Answer:
Lens at a distance = 7.5 cm
Lens at a distance = 6.86 cm (Approx)
Explanation:
Given:
Object distance u = 12 cm
a) Focal length = 20 cm
b) Focal length = 16 cm
Computation:
a. 1/v = 1/u + 1/f
1/v = 1/20 + 1/12
v = 7.5 cm
Lens at a distance = 7.5 cm
b. 1/v = 1/u + 1/f
1/v = 1/16 + 1/12
v = 6.86 cm (Approx)
Lens at a distance = 6.86 cm (Approx)
Answer:
23 electrons
Explanation:
i just know because im a god
Answer:
Look to the explanation
Explanation:
<u><em>Work done</em></u> is is the energy transferred to or from an object by means
of a force acting on the object.
Work is positive if energy transferred to the object, and work is
negative if energy transferred from the object.
<em>Work = Force in the direction motion of object × its displacement</em>

The SI unit of the work is joule (J)
<u><em>Power</em></u> is the rate of work
<em>Power = work done ÷ time taken</em>
Power = 
Displacement (s) ÷ time (t) = velocity (v)
<em>Power = Force × velocity</em>

The SI unit of the power is watt (w)
Answer:
5.791244495 KNm
Explanation:
The height h is given by,
Potential energy, PE is given by
PE=mgh where m is mass of the woman, g is acceleration due to gravity whose value is taken as
and h is already given hence substituting 77 Kg for m we obtain
PE=21.6567095 KNm
We also know that Kinetic energy is given by
where v is the velocity and substituting v for 20.3 we obtain
KE=15.865465 KNm
Friction work is the difference between PE and KE hence
Friction work=21.6567095 KNm-15865.465 Nm=5.791244495 KNm