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andrew11 [14]
3 years ago
5

The world's most powerful tugboats are built in Finland. One of these boats can do 9.8 × 107 J of work through a distance of 35

m. What is the force exerted by the tugboat? 
Mathematics
1 answer:
kkurt [141]3 years ago
3 0
<span>2.8 x 10^6 N We have 9.8 x 10^7 J or 9.8 x 10^7 kg*m^2/s^2 of work done through a distance of 35 meters. Since work is defined as force over distance, that means that: F * 35m = 9.8 x 10^7 kg*m^2/s^2 Solving for F, gives F * 35m = 9.8 x 10^7 kg*m^2/s^2 F = 9.8 x 10^7 kg*m^2/s^2 / 35m F = 2.8 x 10^6 kg*m/s^2 F = 2.8 x 10^6 N So the force exerted by the tugboat is 2.8x10^6 Newtons.</span>
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1. Consider the right triangle ABC given below.
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#1) 
A) b = 10.57
B) a = 22.66; the different methods are shown below.
#2)
A) Let a = the side opposite the 15° angle; a = 1.35.
Let B = the angle opposite the side marked 4; m∠B = 50.07°.
Let C = the angle opposite the side marked 3; m∠C = 114.93°.
B) b = 10.77
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Explanation
#1)
A) We know that the sine ratio is opposite/hypotenuse.  The side opposite the 25° angle is b, and the hypotenuse is 25:
sin 25 = b/25

Multiply both sides by 25:
25*sin 25 = (b/25)*25
25*sin 25 = b
10.57 = b

B) The first way we can find a is using the Pythagorean theorem.  In Part A above, we found the length of b, the other leg of the triangle, and we know the measure of the hypotenuse:
a²+(10.57)² = 25²
a²+111.7249 = 625

Subtract 111.7249 from both sides:
a²+111.7249 - 111.7249 = 625 - 111.7249
a² = 513.2751

Take the square root of both sides:
√a² = √513.2751
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The second way is using the cosine ratio, adjacent/hypotenuse.  Side a is adjacent to the 25° angle, and the hypotenuse is 25:
cos 25 = a/25

Multiply both sides by 25:
25*cos 25 = (a/25)*25
25*cos 25 = a
22.66 = a

The third way is using the other angle.  First, find the measure of angle A by subtracting the other two angles from 180:
m∠A = 180-(90+25) = 180-115 = 65°

Side a is opposite ∠A; opposite/hypotenuse is the sine ratio:
a/25 = sin 65

Multiply both sides by 25:
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#2)
A) Let side a be the one across from the 15° angle.  This would make the 15° angle ∠A.  We will define b as the side marked 4 and c as the side marked 3.  We will use the law of cosines:
a² = b²+c²-2bc cos A
a² = 4²+3²-2(4)(3)cos 15
a² = 16+9-24cos 15
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√a² = √1.82
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sin A/a = sin B/b
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Cross multiply:
4*sin 15 = 1.35*sin B

Divide both sides by 1.35:
(4*sin 15)/1.35 = (1.35*sin B)/1.35
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Take the inverse sine of both sides:
sin⁻¹((4*sin 15)/1.35) = sin⁻¹(sin B)
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Subtract both known angles from 180 to find m∠C:
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b*sin 52 = 12*sin 45

Divide both sides by sin 52:
(b*sin 52)/(sin 52) = (12*sin 45)/(sin 52)
b = 10.77

Find m∠A by subtracting both known angles from 180:
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Use the law of sines to find side a:
sin C/c = sin A/a
sin 52/12 = sin 83/a

Cross multiply:
a*sin 52 = 12*sin 83

Divide both sides by sin 52:
(a*sin 52)/(sin 52) = (12*sin 83)/(sin 52)
a = 15.11
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