to the size of the image produced by the lens compared to the size of the object. For lenses: Magnification "m" is the ratio of image height to object height. The magnification of a lens is defined as the ratio of image height to object height. It is also given by image distance and object distance. equal to the ratio of image distance to object distance.

Learn more about magnification here:-brainly.com/question/15744335

#SPJ4

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Please help me with this question​
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Answer:

1. 12 V

2a. R₁ = 4 Ω

2b. V₁ = 4 V

3a. A = 1.5 A

3b. R₂ = 4 Ω

4. Diagram is not complete

Explanation:

1. Determination of V

Current (I) = 2 A

Resistor (R) = 6 Ω

Voltage (V) =?

V = IR

V = 2 × 6

V = 12 V

2. We'll begin by calculating the equivalent resistance. This can be obtained as follow:

Voltage (V) = 12 V

Current (I) = 1 A

Equivalent resistance (R) =?

V = IR

12 = 1 × R

R = 12 Ω

a. Determination of R₁

Equivalent resistance (R) = 12 Ω

Resistor 2 (R₂) = 8 Ω

Resistor 1 (R₁) =?

R = R₁ + R₂ (series arrangement)

12 = R₁ + 8

Collect like terms

12 – 8 =

4 = R₁

R₁ = 4 Ω

b. Determination of V₁

Current (I) = 1 A

Resistor 1 (R₁) = 4 Ω

Voltage 1 (V₁) =?

V₁ = IR₁

V₁ = 1 × 4

V₁ = 4 V

3a. Determination of the current.

Since the connections are in series arrangement, the same current will flow through each resistor. Thus, the ammeter reading can be obtained as follow:

Resistor 1 (R₁) = 4 Ω

Voltage 1 (V₁) = 6 V

Current (I) =?

V₁ = IR₁

6 = 4 × I

Divide both side by 4

I = 6 / 4

I = 1.5 A

Thus, the ammeter (A) reading is 1.5 A

b. Determination of R₂

We'll begin by calculating the voltage cross R₂. This can be obtained as follow:

Total voltage (V) = 12 V

Voltage 1 (V₁) = 6 V

Voltage 2 (V₂) =?

V = V₁ + V₂ (series arrangement)

12 = 6 + V₂

Collect like terms

12 – 6 = V₂

6 = V₂

V₂ = 6 V

Finally, we shall determine R₂. This can be obtained as follow:

Voltage 2 (V₂) = 6 V

Current (I) = 1.5 A

Resistor 2 (R₂) =?

V₂ = IR₂

6 = 1.5 × R₂

Divide both side by 1.5

R₂ = 6 / 1.5

R₂ = 4 Ω

4. The diagram is not complete

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