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3 years ago

When a 20 kg explosive detonates and sends a 5 kilogram piece traveling to the right at 105 m/s

Physics

1 answer:

weeeeeb [17]3 years ago

7 0

Answer:

speed: 35m/s

direction: left

Explanation:

Assuming the right side is the positive direction:

before explosion:

P = mv = 0

after explosion:

P' = 15P + 5P

(Set the velocity of the 15kg piece after explosion as v1' and the velocity of the 5kg piece after explosion as v2')

P' = 0.75mv1' + 0.25mv2'

P' = (15kg)v' + (5kg)(105m/s)

P' = 525kg/m/s + (15kg)v1'

P = P'

525kg/m/s + (15kg)v1' = 0

(15kg)v1' = -525kg/m/s

v1' = -35m/s

speed = |-35| = 35m/s

direction is to the left since the right side is the positive direction.

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A sphere of radius 5.15 cm and uniform surface charge density +12.1 µC/m2 exerts an electrostatic force of magnitude 35.9 ✕ 10-3

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r is their separation

Re-arranging the equation, we can find the separation between the sphere and the point charge:
$r=\sqrt{ \frac{k_e Q q}{F} }= \sqrt{ \frac{(8.99 \cdot 10^9 Nm^2 C^{-2})(4.03 \cdot 10^{-8} C)(1.75 \cdot 10^{-6}C)}{35.9 \cdot 10^{-3}N} }=0.133 m=13.3 cm$

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