Answer:

Explanation:
Given:
- mass of car,

- distance of skidding after the application of brakes,

- coefficient of kinetic friction,

<u>So, the energy dissipated during the skidding of car:</u>
<em>Frictional force:</em>

where N = normal reaction by ground on the car


<em>Now from the work-energy equivalence:</em>


is the dissipated energy.
Answer:
<em>0.45 mm</em>
Explanation:
The complete question is
a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?
A) 0.45 mm
B) 0.63 mm
C.) 0.68 mm
D) 0.91 mm
Current in the fuse is 1.0 A
Current density of the fuse when it melts is 620 A/cm^2
Area of the wire in the fuse = I/ρ
Where I is the current through the fuse
ρ is the current density of the fuse
Area = 1/620 = 1.613 x 10^-3 cm^2
We know that 10000 cm^2 = 1 m^2, therefore,
1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2
Recall that this area of this wire is gotten as
A = 
where d is the diameter of the wire
1.613 x 10^-7 = 
6.448 x 10^-7 = 3.142 x 
=
d = 4.5 x 10^-4 m = <em>0.45 mm</em>
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 .