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wlad13 [49]
3 years ago
6

When preparing to dock, what is the safest way to stop the forward motion of your boat?

Physics
1 answer:
tiny-mole [99]3 years ago
3 0
When preparing to dock,  the safest way to stop the forward motion of your boat is to shift into reverse gear. The dock should be slowly approached and at a sharp angle.To stop use reverse and at the end p<span>ut the boat in forward gear briefly, and slowly turn the steering wheel hard away from the dock.</span>
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Which image illustrates why dark clothing helps to keep you warm on a cool, sunny day?
xeze [42]

Answer:

Explanation:

I suppose it has to do with the way the diagram is drawn. The heat does not reflect which makes both A and B incorrect.

C would have nothing to do with either reflection or refraction.

That only leaves D which is the answer.

7 0
3 years ago
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Which of these terms means an object's resistance to changing its motion?
Mila [183]

Answer:

Inertia

F=ma

Action, reaction

All of the above

A heavy object requires more force to push than a lighter object.

8 0
2 years ago
When a point charge of q is placed on one corner of a square, an electric field strength of 2 N/C is observed at the center of t
bearhunter [10]

Answer:

Explanation:

Net electric field at the centre will be zero .

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3 0
2 years ago
A supertanker ( = 1.70 × 108 kg) is moving with a constant velocity. Its engines generate a forward thrust of 7.40 × 105 N. Dete
a_sh-v [17]

Answer:

(a) 0 (b) F=1.67\times 10^9\ N

Explanation:

Given that,

Mass of a supertanker, m=1.7\times 10^8\ kg

The engine of a generate a forward thrust of, F=7.4\times 10^5\ N

(a) As the supertanker is moving with a constant velocity. We need to find the magnitude of the resistive force exerted on the tanker by the water. It is given by :

F = ma, a is the acceleration

For constant velocity, a = 0

So, F = 0

(b) The magnitude of the upward buoyant force exerted on the tanker by the water is equal to the weight of the ship.

F = mg

F=1.7\times 10^8\times 9.8\\\\F=1.67\times 10^9\ N

Hence, this is the required solution.

4 0
2 years ago
2 masses of 4kg and 2.5kg have a separation of r =1.5m. Calculate the force of gravitational attraction between them (take g as
Evgesh-ka [11]

Answer:

F = 29.64 × 10-¹¹N

Explanation:

From newton's law of gravitation which states that every object in the universe will attract each other with a force which is directly proportional to the product of their masses and inversely proportional to the square of their distance apart.

That is,   F = (Gm1m2)/ r²

From the question m1=4kg, m2 = 2.5kg, r= 1.5m,G = 6.67×10-¹¹

F=( 6.67×10-¹¹ × 4 × 2.5) / 1.5²

F = (66.7×10-¹¹) / 2.25

F = 66.7/2.25 × 10-¹¹ N

F = 29.64 × 10-¹¹N

F = 29.64 × 10-¹¹N

I hope this was helpful, please mark as brainliest

5 0
3 years ago
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