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3 years ago

6

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

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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Can a body having larger mass be hotter than another of smaller mass.expalain

Rom4ik [11]

Answer:

it would make sense because a larger body could produce more body heat.

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2 years ago

A weightlifter lifts a set of weights a vertical distance of 2.00m.If a constant net force of 350 N is exerted on the weights,wh

MrMuchimi

Answer:

<em>W=700 Joule</em>

Explanation:

<u>Physics Work </u>

Is the dot product of the force vector by the displacement vector

$W=\vec F \cdot \vec r$

When both the force and the displacements are pointed in the same direction, the formula reduces to its scalar version

$W=F.d$

The weightlifter is applying a net force of 350 N to lift the weights a distance of 2 m, thus the net work done is

$W=350\ N\ .\ 2\ m=700\ Joule$

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3 years ago

A car starting from rest accelerated at a rate of 0 .5 m/s up to 2 km what would be the final velocity of the car and how much t

Julli [10]

From equation of motion v^2 = u^2 +2aS

Hence, the final velocity is 40 m/s.

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2 years ago

Read 2 more answers

Isla’s change in velocity is 30 m/s, and Hazel has the same change in velocity. Which best explains why they would have differen

irina [24]

Answer:

B

Explanation:

....

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2 years ago

Read 2 more answers

Block A weighs 14 lb and block B weighs 5 lb If B is moving downward with a velocity (vB)1 = 3 ft/s at t = 0, determine the velo

kolezko [41]

Velocity, va2 = 10.5 ft/s

<u>Explanation:</u>

From the figure:

Length of the cable = Sa + 2Sb = l

∴ vₐ = -2vb

Applying the principle of Impulse and momentum in x-direction

$mv_x_1 + \int\limits^t_t {F_x} \, dt = mv_x_2$

Limit is t1 to t2

$-(\frac{14}{32.2}) (2) (3) - T = \frac{14}{32.2} v_y_2$ -(1)

Applying the principle of Impulse and momentum in y-direction

$mv_y_1 + \int\limits^t_t {F_y} \, dt = mv_y_2$

Limit is t1 to t2

$(\frac{5}{32.2}) (3) - 2T + 3 = -\frac{5}{32.2} \frac{v_y_2}{2}$ -(2)

Solving equation (1) and (2), we obtain

T = 1.6lb

va2 = 10.5 ft/s

8 0

2 years ago