Answer:
μ=0.151
Explanation:
Given that
m= 3.5 Kg
d= 0.96 m
F= 22 N
v= 1.36 m/s
Lets take coefficient of kinetic friction = μ
Friction force Fr=μ m g
Lets take acceleration of block is a m/s²
F- Fr = m a
22 - μ x 3.5 x 10 = 3.5 a ( take g =10 m/s²)
a= 6.28 - 35μ m/s²
The final speed of the block is v
v= 1.36 m/s
We know that
v²= u²+ 2 a d
u= 0 m/s given that
1.36² = 2 x a x 0.96
a= 0.963 m/s²
a= 6.28 - 35μ m/s²
6.28 - 35μ = 0.963
μ=0.151
Answer:
v = 69.82 ms^-1
Explanation:
As we know,
R = vi2 sin2Ꝋ / g
vi2 =R g / sin2 Ꝋ where R is range R = 52m, Ꝋ = 3 Degrees
vi2 = 52 x 9.8 / sin 2(3) = 4875.227
v = 69.82 ms^-1
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