Answer:
Explanation:
Initial angular velocity ω₀ = 151 x 2π / 60
= 15.8 rad /s
final velocity = 0
Angular deceleration α = 2.23 rad / s
ω² = ω₀² - 2 α θ
0 = 15.8² - 2 x 2.23 θ
= 55.99 rad
one revolution = 2π radian
55.99 radian = 55.99 / 2 π no of terns
= 9 approx .
Answer:
Expression of work done is
Work done to move the sled is given as 187.2 J
Explanation:
As we know that the formula of work done is given as
here we know that
F = 12.6 N
d = 15.4 m
so we will have
A)acceleration is in the direction of motion
Answer:
a)Distance traveled during the first second = 4.905 m.
b)Final velocity at which the object hits the ground = 38.36 m/s
c)Distance traveled during the last second of motion before hitting the ground = 33.45 m
Explanation:
a) We have equation of motion
S = ut + 0.5at²
Here u = 0, and a = g
S = 0.5gt²
Distance traveled during the first second ( t =1 )
S = 0.5 x 9.81 x 1² = 4.905 m
Distance traveled during the first second = 4.905 m.
b) We have equation of motion
v² = u² + 2as
Here u = 0, s= 75 m and a = g
v² = 0² + 2 x g x 75 = 150 x 9.81
v = 38.36 m/s
Final velocity at which the object hits the ground = 38.36 m/s
c) We have S = 0.5gt²
75 = 0.5 x 9.81 x t²
t = 3.91 s
We need to find distance traveled last second
That is
S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m
Distance traveled during the last second of motion before hitting the ground = 33.45 m
