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3 years ago

6.

Physics

1 answer:

yulyashka [42]3 years ago

8 0

Answer:

The ribbon will move to the right.

Explanation:

To know the the correct answer to the question, we shall determine the net force and direction. This can be obtained as follow:

Force to the right (Fᵣ) = 120 N

Force to the left (Fₗ) = 80 N

Net force (Fₙ) =?

Fₙ = Fᵣ – Fₗ

Fₙ = 120 – 80

Fₙ = 40 N to the right.

From the calculation made above, the net force is 40 N to the right. Thus, the ribbon will move to the right.

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Two ropes are attached to either side of a 100.0 kg wagon as shown below. The rope on the right is pulled at an angle 40.0° rela

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1. The responses for question 1 are;

x-component of the tension in the rope on the right is approximately 91.93 N

y-component of the tension in the rope on the right is approximately 71.135 N

x-component of the tension in the rope on the left is -80.0 N

y-component of the tension in the rope on the left is 0

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Reasons:

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Mass of the wagon, m = 100.0 kg

Angle of inclination to the horizontal of the rope to the right, θ = 40.0°

Tension in the rope on the right = 120.0 N

Direction in which the rope on the left is pulled = To the west

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1. The x and y component of the tension in the rope on the right are;

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2. The net force in the horizontal direction, Fₓ, is found as follows;

Fₓ = The x-component of the rope on the left + The x-component of the rope on the right

Which gives;

Fₓ = 91.93 N - 80.0 N = 11.93 N

3. The net acceleration of the block is given as follows;

According to Newton's Second Law of motion, we have;

Force in the horizontal direction, Fₓ = Mass of wagon, m × Acceleration of the wagon in the horizontal direction, aₓ

Fₓ = m × aₓ

Therefore;

$\displaystyle a_x = \frac{F_x}{m} \approx \frac{11.93 \, N}{100.0 \, kg} = \mathbf{0.1193 \ m/s^2}$

The acceleration of the wagon in the horizontal direction, aₓ ≈ 0.1193 m/s².

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