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2 years ago

8

A nautical mile is 6076 feet, and 1 knot is a unit of speed equal to 1 nautical mile/hour. How fast is a boat going 8 knots goin

g in feet/s

1 answer:

mezya [45]2 years ago

5 0

Answer:

The speed of the boat is equal to 13.50 ft/s.

Explanation:

given,

1 nautical mile = 6076 ft

1 knot = 1 nautical mile /hour

1 knot = 6076 ft/hr

speed of boat = 8 knots

8 knots = 8 nautical mile /hour

= $8 \times \dfrac{6076\ ft}{1\nautical\ mile}\times \dfrac{1\ hour}{60\times 60\ s}$

= 13.50 ft/s

The speed of the boat is equal to 13.50 ft/s.

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The speed of sound waves

neonofarm [45]

B. Is faster in solids than liquids would be the correct answer because the molecules in solids are much closer and can pass along energy faster and more effectively.

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2 years ago

A city of punjab has 15 percemt chance of wet weather on any given day. What is probability that it will take a week for it thre

sesenic [268]

Answer:

The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447

Explanation:

We are given that A city of Punjab has 15 percent chance of wet weather on any given day.

So, Probability of wet weather = 0.15

Probability of not being a wet weather = 1-0.15 =0.85

We are supposed to find probability that it will take a week for it three wet weather on 3 separate days

Total number of days in a week = 7

We will use binomial over here

n = 7

p =probability of failure = 0.15

q = probability of success=0.85

r=3

Formula : $P(r=3)=^nC_r p^r q ^{n-r}$

$P(r=3)=^{7}C_{3} (0.15)^3 (0.85)^{7-3}\\P(r=3)=\frac{7!}{3!(7-3)!} (0.15)^3 (0.85)^{7-3}\\P(r=3)=0.06166$

Standard deviation = $\sqrt{n \times p \times q}$

Standard deviation = $\sqrt{7 \times 0.15 \times 0.85}$

Standard deviation =0.9447

Hence The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447

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2 years ago

Help with science for A P E X

eimsori [14]

Answer:

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Explanation:

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2 years ago

A train moving west with an initial velocity of 20 m/s accelerates at 4 m/s' for 10 secon

MArishka [77]

Answer:

Distance of 400m.

Explanation:

Use your kinematics equation to solve for distance (we can use kinematics b/c acceleration is constant).

d = (initial velocity x time) + 1/2 at^2

d = (20 x 10) + 1/2 (4) (10)^2

d = 200 + 200

d = 400 m

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1 year ago

A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stre

Brut [27]

Answer:

The time taken is $t = 0.356 \ s$

Explanation:

From the question we are told that

The length of steel the wire is $l_1 = 31.0 \ m$

The length of the copper wire is $l_2 = 17.0 \ m$

The diameter of the wire is $d = 1.00 \ m = 1.0 *10^{-3} \ m$

The tension is $T = 122 \ N$

The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

$t = t_s + t_c$

Where $t_s$ is the time taken to transverse the steel wire which is mathematically represented as

$t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ]$

here $\rho_s$ is the density of steel with a value $\rho_s = 8920 \ kg/m^3$

So

$t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_s = 0.235 \ s$

And

$t_c$ is the time taken to transverse the copper wire which is mathematically represented as

$t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ]$

here $\rho_c$ is the density of steel with a value $\rho_s = 7860 \ kg/m^3$

So

$t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_c =0.121$

So

$t = t_c + t_s$

$t = 0.121 + 0.235$

$t = 0.356 \ s$

4 0

2 years ago