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Andre45 [30]
3 years ago
14

If the wind or current is pushing your boat away from the dock as you prepare to dock, which line should you secure first?

Physics
2 answers:
sveta [45]3 years ago
5 0

The line that ought to be verified first in pushing the vessel

away from the dock in readiness to dock is the bow line. At the point when the bow line is  

Further explanation:

verified, it is ideal to switch it and go to the dock, this will draw in the  

line to fix such that will enable it to swing back in the dock.

Docking With Wind or Current from dock

Approach the dock gradually at a sharp edge (around 40 degrees).  

Utilize turn around to stop when near the dock. Secure the bow line.  

Put the pontoon in forward apparatus quickly, and gradually turn the guiding wheel hard away from the dock—this will swing in the stern. Secure the stern line.

Plane:

A point exists in zero measurements. A line exists in one measurement, and we indicate a line with two. A plane exists in two measurements. We determine a plane with three. Any two of the focuses determine a linea.

Thank you so much Hope will understand how this process will happen.

Answer Details:

Subject: Physics  

Level: High School

Key Words:

Further explanation:

Docking With Wind or Current From the Dock:

Plane:

For further Evaluation

brainly.com/question/1156977

brainly.com/question/10494507

Sphinxa [80]3 years ago
4 0

The line that should be secured first in pushing the boat away from the dock in preparation to dock is the bow line. When the bow line is secured, it is best to reverse it and turn to the dock, this will engage the line to tighten in a way that will help it swing back in the dock.

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What is the total momentum of a 30 kg object traveling left at 3 m/s and a 50 kg object traveling at 2 m/s to the right?
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Answer:

there are 25 kg objective travelling at 2m/s to the right.

4 0
2 years ago
Determine the gravitational field 300km above the surface of the earth. How does this compare to the field on the earth's surfac
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The strength of the gravitational field is given by:
g= \frac{GM}{r^2}
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.

In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
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And now we can use the previous equation to calculate the field strength at that altitude:
g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2}  = 8.95 m/s^2

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4 0
3 years ago
Space scientists have a large test chamber from which all the air can be evacuated and in which they can create a horizontal uni
nexus9112 [7]

Answer:

the magnitude of the electric force on the projectile is 0.0335N

Explanation:

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= (2 * 6.0m/s * sin35º) / 9.8m/s²

= 0.702 s

The body travels for this much time and cover horizontal displacement x from the point of lunch

So, use kinematic equation for horizontal motion

horizontal displacement

x = Vcosθ*t + ½at²

2.9 m = 6.0m/s * cos35º * 0.702s + ½a * (0.702s)²

a = -2.23 m/s²

This is the horizontal acceleration of the object.

Since the object is subject to only electric force in horizontal direction, this acceleration is due to electric force only

Therefore,the magnitude of the electric force on the projectile will be

F = m*|a|

= 0.015kg * 2.23m/s²

= 0.0335 N

Thus, the magnitude of the electric force on the projectile is 0.0335N

3 0
3 years ago
Read 2 more answers
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