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
The cheetah can run 28,25 km
Answer:
Heat of vaporization will be 22.59 j
Explanation:
We have given mass m = 10 gram
And heat of vaporization L = 2.259 J/gram
We have to find the heat required to vaporize 10 gram mass
We know that heat of vaporization is given by
, here m is mass and L is latent heat of vaporization.
So heat of vaporization Q will be = 10×2.259 = 22.59 J
Energy flows with kinetic energy
<span>The normal force acting on the bag by the floor is calculated using th data on the force used to push the bag and the coefficient of friction between the bag and the floor. The normal force is determined through the formula Force / kin.coeff. Substituting, 27 N / 0.23 is equal to 117. 39 N. The answer is D.</span>