A 300 g bird flying along at 5.5 m/sm/s sees a 10 g insect heading straight toward it with a speed of 26 m/s (as measured by an observer on the ground, not by the bird). The bird opens its mouth wide and enjoys a nice lunch. Required:
What is the bird's speed immediately after swallowing?

the bird's speed immediately after swallowing is 4.48 m/s

Explanation:

Given;

mass of the bird, m₁ = 300 g = 0.3 kg

initial velocity of the bird, v₁ = 5.5 m/s

mass of the insect, m₂ = 10 g = 0.01 kg

initial velocity of the insect, v₂ = - 26 m/s (negative because it is moving in opposite direction to the bird)

let the final velocity of the bird-insect system = v

Apply the principle of conservation of linear momentum for inelastic collision to determine the final velocity of the bird after swallowing.

m₁v₁ + m₂v₂ = v(m₁ + m₂)

(0.3 x 5.5) + (0.01 x - 26) = v(0.3 + 0.01)

1.65 - 0.26 = v(0.31)

1.39 = v(0.31)

v = 1.39 / 0.31

v = 4.48 m/s

Therefore, the bird's speed immediately after swallowing is 4.48 m/s

i think it is non rem sleep

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂