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
<span>D is at rest at the top of a 2 m high slope. The sled has a mass of 45 kg. The sled's potential energy is J?
</span>Answer: The sled's potential energy is 882 Joules
B. Kinetic
Kinetic energy depends on motion and mass
Answer:
N2- N1 = 
V g
Explanation:
Let's use for this exercised the equilibrium relationship
∑ F = 0
N -W = 0
N = W = mg
Let's use density redefinition
ρ = m / v
m = ρ V
N₁ = (ρ V) g
Repeat for when the liquid is submerged, in this case an upward thrust appears as a result of the water dispensed (rgow)
N₂ + B - W = 0
N₂ = W - B
B = 
g
= m / V
= V
N = ρ V g - Vg
N2 = N1 + V g
N2- N1 = V g
True because they will work good with eachother
