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Alexeev081 [22]
3 years ago
15

a boat travels 40 miles in two hours, speeds up to travel the next 80 miles in three hours then slows down to travel the last 40

miles in three hours. What is the average speed of the boat the entire trip?
Physics
2 answers:
Bumek [7]3 years ago
6 0

Answer:

Average speed, v = 2 miles/hour

Explanation:

It is given that, a boat travels 40 miles in two hours, speeds up to travel the next 80 miles in three hours then slows down to travel the last 40 miles in three hours.

Total distance travelled, d = 40 + 80 + 40 = 160 miles

Total time taken by the boat, t = 2 + 3 + 3 = 8 hours

We need to find the average speed of the boat the entire trip. It is given by total distance travelled by the boat divided by total time taken. Mathematically,

v=\dfrac{d}{t}

v=\dfrac{160\ miles}{8\ hours}

v = 2 miles/hour

So, the average speed of the boat the entire trip is 2 miles/hour. Hence, this is the required solution.

Over [174]3 years ago
5 0

Answer:

The average speed of the boat the entire trip is 20 miles/hour.

Explanation:

Given that,

A boat travels 40 miles in 2 hours next 80 miles in three hours and last 40 miles in three hours.

We need to calculate the average speed of the boat the entire trip

Using formula of average speed

v_{avg}=\dfrac{D}{T}

Where, D = total distance

T = total time

Put the value into the formula

v_{avg}=\dfrac{40+80+40}{2+3+3}

v_{avg}=\dfrac{160}{8}

v_{avg}=20\ miles/hour

Hence, The average speed of the boat the entire trip is 20 miles/hour.

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∴                                         t = d/v

Substituting in the above values in the given equation

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The acceleration is given by the formula

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Two spherical asteroids have the same radius R. Asteroid 1 has mass M and asteroid 2 has mass 1.97·M. The two asteroids are rele
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0.536\sqrt{\frac{GM}{R}}

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We have to find the speed of second asteroid just before they collide.

According to law of conservation of momentum

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According to law of conservation of energy

Gm_1m_2(\frac{1}{d'}-\frac{1}{d})=\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2

GM(1.97M)(\frac{1}{2R}-\frac{1}{13.63R})=\frac{1}{2}M(-1.97v_2)^2+\frac{1}{2}(1.97M)v^2_2

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v_2=\sqrt{\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}}

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