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kotykmax [81]
3 years ago
5

A boat has a mass of 4900 kg. Its engines generate a drive force of 4600 N due west, while the wind exerts a force of 670 N due

east and the water exerts a resistive force of 1170 N due east. Take west to be the positive direction. What is the boat's acceleration, with correct sign?
Physics
1 answer:
melomori [17]3 years ago
4 0

Answer:

a = +0.56 m/s²  (due west)

Explanation:

We apply Newton's second law:  

∑Fx = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)  

m : mass in kilograms (kg)  

a : acceleration in meters over second square (m/s²)  

Data:  

m=  4900 kg

Fd= 4600 N : drive force

Fw = 670 N :  wind force

Fr = 1170 N : water resistive force

Problem development

We apply the formula (1)

∑Fx = m*a

Fd-Fw-Fr = m*a

4600 - 670 - 1170 =  4900 *a

2760 =  4900 *a

a= (2760) / ( 4900 )

a = +0.56 m/s² (due west)

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Answer:

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b. The first order fringe is the fringe located between the first minimum and the second minimum. From dsinθ = mλ and tanθ = y/D when θ is small, sinθ ≈ θ ≈ tanθ. So, y = mλD/d. Let m= 1 and m=2 be the first and second minima respectively. So,y₁ =  λD/d and y₂ =  2λD/d. The difference Δy₁ = y₂ - y₁ is the width of the first order fringe. Therefore, Δy₁ = 2λD/d - λD/d= λD/d. Substituting the values from above, we have

λD/d= 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴= 1.95 × 10⁻³ m = 1.95 mm

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