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3 years ago

A bullet 2cm log is fired at 420m/s and passes straight a 10cm thick board exiting at 280m/s

2 answers:

7 0

Solving for the acceleration of the bullet

acceleration = (vf^2 – vi^2) / 2d

acceleration = ((280 m/s)^2 – (420 m/s)^2) / (2 * 0.12 m)

acceleration = (78400 - 176400) / 0.24 m

acceleration = -98000 / 0.24

acceleration = -408333 m/s^2

Solving for contact time with board

t^2 = 2d/a

t^2 = 2 * 0.12 m / 408333 m/s^2

t^2 = 0.24 m / 408333 m/s^2

t^2 = 5.8775558 x 10^-7

t = 0.0007666 s or 767 microseconds

(I was only able to do A and B)

Alexeev081 [22]3 years ago

6 0

Answer:

Explanation:

(a)Solving for the acceleration of the bullet

acceleration = (vf^2 – vi^2) / 2d

acceleration = ((280 m/s)^2 – (420 m/s)^2) / (2 * 0.12 m)

acceleration = (78400 - 176400) / 0.24 m

acceleration = -98000 / 0.24

acceleration = -408333 m/s^2

(a)Solving for contact time with board

t^2 = 2d/a

t^2 = 2 * 0.12 m / 408333 m/s^2

t^2 = 0.24 m / 408333 m/s^2

t^2 = 5.8775558 x 10^-7

t = 0.0007666 s or 767 microseconds

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Answer:

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Explanation:

From the question we are told that

The wavelength of the light source is $\lambda = 700 \ nm = 700 *10^{-9} \ m$

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3 years ago

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Answer:

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3 years ago

Ten high-technology batteries are tested for 200 hours each. One failed at 20 hours; another failed at 140 hours; all others com

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Answer:

Failure rate = 20%

MTBF = 880 hours

Explanation:

given data

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tested = 200 hours

one failed = 20 hours

another fail at = 140 hours

solution

we know that Mean Time between Failures is express as = (Total up time) ÷ (number of breakdowns) ....................1

so here Total up time will be

Total up time = 200 × 10

Total up time = 2000

and here

Number of breakdown = 1 at 20 hour and another at 140 hour = 2

so it will be = (Total up time) ÷ (number of breakdowns) .......2

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gap between occurrences = 120 hour

and

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and

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3 years ago

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Answer:

t = 1.58 s

Explanation:

given,

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v = 56 x 0.278 = 15.57 m/s

distance, d = 65 m

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