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3 years ago

There are two pendulums with periods 1.5 s and 1 s, which one is longer, and which one is shorter.

Physics

1 answer:

Stells [14]3 years ago

3 0

Answer:

The greater the length, the greater the period. Or vice versa

Explanation:

T = 2π * √(L/g)

Given that T = 1s , g = 9.8m/s^2 , π = 3.14

1 = 2× 3.14 √(L/9.8)

1 = 6.28 ×L /9.8

L= 9.8/6.28

L= 1.56m

When T = 1.5s

1.5 = 2×3.14√(L/9.8)

1.5^2 = 6.28×L/9.8

2.25 = 6.28L/9.8

2.25×9.8 = 6.28L

22.05 = 6.28L

L = 22.05/6.28

L = 3.51m

From the above mathematical solution, it is clear that the greater the length, the greater the period. Or vice versa

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Answer:

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3 years ago

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

NikAS [45]

The acceleration of the wagon is found by applying Newton's Second Law of motion.

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

2. The net force in the x-direction is approximately 11.93 N

3. The net acceleration of the wagon in the horizontal direction is approximately 0.1193 m/s².

Reasons:

The given parameters are;

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

Tension in the rope on the left = 80.0 N

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

x-component = 120.0 N × cos(40.0°) ≈ 91.93 N

y-component = 120.0 N × sin(40.0°) ≈ 77.135 N

The x and y component of the tension in the rope on the left are;

x-component = 80.0 N × cos(180°) = -80.0 N

y-component = 80.0 N × sin(180°) = 0.0 N

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.

Learn more here:

brainly.com/question/20357188

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2 years ago

The weight of a block on the inclined plane is 500 N and the angle of incline is 30 degrees. What is the magnitude of the force

yanalaym [24]

Answer:

250 N

433 N

Explanation:

N = Normal force by the surface of the inclined plane

W = Weight of the block = 500 N

f = static frictional force acting on the block

Parallel to incline, force equation is given as

f = W Sin30

f = (500) Sin30

f = 250 N

Perpendicular to incline force equation is given

N = W Cos30

N = (500) Cos30

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3 years ago

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2 years ago

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maksim [4K]

Answer:

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