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mote1985 [20]
3 years ago
13

Two astronauts on opposite ends of a spaceship are comparing lunches. One has an apple, the other has an orange. They decide to

trade. Astronaut 1 tosses the 0.132 kg apple toward astronaut 2 with a speed of 1.25 m/s. The 0.143 kg orange is tossed from astronaut 2 to astronaut 1 with a speed of 1.14 m/s. Unfortunately, fruits collide, sending the orange off with a speed 1.03 m/s and an angle of 43.0° with respect to its original direction of motion. Using conservation of linear momentum, find the final speed and direction of the apple. Assume an elastic collision occurs. Give the apple’s direction relative to its original direction of motion.
Physics
1 answer:
lbvjy [14]3 years ago
6 0

Answer:

Explanation:

We shall consider direction towards left as positive Let the required velocity be v and let v makes an angle φ

Applying law of conservation of momentum along direction of original motion

m₁ v₁  - m₂ v₂ = m₂v₃ - m₁ v₄

0.132 x 1.25 - .143 x 1.14 = 1.03 cos43 x .143 - v cos θ

v cos θ = .8

Applying law of conservation of momentum along direction perpendicular to direction of original motion

1.03 sin 43 x .143 = .132 x v sinθ

v sinθ = .76

squaring and adding

v² = .76 ² + .8²

v = 1.1 m /s

Tan θ = .76 / .8

θ = 44°

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Answer:

0.71 kg

Explanation:

L = length of the steel wire = 3.0 m

d = diameter of steel wire = 0.32 mm = 0.32 x 10⁻³ m

Area of cross-section of the steel wire is given as

A = (0.25) πd²

A = (0.25) (3.14) (0.32 x 10⁻³)²

A = 8.04 x 10⁻⁸ m²

ΔL = change in length of the wire = 1.3 mm = 1.3 x 10⁻³ m

Y = Young's modulus of steel = 20 x 10¹⁰ Nm⁻²

m = mass hanging

F = weight of the mass hanging

Young's modulus of steel is given as

Y = \frac{FL}{A\Delta L}

20\times 10^{10} = \frac{F(3)}{(8.04\times 10^{-8})(1.3\times 10^{-3})}

F = 6.968 N

Weight of the hanging mass is given as

F = mg

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A spinning top initially spins at 16rad/s but slows down to 12rad/s in 18s, due to friction. If the rotational inertia of the to
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Answer:

The  change  in angular momentum is \Delta  L  = 0.0016 \ kgm^2/s

Explanation:

From the question we are told that

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      The angular velocity after it slow down is  w_2 =  12 \ rad/s

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