When the object is at the focal point the angular magnification is 2.94.
Angular magnification:
The ratio of the angle subtended at the eye by the image formed by an optical instrument to that subtended at the eye by the object when not viewed through the instrument.
Here we have to find the angular magnification when the object is at the focal point.
Focal length = 6.00 cm
Formula to calculate angular magnification:
Angular magnification = 25/f
= 25/ 8.5
= 2.94
Therefore the angular magnification of this thin lens is 2.94
To know more about angular magnification refer:: brainly.com/question/28325488
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Stan looga
this is correct answer
This can be seen as a trick question because heat engines can typically never be 100 percent efficient. This is due to the presence of inefficiencies such as friction and heat loss to the environment. Even the best heat engines can only go up to around 50% efficiency.
<span>The rate at which energy is transferred is called "power."</span>
Answer:
0.26
Explanation:
Given that :
Diameter of ball = 3.81 cm = 3.81/100 = 0.0381 m
Radius (r) = 0.0381 / 2 = 0.01905 m
Average density of ball (Db) = 0.0842 g/cm³ = (0.0842 / 1000)kg / 10^-6 = 0.0842/ 1000 * 10^6 = 84.2kg/m³
Density of water (Dw) = 1000kg/m³
Volume of hollow ball: (4/3) * pi * r³
V = (4/3) * π * 0.01905^3
V = 0.0000289583 m³
Required force = (Dw * V * g) - (Db * V * g)
= (1000 * 0.0000289583 * 9.8) - (84.2 * 0.0000289583 * 9.8)
= 0.259896109172
= 0.2598 N
