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vlada-n [284]
3 years ago
10

A rope is attached to a block. The rope pulls on the block with a force of 240 N, at an angle of 40 degrees to the horizontal (t

his force is equal to the tension in the rope). What is the x-component of the force on the block due to the rope?

Physics
2 answers:
Tom [10]3 years ago
3 0

Answer:

X Component is 183.85N

Explanation:

The x component of the force on the block due to the rope;

X = F cos @ where if is the force, @ is the angle mad with the block.

X = F cos @

X = 240 cos 40

Cos 40= 0.7660, so

X = 240 × 0.7660

X component= 183.85N// rounded to two decimal places.

LenKa [72]3 years ago
3 0

Answer:

The x-component of the force on the block due to the rope is 183.8 newtons

Explanation:

See the picture attached.

Because the force F on the block due the rope is applied at an angle of 40 degrees respect the horizontal, it has a component on x (Fx) and a component on y (Fy) we should find the x component. We should use trigonometric relations for right tringle. We need the equation that relates hypotenuse (F) and adjacent side (Fx), this relation is:

cos 40 =\frac{F_x}{F}

solving for Fx:

F_x = F*cos 40 = 240*cos(40)

F_x = 183.8 N

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\displaystyle F_N = F_g = (5\text{ kg})(10 \text{ N/kg}) = 50 \text{ N}

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Answer:

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(P₀/P) = (T₀/T)^(k/(k-1))

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T₀ = initial temperature of air = 30°C = 303.15 K

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T = 303.1758 K

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Δρ = ρ - ρ₀

P₀ = ρ₀RT₀

ρ₀ = P₀/RT₀

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ρ₀ = 101000/(287 × 303.15) = 1.1608655 kg/m³

ρ = P/RT

P = 101030 Pa, T = 303.1758K

ρ = 101030/(287×303.1758) = 1.1611115 kg/m³

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Second method

(ρ₀/ρ) = (T₀/T)^(1/(k-1))

Where ρ₀ is initially calculated from ρ₀ = P₀/RT₀, then ρ is then computed and the diff taken.

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c₀ = √(kRT₀) = √(1.4 × 287 × 303.15) = 349.00669 m/s

c = √(kRT) = √(1.4 × 287 × 303.1758) = 349.0215415 m/s

Δc = c₀ - c = 349.0215415 - 349.00669 = 0.0148 m/s

QED!

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