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zloy xaker [14]
3 years ago
11

A 0.400 kg bead slides on a straight frictionless wire and moves with a velocity of 3.50 cm/s to the right, as shown below. The

bead collides elastically with a larger 0.600 kg bead that is initially at rest. After the collision, the smaller bead moves to the left with a velocity of 0.70 cm/s.
What is the total kinetic energy of the system of beads after the collision?
A 1.40 × 10−4 J
B 2.45 × 10−4 J
C 4.70 × 10−4 J
D 4.90 × 10−4 J
Physics
1 answer:
irakobra [83]3 years ago
6 0

Answer:

B

Explanation:

KE = 1/2 m v^2 = 1/2 (0.400) (3.50^2)

= 2.45 J

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Coherent light of wavelength 525 nm passes through two thin slits that are 4.15×10^(−2) mm apart and then falls on a screen 7
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B) 8.1 cm

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sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}

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y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm

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