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Setler79 [48]
3 years ago
7

Imagine you leave the dock on a fishing boat that travels east for 30 kilometers when the captain realizes he left the fishing r

ods on the dock. It takes him 40 minutes to travel out against the current but only 20 minutes to return to the dock.
Physics
1 answer:
telo118 [61]3 years ago
3 0

Answer:

Average speed is 60 km/hr whereas average velocity is 0 km/ hr.

Explanation:

The average speed of the boat is 60 km/ hr while on the other hand, the average velocity of the boat is 0 km/ hr because average speed is the total distance covered by the boat in total time and average velocity is the displacement covered by the boat in total time. The total distance covered by the boat is 60 km and the total time is 60 minutes which is equals to one hour so the answer is 60 km/hr whereas the displacement is the shortest distance between initial and final position which is 0 in this case so 0 divided by 60 minutes or one hour is also 0.

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Speed training increases one's maximum velocity.<br> a. True<br> b. False
Harrizon [31]
That is absolutely true.
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3 years ago
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The energy a sled has when you pull it back up a hill is best described as
yawa3891 [41]
The answer is d because you are using energy to pull the sled back up, which is mechanical energy
5 0
3 years ago
What would be your estimate of the age of the universe if you measured a value for Hubble's constant of H0 = 30 km/s/Mly ? You c
Maksim231197 [3]

Answer:

The age of the universe would be 9.9 billion years

Explanation:

We can calculate an estimate for the age of the Universe from Hubble's Law. Let's suppose the distance between two galaxies is D and the apparent velocity with which they are separating from each other is v. At some point, the galaxies were touching, and we can consider that time the moment of the Big Bang.

Thus, the time it has taken for the galaxies to reach their current separations is:

\displaystyle{t=D/v}

and from Hubble's Law:

v =H_0D

Therefore:

\displaystyle{t=D/v=D/(H_0\times D)=1/H_0}

With the given value for the Hubble's constant we have:

H_0=(30\ km/s/Mly) \times (1 Mly/ 9.461 \times 10^{18} km) = 3.17\times 10^{-18}\ 1/s

and thus,

t=1/H_0 = 1/(3.17\times 10^{-18} 1/s) = 0.315 \times 10^{18}\ s \approx 9988584474.8858\ years \approx 9.9\ billion\ years

6 0
2 years ago
Newton's First and Second Laws of Motion:Question 3Kepler-10b, the first confirmed terrestrial extrasolar planet, is about 564 l
OleMash [197]

Answer:

F = 85696.5 N = 85.69 KN

Explanation:

In this scenario, we apply Newton's Second Law:

Unbalanced\ Force = ma\\Upthrust - Weight\ of\ Space\ Craft = ma\\F - W = ma\\F - mg = ma\\F = m(g + a)\\

where,

F = Upthrust = ?

m = mass of space craft = 5000 kg

g = acceleration due to gravity on surface of Kepler-10b = (1.53)(9.81 m/s²)

g = 15.0093 m/s²

a = acceleration required = 2.13 m/s²

Therefore,

F = (5000\ kg)(15.0093\ m/s^{2} + 2.13\ m/s^{2})\\

<u>F = 85696.5 N = 85.69 KN</u>

8 0
3 years ago
What is your displacement if you travel 5 miles north to the store
kumpel [21]

Answer:

5miles North

Explanation:

The displacement to the store is 5miles northward.

Displacement is the distance traveled in a specific direction. Displacement is a vector quantity.

A vector has both magnitude and direction.

The magnitude is the amount of that quantity

The direction is its orientation from a reference.

 Therefore, the displacement is 5miles north

3 0
2 years ago
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