"If one increases the force on an object, its acceleration increases too because the push it feels is greater"
We have the 2nd law of Newton that relates the 3 concepts; F=m*a. We have that if the mass of an object increases (put weight in luggage), the accelearation decreases; in fact it is inversely proportional to the mass. Hence if the mass is doubled, acceleration is halved. Accelerations is proportional to force; if one doubles the force, the acceleration doubles too.
Electric potential = work done/charge of electron = 2.18×10⁻¹⁸/1.6×10⁻¹⁹
= 13.625 V
The complete and comprehensive solution is attached.
Answer:
The distribution is as depicted in the attached figure.
Explanation:
From the given data
- The plane wall is initially with constant properties is initially at a uniform temperature, To.
- Suddenly the surface x=L is exposed to convection process such that T∞>To.
- The other surface x=0 is maintained at To
- Uniform volumetric heating q' such that the steady state temperature exceeds T∞.
Assumptions which are valid are
- There is only conduction in 1-D.
- The system bears constant properties.
- The volumetric heat generation is uniform
From the given data, the condition are as follows
<u>Initial Condition</u>
At t≤0

This indicates that initially the temperature distribution was independent of x and is indicated as a straight line.
<u>Boundary Conditions</u>
<u>At x=0</u>
<u />
<u />
This indicates that the temperature on the x=0 plane will be equal to To which will rise further due to the volumetric heat generation.
<u>At x=L</u>
<u />
<u />
This indicates that at the time t, the rate of conduction and the rate of convection will be equal at x=L.
The temperature distribution along with the schematics are given in the attached figure.
Further the heat flux is inferred from the temperature distribution using the Fourier law and is also as in the attached figure.
It is important to note that as T(x,∞)>T∞ and T∞>To thus the heat on both the boundaries will flow away from the wall.
Answer:
Explanation:
b ) First is concave lens with focal length f₁ = - 12 cm .
object distance u = - 20 cm .
Lens formula
1 / v - 1 / u = 1 / f
1 / v + 1 / 20 = -1 / 12
1 / v = - 1 / 20 -1 / 12
= - .05 - .08333
= - .13333
v = - 1 / .13333
= - 7.5 cm
first image is formed before the first lens on the side of object.
This will become object for second lens
distance from second lens = 7.5 + 9 = 16.5 cm
c )
For second lens
object distance u = - 16.5 cm
focal length f₂ = + 12 cm ( lens is convex )
image distance = v
lens formula ,
1 / v - 1 / u = 1 / f₂
1 / v + 1 / 16.5 = 1 / 12
1 / v = 1 / 12 - 1 / 16.5
= .08333- .0606
= .02273
v = 1 / .02273
= 44 cm ( approx )
It will be formed on the other side of convex lens
distance from first lens
= 44 + 9 = 53 cm .
magnification by first lens = v / u
= -7.5 / -20 = .375 .
magnification by second lens = v / u
= 44 / - 16.5
= - 2.67
d )
total magnification
= .375 x - 2.67
= - 1.00125
height of final image
= 2.50 mm x 1.00125
= 2.503mm
e )
The final image will be inverted with respect to object because total magnification is negative .