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Alexxx [7]
3 years ago
10

A 200-gr (7000 gr = 1 lb) bullet goes from rest to 3300 ft/s in 0.0011 s. Determine the magnitude of the impulse imparted to the

bullet during the given time interval. In addition, determine the magnitude of the average force acting on the bullet.
Engineering
1 answer:
OlgaM077 [116]3 years ago
3 0

Answer:

The force acting on the bullet F = 84000 \frac{lb ft}{s^{2} }

The value of impulse on the bullet in the given time interval P = 92.4 \frac{lb ft}{sec}

Explanation:

Mass of the bullet ( m ) = 200 gr = 0.028 lb

Initial velocity ( U ) = 0

Final Velocity ( V ) = 3300 \frac{ft}{sec}

Force acting on the bullet F = \frac{m ( V - U )}{t}

⇒ F = \frac{ 0.028 ( 3300 - 0 )}{0.0011}

⇒ F = 84000 \frac{lb ft}{s^{2} }

This is the force acting on the bullet.

Magnitude of the impulse imparted on the bullet  P = F dt  -------- (1)

Put the value of F & dt in above equation we get,

P = 84000 × 0.0011

P = 92.4 \frac{lb ft}{sec}

This is the value of impulse on the bullet in the given time interval.

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How are isometric drawings and orthographic drawings similar?
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The horizontal lines of the orthographic drawing

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4 years ago
Suppose that the weights for newborn kittens are normally distributed with a mean of 125 grams and a standard deviation of 15 gr
kherson [118]

(a) If a kitten weighs 99 grams at birth, it is at 5.72 percentile of the weight distribution.

(b) For a kitten to be at 90th percentile, the minimum weight is 146.45 g.

<h3>Weight distribution of the kitten</h3>

In a normal distribution curve;

  • 2 standard deviation (2d) below the mean (M), (M - 2d) is at 2%
  • 1 standard deviation (d) below the mean (M), (M - d) is at 16 %
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  • 2 standard deviation (2d) above the mean (M), (M + 2d) is at 98%

M - 2d = 125 g - 2(15g) = 95 g

M - d = 125 g - 15 g = 110 g

95 g is at 2% and 110 g is at 16%

(16% - 2%) = 14%

(110 - 95) = 15 g

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From 95 g to 99 g:

99 g - 95 g  = 4 g

4g x 0.93%/g = 3.72%

99 g will be at:

(2% + 3.72%) = 5.72%

Thus, if a kitten weighs 99 grams at birth, it is at 5.72 percentile of the weight distribution.

<h3>Weight of the kitten in the 90th percentile</h3>

M + d = 125 + 15 = 140 g      (at 84%)

M + 2d = 125 + 2(15) = 155 g   ( at 98%)

155 g - 140 g = 15 g

14% / 15g = 0.93%/g

84% + x(0.93%/g) = 90%

84 + 0.93x = 90

0.93x = 6

x = 6.45 g

weight of a kitten in 90th percentile = 140 g + 6.45 g  = 146.45 g

Thus, for a kitten to be at 90th percentile, the approximate weight is 146.45 g

Learn more about standard deviation here: brainly.com/question/475676

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b. i) Complete condensation of the vapor at the condenser to saturated liquid for pumping to the boiler

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