Answer:
.012
Explanation:
Take the mass of the fish and divide it by the mass of the water:
65/.30=216.667
Divide the given speed by the value we found above:
2.5/216.667=.0115
Answer can be rounded up to .012
The speed of the archerfish immediately after it expels the drop of water is about (b) 0.012 m/s
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Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.
[tex]\large {\boxed {F = ma }[/tex]
F = Force ( Newton )
m = Object's Mass ( kg )
a = Acceleration ( m )
Let us now tackle the problem !
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The Question:
What is the speed of the archerfish immediately after it expels the drop of water?
(a) 0.0025 m/s; (b) 0.012 m/s; (c) 0.75 m/s; (d) 2.5 m/s.
Given:
mass of fish = m₁ = 65 g
mass of drop of water = m₂ = 0.30 g
speed of water = v₂ = 2.5 m/s
initial speed of water = initial speed of archerfish = u₁ = u₂ = 0 m/s
Asked:
the speed of the archerfish = v₁ = ?
Solution:
We will use Conservation of Momentum Law as follows:
[tex]m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2[/tex]
[tex]65(0) + 0.30(0) = 65v_1 + 0.30(2.5)[/tex]
[tex]0 + 0 = 65v_1 + 0.75[/tex]
[tex]65v_1 = -0.75[/tex]
[tex]v_1 = -0.75 \div 65[/tex]
[tex]\boxed{v_1 \approx -0.012 \texttt{ m/s}}[/tex]
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The speed of the archerfish immediately after it expels the drop of water is about 0.012 m/s
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Grade: High School
Subject: Physics
Chapter: Dynamics
