Answer:4A
Explanation:
Given
Mass is displace x= A units from its mean position x=0'
When it is set to free it will oscillate about its mean position with maximum amplitude A i.e. from x=-A to x=A
One cycle is completed when block returns to its original position
so first block will go equilibrium position x=0 and then to x=-A
from x=-A it again moves back to x=0 and finally back to its starting position x=A
so it travels a distance of A+A+A+A=4A
Answer:
a) w = 9.599 10⁴ rad / s
, b) v = 3.35 10¹⁶ m / s
, c) a = 3.22 10²¹ m / s²
Explanation:
For this exercise we must use the relation of angular kinematics
a) angular velocity, the distance remembered in orbit between time (period)
w = 2π r / T
w = 2 π 3.59 10¹¹ / 2.35 10⁷
w = 9.599 10⁴ rad / s
b) linear and angular velocity are related by the equation
v = w r
v = 9,599 10⁴ 3.49 10¹¹
v = 3.35 10¹⁶ m / s
c) the centripetal acceleration is
a = v² / r = w² r
a = (9,599 10⁴)² 3.49 10¹¹
a = 3.22 10²¹ m / s²
For a flower to be pollinated, pollen from an anther (which is located at the top of the stamen) needs to reach a stigma (at the top of the pistle.) Some plants are genetically capable of pollinating themselves if their own pollen reaches their own stigma; some plants are not capable of self pollination under any circumstances.
For plants that can genetically self pollinate, but would prefer not to, they can avoid this by having their pistil and pollen/stamens mature at different times. If the stamens mature first, the pollen will be dispersed by animals or wind or whatever dispersal mechanism it relies on. Then by the time the pistil is ready to be pollinated, there is no pollen left in that flower to land on the stigma.
Answer:
the final angular velocity of the platform with its load is 1.0356 rad/s
Explanation:
Given that;
mass of circular platform m = 97.1 kg
Initial angular velocity of platform ω₀ = 1.63 rad/s
mass of banana
= 8.97 kg
at distance r = 4/5 { radius of platform }
mass of monkey
= 22.1 kg
at edge = R
R = 1.73 m
now since there is No external Torque
Angular momentum will be conserved, so;
mR²/2 × ω₀ = [ mR²/2 +
(
R)² +
R² ]w
m/2 × ω₀ = [ m/2 +
(
)² +
]w
we substitute
w = 97.1/2 × 1.63 / ( 97.1/2 + 8.97(16/25) + 22.1
w = 48.55 × [ 1.63 / ( 48.55 + 5.7408 + 22.1 )
w = 48.55 × [ 1.63 / ( 76.3908 ) ]
w = 48.55 × 0.02133
w = 1.0356 rad/s
Therefore; the final angular velocity of the platform with its load is 1.0356 rad/s
Answer:
Explanation:
a) Force of friction = μ R where μ is coefficient of kinetic friction and R is reaction force
R = mg where m is mass of the block
Force of friction F = μ x mg
= .173 x 12.2 x 9.8
= 20.68 N
b ) Only force of friction is acting on the body so
deceleration = force / mass = 20.68 / 12.2 = 1.7 m /s²
acceleration = - 1.7 m /s²
c )
v² = u² - 2 a s
v = 0 , u = 3.9 m /s
a = 1.7 m /s
0 = 3.9² - 2 x 1.7 x s
s = 4.47 m