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
so as this ant moves
5 cm every second you multiply 5 by 120 (60 per minute as there are 60 seconds in a minute)
this is 600 cm
or
6 meters
Answer:
the aircraft must travel at a speed of <em>73.4 m/s</em> in order to create the ideal lift.
Explanation:
We will use Bernoulli's theorem in order to determine the pressure lift:
ΔP = 1/2 (ρ)(v₂² - v₁²)
the generated pressure lift is ΔP = 1000 N/m²
Therefore,
1000 = 1/2(ρ)(v₂² - v₁²)
v₂² - v₁² = 2000 / ρ
v₂² = (2000 N/m² / 1.29 kg/m³) + (62 m/s)²
v₂ = √[ (2000 N/m² / 1.29 kg/m³) + (62 m/s)² ]
<em>v₂ = 73.4 m/s </em>
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Therefore, the aircraft must travel at a speed of <em>73.4 m/s</em> in order to create the ideal lift.
Answer:
push up agaisnt you with equal force.
