3 years ago

Can someone please help me with this problem. I’ve tried but I can’t seem to figure it out.

1 answer:

Zolol [24]3 years ago

7 0

The last one, Tension = 30N, Normal force = 20 N

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Montano1993 [528]

An unbalanced force.
Newton's first law states that an object in motion will stay in motion until a unbalanced force acts upon it.

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3 years ago

BaLLatris [955]

Answer:

The force exerted by the ocean tide is directed to the right (east).

The force exerted by the wind is directed to the northwest (45° N of W).

We should separate the x- and y- components of the wind force, and evaluate each component separately.

$F_x = 6000\cos(\pi/4)(-\^{x})\\F_y = 6000\sin(\pi/4)(+\^{y})$

We denote the direction to the right as the positive direction, so the x-component of the wind force is in the negative direction.

The resultant force is as follows:

$F_R = [2000 - 4242.64](\^{x}) + [4242.64](\^{y})\\F_R = -2242.64\^{x} + 4242.64\^{y}$

The resultant acceleration can be found by Newton's Second Law:

$F = ma\\\\a_R = F_R/m = -1.12\^{x} + 2.12\^{y}$

The magnitude of the resultant acceleration is

$|a_R| = \sqrt{(1.12)^2 + (2.12)^2} = 2.4 ~m/s^2$

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3 years ago

Alecsey [184]

Answer:

$F {(0.24)}^{2} = 13.6 \times {10}^{ - 8} \\ F = \frac{13.6 \times {10}^{ - 8}}{ {(0.24)}^{2} }N$

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3 years ago

ryzh [129]

Acceleration

a=v/t

Thanks

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4 years ago

Mumz [18]

No that is not normal.

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3 years ago