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Westkost [7]
3 years ago
12

A child of mass M is swinging on a swing set. The ropes attaching the swing to the top bar have length L. Find the gravitational

potential energy of the child relative to her lowest position when the ropes are (a) hanging straight down; (b) exactly horizontal; (c) at an angle ϴ from the vertical; and (d) at an angle phi from the horizontal.
Physics
1 answer:
myrzilka [38]3 years ago
5 0

Answer:

(a) 0

(b) 10ML

(c) 10ML(1 - cos(\theta))

(d) 10ML(1 + sin(\phi))

Explanation:

(a) When hanging straight down. The child is at the lowest position. His potential energy with respect to this point would also be 0.

(b) Since the rope has length L m. When the rope is horizontal, he is at L (m) high with respect to the lowest swinging position. His potential energy with respect to this point should be

E_h = mgh = 10ML

where g = 10m/s2 is the gravitational acceleration.

(c) At angle \theta from the vertical. Vertically speaking, the child should be at a distance of Lcos(\theta) to the swinging point, and a vertical distance of L - Lcos(\theta) to the lowest position. His potential energy to this point would be:

E_{\theta} = mgh = 10M(L - Lcos(\theta)) = 10ML(1 - cos(\theta))

(d) at angle \phi from the horizontal. Suppose he is higher than the horizontal line. This would mean he's at a vertical distance of Lsin(\phi) from the swinging point and higher than it. Therefore his vertical distance to the lowest point is L + Lsin(\phi) = L(1 + sin(\phi))

His potential energy to his point would be:

E_{\phi} = mgh = 10ML(1 + sin(\phi))

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A bionic man running at 6.5 m/sec , east is acceleration at a uniform rate of 1.5 m/sec^2 east over a displacement of 100.0 m ea
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Given:
u = 6.5 m/s, initial velocity 
a = 1.5 m/s², acceleration
s = 100.0 m, displacement

Let v =  the velocity attained after the 100 m displacement.
Use the formula
v² = u² + 2as
v² = (6.5 m/s)² + 2*(1.5 m/s²)*(100 m) = 342.25 (m/s)²
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3 years ago
Describe the sequence of mechanical energy events that lets you hear the
lina2011 [118]

Answer:

Starting from the beginning.

There is a radio signal that is received by the radio.

The radio interprets the signal and produces a current in response to it.

That current goes to a membrane that oscillates producing sound, the oscillation of the membrane is the first mechanical energy event here.

These oscillations can travel in material mediums, for example, the air. Then there is a production of waves (soundwaves) that travel in the air (second event).

Those waves now hit the wall that separates you and your neighbor, as the wall is made of a material, the soundwaves can travel through it, but they will be dispersed (a part of the waves rebounds on the wall, and another part is dissipated as the wave travels through the wall), there is also a transmitted part of the wave, that is now in your house. (this change of medium will be the third event). Now only the lower frequencies survive, this is why the sound is "muffled".

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3 0
2 years ago
point) A circular swimming pool has a diameter of 12 m. The circular side of the pool is 3 m high, and the depth of the water is
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Answer:

(a) 86.65 J

(b) 149.65 J

Solution:

As per the question:

Diameter of the pool, d = 12 m

⇒ Radius of the pool, r = 6 m

Height of the pool, H = 3 m

Depth of the pool, D = 2.5 m

Density of water, \rho_{w} = 1000\ kg//m^{3}

Acceleration due to gravity, g = 9.8\ m/s^{2}

Now,

(a) Work done in pumping all the water:

Average height of the pool, h = \frac{H + D}{2}

h = \frac{3 + 2.5}{2} = 2.75\ m

Volume of water in the pool, V = \pi r^{2}h = \pi \times 6^{2}\times 2.75 = 311.02\ m^{3}

Mass of water, m_{w} = \frac{\rho_{w}}{V}

m_{w} = \frac{1000}{311.02} = 3.215\ kg

Work done is given by the potential energy of the water as:

W = m_{w}gh = 3.215\times 9.8\times 2.75 = 86.65\ J

(b) Work done to pump all the water through an outlet of 2 m:

Now,

Height, h = 2.75 + 2 = 4.75

Work done,W = m_{w}gh = 3.215\times 9.8\times 4.75 = 149.65\ J

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