Work = (500N)(15m)=7500J Power= (7500J)/(20s) =375 W
Hopefully this helps
Answer: a) 6.67cm/s b) 1/2
Explanation:
According to law of conservation of momentum, the momentum of the bodies before collision is equal to the momentum of the bodies after collision. Since the second body was initially at rest this means the initial velocity of the body is "zero".
Let m1 and m2 be the masses of the bodies
u1 and u2 be their velocities respectively
m1 = 5.0g m2 = 10.0g u1 = 20.0cm/s u2 = 0cm/s
Since momentum = mass × velocity
The conservation of momentum of the body will be
m1u1 + m2u2 = (m1+m2)v
Note that the body will move with a common velocity (v) after collision which will serve as the velocity of each object after collision.
5(20) + 10(0) = (5+10)v
100 + 0 = 15v
v = 100/15
v = 6.67cm/s
Therefore the velocity of each object after the collision is 6.67cm/s
b) kinectic energy of the 10.0g object will be 1/2MV²
= 1/2×10×6.67²
= 222.44Joules
kinectic energy of the 5.0g object will be 1/2MV²
= 1/2×5×6.67²
= 222.44Joules
= 111.22Joules
Fraction of the initial kinetic transferred to the 10g object will be
111.22/222.44
= 1/2
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:
v = 6t² + t + 2, s = 2t³ + ½ t² + 2t
59 m/s, 64.5 m
Explanation:
a = 12t + 1
v = ∫ a dt
v = 6t² + t + C
At t = 0, v = 2.
2 = 6(0)² + (0) + C
2 = C
Therefore, v = 6t² + t + 2.
s = ∫ v dt
s = 2t³ + ½ t² + 2t + C
At t = 0, s = 0.
0 = 2(0)³ + ½ (0)² + 2(0) + C
0 = C
Therefore, s = 2t³ + ½ t² + 2t.
At t = 3:
v = 6(3)² + (3) + 2 = 59
s = 2(3)³ + ½ (3)² + 2(3) = 64.5
