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4vir4ik [10]
3 years ago
15

A 27.3 g marble sliding to the right at 21.0 cm/s overtakes and collides with a 11.7 g marble moving in the same direction at 12

.6 cm/s. After the collision, the 11.7 g marble moves to the right at 23.7 cm/s. Find the velocity of the 27.3 g marble after the collision. Answer in units of cm/s.
Chemistry
1 answer:
NISA [10]3 years ago
4 0

Answer:

The answer to this is

The velocity of the 27.3Kg marble after collision is = 16.24 cm/s

Explanation:

To solve the question, let us list out the given variables and their values

Mass of first marble m1 = 27.3g

Velocity of the first marble v1 = 21.0 cm/s

Mass of second marble m2 = 11.7g

Velocity of the second marble v2 = 12.6 cm/s

After collision va1 = unknown and va2 = 23.7 cm/s

From Newton's second law of motion, force = rate of change of momentum produced

Hence m1v1 + m2v2 = m1va1 + m2va2 or

va1 = (m1v1 + m2v2 - m2va2)÷m2 or (720. 72-277.29)÷m1 → va1 = 16.24 cm/s

The velocity of the 27.3Kg marble after collision is = 16.24 cm/s

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Answer:

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

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Length of the base = l

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V=\frac{1}{3}\times (\frac{6h}{13})^2\times h

V=\frac{12}{169}h^3

Differentiating V with respect to dt:

\frac{dV}{dt}=\frac{d(\frac{12}{169}h^3)}{dt}

\frac{dV}{dt}=3\times \frac{12}{169}h^2\times \frac{dh}{dt}

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\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times (4 cm)^2}

=13.20 cm/s

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

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