Each BB 0.349 grams and 5.386 grains. 349 mg & 5.386 grains each steel BBs.
(a)
KE = m v^2 / 2 = (1200 kg)(20 m/s)^2 / 2 = 240,000 J
(b)
The energy is entirely dissipated by the force of friction in the brake system.
(c)
W = delta KE = KEf - KEi = (0 - 240,000) J = -240,000 J
(d)
Fd = delta KE
F = (delta KE) / d = (-240,000 J) / (50 m) = -4800 N
The magnitude of the friction force is 4800 N.
Answer:
Explanation:
Let the first height be h . second height .75h
third height .75h . fourth height .75²h
fifth height .75²h , sixthth height .75³ and so on
Total distance consists of two geometric series as follows
1 ) first series
h + .75h + .75²h + .75³h......
2 ) second series
.75h +.75²h +.75³h + .75⁴h .......
Sum of first series :
first term a = h , commom ratio r = .75
sum = a / (1 - r )
= h / 1 - .75
= h / .25
4h
sum of second series :--
first term a = .75 h , commom ratio r = .75
sum = a / (1 - r )
= .75h / 1 - .75
= .75h / .25
3h
Total of both the series
= 4h + 3h
= 7h .
h = 1 m
Total distance = 7 m
Answer: D) dedicate as many hours as needed to the work.
Explanation:
edge 2020
