The focal length of given concave lens will be -26.85 cm
The height of an image to the height of an object is the ratio that is used to determine a lens' magnification. Additionally, it is provided in terms of object and image distance. It is equivalent to the object distance to image distance ratio.
Given concave lens creates a virtual image at -47.0 cm and a magnification of +1.75.
We have to find focal length
The focal length can be found out by following way:
Magnification = m = +1.75
m = hi/h
hi = -47 cm
1.75 = -47/h
h = -26.85 cm
So the focal length of given concave lens will be -26.85 cm
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The mass m of the object = 5.25 kg
<h3>Further explanation</h3>
Given
k = spring constant = 3.5 N/cm
Δx= 30 cm - 15 cm = 15 cm
Required
the mass m
Solution
F=m.g
Hooke's Law
F = k.Δx
Answer:
I = Δq / t
Explanation:
The quantity of electricity i.e charge is related to current and time according to the equation equation:
Q = It
Δq = It
Where:
Q => is the quantity of electricity i.e charge
I => is the current.
t => is the time.
Thus, we can rearrange the above expression to make 'I' the subject. This is illustrated below:
Δq = It
Divide both side by t
I = Δq / t
If one of two interacting charges is doubled, the force between the charges will double.
Explanation:
The force between two charges is given by Coulomb's law
K=constant= 9 x 10⁹ N m²/C²
q1= charge on first particle
q2= charge on second particle
r= distance between the two charges
Now if the first charge is doubled,
we get 
F'= 2 F
Thus the force gets doubled.
