0.013*30a 340 g bird flying along at 6.0 m/s sees a 13 g insect heading straight toward it with a speed of 30 m/s. the bird opens its mouth wide and enjoys a nice lunch. what is the bird's speed immediately after swallowing?

The speed of the bird immediately after swallow = 4.67 m/s

Explanation:

According to the law of momentum,

initial momentum before collision(swallow) = final momentum after collision (swallow)

Note: The collision between the bird and the insect is inelastic, as both moves with the same velocity after collision (swallow)

m₁u₁ - m₂u₂ =V (m₁ + m₂) Equation 1

V = (m₁u₁-m₂u₂)/(m₁+m₂) Equation 2

Note: The bird and the insect moves in opposite direction

where m₁ = mass of the bird, m₂ = mass of the insect, u₁ = initial velocity of the bird, u₂ = initial velocity of the insect, V = common velocity of the bird and the insect.

Given: m₁ = 340 g = (340/1000) kg = 0.34 kg, m₂ = 13 g = 0.013 kg, u₁ = 6 m/s, u₂ = 30 m/s

Substituting these values into equation 2

V = [0.34(6) -0.013(30)]/(0.34+0.013)

V = (2.04-0.39)/0.353

V = 1.65/0.353

V = 4.67 m/s

Thus the speed of the bird immediately after swallow = 4.67 m/s

answer: option c. i.e. electric current flowing “downhill” from the negative electrode to the positive electrode in a battery

explanation:

the water analogy to the circuits define the pipes (water flows) to the electric current. now, we know that the electric current flows from the positive terminal to the circuit (as shown in the diagram attached).

hence, option c is the correct answer.