Answer:
a= 23.65 ft/s²
Explanation:
given
r= 14.34m
ω=3.65rad/s
Ф=Ф₀ + ωt
t = Ф - Ф₀/ω
= (98-0)×
/3.65
98°= 1.71042 rad
1.7104/3.65
t= 0.47 s
r₁(not given)
assuming r₁ =20 in
r₁ = r₀ + ut(uniform motion)
u = r₁ - r₀/t
r₀ = 14.34 in= 1.195 ft
r₁ = 20 in = 1.67 ft
= (1.667 - 1.195)/0.47
0.472/0.47
u= 1.00ft/s
acceleration at collar p
a=rω²
= 1.67 × 3.65²
a = 22.25ft/s²
acceleration of collar p related to the rod = 0
coriolis acceleration = 2ωu
= 2× 3.65×1 = 7.3 ft/s²
acceleration of collar p
= 22.5j + 0 + 7.3i
√(22.5² + 7.3²)
the magnitude of the acceleration of the collar P just as it reaches B in ft/s²
a= 23.65 ft/s²
Answer:
Multiple so you can test multiple hypothesis at once
Explanation:
because
<span>motion of truck constitutes of 3 travels.
1. accelerating uniformly with acceleration a1 = 2 m/s^2 until its velocity reached 20 m/s travelling a
distance of 's1' meters.
2. uniform motion with 20 m/s for a time duration t1 = 20s travelling a distance of 's2' meters.
3. uniform deceleration for t2 = 5 sec which stops the truck after travelling a distance of 's3' meters.. </span>
Answer:
42.69 N and 18.07 N
Explanation:
We are given that
Mass of ladder=6.2 kg
Length of ladder=1.97 m
Distance of Sawhorse A from one end=0.64 m
Distance of sawhorse B from other end=0.17 m
Let center of Ladder=
Now, the distance of sawhorse A from center=r=0.985-0.64=0.345 m
Distance of sawhorse B from center of ladder=0.985-0.17=0.815 m
Force one ladder due to gravity=mg=
Where 
Torque applied on Sawhorse A=
Torque applied on Sawhorse B=
In equilibrium
Total force=
