Please answer this question.

What is the magnification of a virtual image if the image is 10.0 cm from a lens and the object is 30 cm from the lens

7nadin3 [17]

Answer:

The magnification of a virtual image is -0.33

Explanation:

We can calculate the magnification as the ratio of height of the image to the height of the object. It can be also stated as the ratio of size of the image obtained to the original size of the object. Also this ratio can be equated to the ratio of image distance to object distance. So magnification can be termed as ratio of image distance to object distance. As in the present case, the image distance or the distance of the formation of image from the lens is 10 cm and the distance of object from the lens or the object distance is 30 cm. Then, $Magnification =-\frac{Image distance}{Object distance}=-\frac{10}{30}=-0.33$ .

So, the image will be diminished to 0.33 times the original size as the magnification of the virtual image is obtained as -0.33.

5 0

3 years ago

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By what factor is the intensity of sound at a rock concert louder than that of a whisper when the two intensity levels are 120 d

lyudmila [28]

Answer:

d) $7.94\times 10^{9}$

Explanation:

β₁ = sound level of sound at rock concert = 120 dB

β₂ = sound level of sound due to whisper = 21 dB

I₁ = Intensity of sound at rock concert

I₂ = Intensity of sound due to whisper

sound level of sound at rock concert is given as

$\beta _{1} = 10 log\left ( \frac{I_{1}}{10^{-12}} \right )$

$120 = 10 log\left ( \frac{I_{1}}{10^{-12}} \right )$

$12 = log\left ( \frac{I_{1}}{10^{-12}} \right )$ Eq-1

sound level due to whisper is given as

$\beta _{2} = 10 log\left ( \frac{I_{2}}{10^{-12}} \right )$

$21 = 10 log\left ( \frac{I_{2}}{10^{-12}} \right )$

$2.1 = log\left ( \frac{I_{2}}{10^{-12}} \right )$ Eq-2

subtracting Eq-2 from Eq-1

$12 - 2.1 = log\left ( \frac{I_{1}}{10^{-12}} \right ) - log\left ( \frac{I_{2}}{10^{-12}} \right )$

$9.9 = log\left ( \frac{I_{1}}{I_{2}} \right )$

$\left ( \frac{I_{1}}{I_{2}} \right ) = 7.94\times 10^{9}$

6 0

3 years ago

What was the resistance of the resistor in Part 1? Include your current-voltage plot with the measured slope in your response. E

Karo-lina-s [1.5K]

Answer:

calculate the slope of the line and the slope is equal to the resistance of the circuit m=R

Explanation:

One method for measuring the resistence is to use ohm's law

V = R I

y = mx +b

b=0

m = R

Where we measure various values of voltage and current, with these values we can make a graph of V vs I where we calculate the slope of the line and the slope is equal to the resistance of the circuit

7 0

4 years ago

A 2.5-kg book slides horizontally and falls from a shelf 3.0 m above the floor. How much work does the force of gravity do on th

frutty [35]

Answer:

73.5 N

Explanation:

W = PE = mgh = 2.5 * 9.8 * 3.0 = 73.5 N

7 0

3 years ago

Example 2: a horizontal cylindrical drum is 2.00 m in diameter and 4.00 m in length. the drum is partially filled with benzene (

sesenic [268]

To find the mass of benzene, we must first solve for the volume of benzene inside the cylindrical drum. To calculate the volume of the partially filled cylindrical drum, refer to the working formula below:

Volume = L {(R^2)cos-1[(R-H)/R)] - (R-H)(2RH-H^2)^0.5}

Where:
L = length of the drum = 2 meters
R = radius of the drum = 4 meters
H = height of the liquid = 0.85 meter

substituting the given to the formula, we get

Volume = 5.087 m3

To solve for the mass of benzene, we must multiply the volume with the density.

Density = 0.879 g/cm3 or 879 kg/m3

Mass of benzene = Volume x Density
Mass of benzene = 5.087 x 879 = 4,471.5 kg

<em>ANSWER: Mass of benzene = 4,471.5 kg</em>

7 0

3 years ago

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