When a bullet gets embbed in a solid block the whole kinetic energy gets lost . This will be perfectly inelastic c ol lision in which only momentum is conserved.
Correct Answer - A
`(a)` At event `A` he gets a number greater than `3`
`:.P(A)=1//2`
At event `B` he gets `5` in last throw
`:.P(B)=1//6`
`:.` Required probability is
`P(E)=(1)/(6)+(1)/(2)*(1)/(6)+(1)/(2)*(1)/(2)*(1)/(6)+…..oo`...
Given: μS = 0.4, m = 0.25 kg, g = 9.8 m/s2
To find: Force (FL)
Formula: FL = μSN = μS(mg)
Calculation: From formula,
FL = 0.4 × 0.25 × 9.8 = 0.98 N
The horizontal force applied...
Given: m = 20 kg, μS = 0.3, g = 9.8 m/s2
To find: Force required (F)
Formula: FS = μSN = μSmg
Calculation: From formula,
FS = 3 × 20 × 9.8 = 58.8N
The force required to move...
Given: FL = 120N, Fk = 80N, m = 40 kg, g = 9.8 m/s2
To find:
i. Coefficient of static friction (μS)
ii. Coefficient of kinetic friction (μk)
Formulae:
i. μs = \(\frac{F_L}{N}\)
ii. μk = \(\frac{F_K}{N}\)
Calculation:
From formula (i) we get.
N =...
Given: m = 20 kg, FL = 100 N, μk = 0.4
To find:
i. Coefficient of static friction (μS)
ii. Minimum force required (Fk)
Formulae:
i. μs = \(\frac{F_L}{N}=\frac{F_L}{mg}\)
ii. μS = \(\frac{F_K}{N}\)
Calculation:
From formula (i),
μs = \(\frac{100}{20\times9.8}\) = 0.5102
From formula (ii),
Fk = μkN...
