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icang [17]
3 years ago
13

A 1.45 kg 1.45 kg falcon catches a 0.415 kg 0.415 kg dove from behind in midair. What is their velocity after impact if the falc

on's velocity is initially 26.5 m/s 26.5 m/s and the dove's velocity is 4.35 m/s 4.35 m/s in the same direction?
Physics
1 answer:
jarptica [38.1K]3 years ago
7 0

Answer: The velocity is 21.5m/s

Explanation:

Let's call:

M1 and V1 as the mass and velocity of the falcon:

M1 = 1.45kg

V1 = 26.5m/s

M2 and V2 as the mass and velocity of the dove:

M2 = 0.415kg

V2 = 4.35m/s

Where both velocities are positive because both animals move in the same direction.

We can think that the interaction between both animals is a perfectly inelastic collision, because afther the interaction they move as one. Then, we have that the final velocity of both animals togheter is:

V = (V1*M1 + V2*M2)/(M1 + M2)

V = (1.45kg*26.5m/s + 0.415kg*4.35m/s)/(1.45kg + 0.415kg) = 21.5m/s

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2 years ago
The cavity within a copper [β = 51 × 10-6 (C°)-1] sphere has a volume of 1.180 × 10-3 m3. Into this cavity is placed 1.100 × 10-
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Cavity and benzene should be extended in equal quantities.

\to 1.18 \times 10^{-3}\times (1+ \Delta T \times 0.000051) = 1.1\times 10^{-3} \times (1+ \Delta T \times 0.00124)\\\\\to  (\frac{1.18}{1.1})\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 - 1- \Delta T \times 0.00124=0\\\\

\to 0.072+ \Delta T \times 0.000054672 - \Delta T \times 0.00124=0\\\\ \to 0.072+ \Delta T ( 0.000054672 -0.00124)=0\\\\ \to \Delta T ( 0.000054672 -0.00124)= -0.072\\\\ \to \Delta T = -\frac{0.072}{( 0.000054672 -0.00124)}\\\\ \to \Delta T = -\frac{0.072}{-0.001185328 }\\

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