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3 years ago

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A 10 m long clothesline is strung so that it is perfectly horizontal. when a shirt is hung in the exact center the line sags to

create an angle that is 1 degree below the horizontal on each end. how far below the original height of the line is the center of the clothesline

1 answer:

Leto [7]3 years ago

3 0

Answer:

x = 0.0873 m

Explanation:

given,

length of clothesline = 10 m

the line sags to create an angle that is 1 degree below the horizontal on each end.

As mass is hang at the center the angle made let the deflection be 'x'

as the shirt is hang at the center distance will be equal to 5 m.

now

$tan \theta = \dfrac{d_e}{d}$

$tan \theta = \dfrac{x}{5}$

$tan 1^0 = \dfrac{W}{5}$

$W = 5 \times 0.0175$

x = 0.0873 m

Hence, the cloth line will be x = 0.0873 m from the original length of the clothesline.

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A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha

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Answer:

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Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

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$\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss}$ (2)

Where:

$m$ - Mass of the roller coaster car, in kilograms.

$v_{1}$ - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

$g$ - Gravitational acceleration, in meters per square second.

$h_{1}$ - Height of the first top of the hill with respect to the bottom, in meters.

$W_{1, loss}$ - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

$v_{2}$ - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

$h_{2}$ - Height of the second top of the hill with respect to the bottom, in meters.

$W_{2, loss}$ - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

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$m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss}$ (3)

Where $W_{loss}$ is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

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$W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]$

$W_{loss} = 6574.75\,J$

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

8 0

3 years ago

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Marta_Voda [28]

Answer:

The angular velocity of the platform is 1.114 rad/s.

Explanation:

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Step 6: Total moment of inertia = 125.80 + 15.73 + 56.06 = 197.59

Step 7: Calculate final angular momentum = 197.59 * ω

197.59 * ω = 220.15

ω = 220.15 / 197.59

This is approximately 1.114 rad/s.

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