Answer:
Kinectic friction is the answer to the lower part of the question
Answer:
Explanation:
For image formation in objective lens
object distance u = 14 +1 = 15 mm
focal length f = 14 mm .
image distance v = ?
lens formula

Putting the values

v = 210 mm .
B )
magnification = v / u
= 210 / 15
= 14
size of image = 14 x 1.1 mm
= 15.4 mm
= 15 mm approx
C )
For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .
21 mm is the answer .
D )
overall magnification =

D = 25 cm , f_e = focal length of eye piece
= 14 x 250 / 21
= 166.67
= 170 ( in two significant figures )
Answer:
a) 1.3 rad/s
b) 0.722 s
Explanation:
Given
Initial velocity, ω = 0 rad/s
Angular acceleration of the wheel, α = 1.8 rad/s²
using equations of angular motion, we have
θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²
where
θ2 - θ1 = 53.2 rad
t2 - t1 = 7s
substituting these in the equation, we have
θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²
53.2 =ω(0) * 7 + 1/2 * 1.8 * 7²
53.2 = 7.ω(0) + 1/2 * 1.8 * 49
53.2 = 7.ω(0) + 44.1
7.ω(0) = 53.2 - 44.1
ω(0) = 9.1 / 7
ω(0) = 1.3 rad/s
Using another of the equations of angular motion, we have
ω(0) = ω(i) + α*t1
1.3 = 0 + 1.8 * t1
1.3 = 1.8 * t1
t1 = 1.3/1.8
t1 = 0.722 s
Answer:
Distance, d = 61.13 ft
Explanation:
It is given that,
Initial speed of the car, u = 50 mi/h = 73.34 ft/s
Finally, it stops i.e. v = 0
Deceleration of the car, 
We need to find the distance covered before the car comes to a stop. Let the distance is s. It can be calculated using third law of motion as :



s = 61.13 ft
So, the distance covered by the car before it comes to rest is 61.13 ft. Hence, this is the required solution.