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Gre4nikov [31]
2 years ago
5

A magnifier has a magnification of 8 x. How far from the lens should an object be placed so that it’s virtual image is at the ne

ar-point distance of 25 cm?
Physics
1 answer:
larisa86 [58]2 years ago
8 0

The distance of the object from the lens is 3.12 cm.

<h3>What is magnification?</h3>

Magnification is the process of of enlarging an apparent size of an object.

<h3>Object distance</h3>

The object distance is calculated using the following lens formula;

M = \frac{V}{U}

where;

  • M is the magnification
  • V is the image distance
  • U is object distance

- 8 = \frac{-25}{U} \\\\U = \frac{-25}{-8} \\\\U = 3.12 \ cm

Thus, the distance of the object from the lens is 3.12 cm.

Learn more about object distance here: brainly.com/question/24894435

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3 years ago
Tina has been dieting for a total of 13 weeks she lost 3 lb on the first week of her diet but gain the back of pound on the seco
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D. 24 lb

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6 0
3 years ago
A heavy piece of hanging sculpture is suspended by a 90 cm-long, 5.0 g steel wire. When the wind blows hard, the wire hums at it
kupik [55]

Answer: The mass of the sculpture is 11.8kg

Explanation:

Using the equation of fundamental frequency of a taut string.

f = (1/2L)*√(T/μ) .... (Eqn1)

Where

f= frequency in Hertz =80Hz

T = Tension in the string = Mg

M represent the mass of the substance (sculpture) =?

g= 9.8m/s^2

L= Length of the string=90cm=0.9m

μ= mass density = mass of string /Length of string

mass of string =5g=0.005kg

L=0.9m

μ=0.005/0.9 = 0.0056kg/m

Using (Eqn1)

80= 1/(2*0.9) √(T/0.0056)

144= √(T/0.0056)

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Recall that T =Mg

116.12= M * 9.8

M=116.12/9.8

M= 11.8kg

Therefore the mass of the sculpture is 11.8kg

4 0
3 years ago
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