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3 years ago

6

A gymnast is swinging on a high bar. The distance between his waist and the bar is 0.829 m, as the drawing shows. At the top of

the swing his speed is momentarily zero. Ignoring friction and treating the gymnast as if all of his mass is located at his waist, find his speed at the bottom of the swing.

1 answer:

vivado [14]3 years ago

6 0

To find the speed with the given values we can apply the energy conservation equations for which we have that the increase in potential energy is compensated in the decrease of the kinetic energy or vice versa.

Since there is conservation and part of the balance we have to

$KE = PE$

Where,

KE = Kinetic Energy

PE = Potential Energy

The values for this energy are given as

$\frac{1}{2}mv^2 = mgh$

$\frac{1}{2} v^2 = gh$

$v = \sqrt{2gh}$

$v = \sqrt{2(9.8)(0.829)}$

$v = 4.03m/s$

Therefore the speed at the bottom of the swing is 4.03m/s

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g In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period

NeTakaya

The question is incomplete. The complete question is :

In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period of the oscillations is 1.2 microseconds (1.2*10^-6). What is Q?

Solution :

The underdamped RLC circuit

$v_{t} = ve^{-\frac{R}{2L}t} \cos \omega t$

$\omega = \sqrt{\frac{1}{LC}-\frac{R^2}{4L^2}}= \frac{2 \pi}{T}$

We know in one time period, v = 2v, at t = T, $v_t = 3.8 v$

so, $9.8 = 5 e^{-\frac{R}{2L}T} \cos \frac{2 \pi}{T}T$

$e^{-\frac{R}{2L}T} = \frac{3.8}{5} \times 1$

$\frac{R}{2L}T= \ln \frac{5}{3.8}$

$\frac{R}{L}= \frac{2}{1.2 \times 10^{-6}} \ln \frac{5}{3.8}$

$\frac{R}{L} = 457.3 \times 10^3$

Now, Q value $= \frac{1}{R}\sqrt{\frac{L}{C}}$

$=\sqrt{\frac{L}{R^2C}\times \frac{L}{L}}$

$=\sqrt{(\frac{L}{R})^2 \times \frac{1}{LC}}$

$\frac{1}{LC}=27.43 \times 10^{12}$

∴ $Q=\sqrt{\left(\frac{1}{457.3 \times 10^3}\right)^2 \times 27.43 \times 10^{12}}$

$Q=\sqrt{131.166}$

= 11.45

4 0

2 years ago

Red Riding Hood sat down to rest and placed her 1.20-kg basket of goodies beside her. A wolf came along, spotted the basket, and

anygoal [31]

Answer:

117.6°

Explanation:

The vertical component of a force directed at some angle α from the vertical is ...

F·cos(α)

We want the vertical components of the wolf's force (Fw) and Red's force (Fr) to total zero. So for some angle from vertical α, Red's force will satisfy ...

Fw·cos(25°) + Fr·cos(α) = 0

cos(α) = -Fw/Fr·cos(25°) ≈ -(6.4 N)/(12.5 N)·0.906308 ≈ -0.464030

α ≈ arccos(-0.464030) ≈ 117.6°

Red was pulling at an angle of about 117.6° from the vertical.

_____

<em>Additional comment</em>

That's about 27.6° below the horizontal.

3 0

2 years ago

The flywheel of a steam engine begins to rotate from rest with a constant angular acceleration of 1.33 rad/s2. It accelerates fo

Svetradugi [14.3K]

Answer:

We first to know that if the wheel rotates from rest means that at t=0 the velocity and the angle rotated is 0.

Then, we know:

$\alpha = 1.33 = \frac{dw}{dt}$

Integrating 2 times, we have:

$w = 1.33t\\angle =0.665t^{2}$

For the first 27.9 s, we have:

w = 37.107 rad/s

angle = 517.6426 rad

For the next seconds, according to the text, the angular velocity is constant so

w = 37.107 rad/s and hence, integrating:

$angle =37.107t$

Then, the time remaining is:

53.5 - 27.9 = 25.6

So for the next 25.6 seconds we have:

$angle = 37.107*25.6=949.9392 rad$

Finally, we add the 2 angles and we have as a result:

$angle = 517.6426+949.9392=1467.5818$

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If a car crashes into another car like this, the wreck should go nowhere. Besides this being an unrealistic question, the physics of it would look like this:

Momentum before and after the collision is conserved.

Momentum before the collision:
p = m * v = 50000kg * 24m/s + 55000kg * 0m/s = 50000kg * 24m/s
Momentum after the collision:
p = m * v = (50000kg + 55000kg) * v

Setting both momenta equal:
50000kg * 24m/s = (50000kg + 55000kg) * v

Solving for the velocity v:
v = 50000kg * 24m/s/(50000kg + 55000kg) = 11,43m/s

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3 years ago

During what stage of engine operation does the piston move upward in the cylinder and force the burned gases out of the cylinder

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3 years ago