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cricket20 [7]
3 years ago
13

A torsional pendulum consists of a 5 kg uniform disk with a diameter of 50 cm attached at its center to a rod 1.5 m in length. T

he torsional spring constant is 0.625 N-m/rad. Disregarding the mass of the rod, what is the natural frequency of the torsional pendulum?
Engineering
1 answer:
IRISSAK [1]3 years ago
7 0

Answer:

natural frequency of the torsional pendulum =  1.4 rad/s

Explanation:

given data

mass = 5 kg

diameter = 50 cm = 0.50 m

so radius R = 0.25 m

length =  1.5 m

torsional spring constant = 0.625 N-m/rad

solution

first we get here moment of inertia that is

moment of inertia = mR²  .........1

moment of inertia = 5 × 0.25²

moment of inertia  = 0.3125 kg-m²

so now we get here natural frequency of the torsional pendulum that is

natural frequency of the torsional pendulum = \sqrt{\frac{Kt}{I} }    ..................2

here Kt is torsional spring constant  and I is moment of inertia  

natural frequency of the torsional pendulum = \sqrt{\frac{0.625}{3125} }  

natural frequency of the torsional pendulum =  1.4 rad/s

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aliina [53]

Answer:

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  • Remove all leaves of T. Let the remaining tree be T1.
  • Remove all leaves of T1. Let the remaining tree be T2.
  • Repeat the "remove" operation as follows: Remove all leaves of Ti. Let remaining tree be Ti+1.
  • When the remaining tree has only one node or two nodes, stop! Suppose now the remaining tree is Tk.
  • If Tk has only one node, that is the center of T. The diameter of T is 2k.
  • If Tk has two nodes, either can be the center of T. The diameter of T is 2k+1.

Explanation:

We can compute the diameter of the tree T by a pruning procedure, starting at the leaves (external nodes).

  • Remove all leaves of T. Let the remaining tree be T1.
  • Remove all leaves of T1. Let the remaining tree be T2.
  • Repeat the "remove" operation as follows: Remove all leaves of Ti. Let remaining tree be Ti+1.
  • When the remaining tree has only one node or two nodes, stop! Suppose now the remaining tree is Tk.
  • If Tk has only one node, that is the center of T. The diameter of T is 2k.
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4 0
3 years ago
In the circuit given below, R1 = 17 kΩ, R2 = 74 kΩ, and R3 = 5 MΩ. Calculate the gain 1formula58.mml when the switch is in posit
Elenna [48]

Answer:a

a) Vo/Vi = - 3.4

b) Vo/Vi = - 14.8

c) Vo/Vi = - 1000

Explanation:

a)

R1 = 17kΩ

for ideal op-amp

Va≈Vb=0 so Va=0

(Va - Vi)/5kΩ + (Va -Vo)/17kΩ = 0

sin we know Va≈Vb=0

so

-Vi/5kΩ + -Vo/17kΩ = 0

Vo/Vi = - 17k/5k

Vo/Vi = -3.4

║Vo/Vi ║ = 3.4    ( negative sign phase inversion)

b)

R2 = 74kΩ

for ideal op-amp

Va≈Vb=0 so Va=0

so

(Va-Vi)/5kΩ + (Va-Vo)74kΩ = 0

-Vi/5kΩ + -Vo/74kΩ = 0

Vo/Vi = - 74kΩ/5kΩ

Vo/Vi = - 14.8

║Vo/Vi ║ = 14.8  ( negative sign phase inversion)

c)

Also for ideal op-amp

Va≈Vb=0 so Va=0

Now for position 3 we apply nodal analysis we got at position 1

(Va - Vi)/5kΩ + (Va - Vo)/5000kΩ = 0           ( 5MΩ = 5000kΩ )

so

-Vi/5kΩ + -Vo/5000kΩ = 0

Vo/Vi = - 5000kΩ/5kΩ

Vo/Vi = - 1000

║Vo/Vi ║ = 1000  ( negative sign phase inversion)

3 0
3 years ago
Do the coil resistances have any effect on the plots?
PolarNik [594]
Because of the skin depth effect, the current at high frequency tends to flow at very low depth from radius. Then at high frequency the effective cross section of the wire is narrower than at DC.

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8 0
2 years ago
A rigid tank having 25 m3 volume initially contains air having a density of 1.25 kg/m3, then more air is supplied to the tank fr
Hoochie [10]

Answer:

\Delta m = 102.25\,kg

Explanation:

The mass inside the rigid tank before the high pressure stream enters is:

m_{o} = \rho_{air}\cdot V_{tank}

m_{o} = (1.25\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})

m_{o} = 31.25\,kg

The final mass inside the rigid tank is:

m_{f} = \rho \cdot V_{tank}

m_{f} = (5.34\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})

m_{f}= 133.5\,kg

The supplied air mass is:

\Delta m = m_{f}-m_{o}

\Delta m = 133.5\,kg-31.25\,kg

\Delta m = 102.25\,kg

4 0
3 years ago
Consider an aircraft powered by a turbojet engine that has a pressure ratio of 9. The aircraft is stationary on the ground, held
77julia77 [94]

Answer:

The break force that must be applied to hold the plane stationary is 12597.4 N

Explanation:

p₁ = p₂, T₁ = T₂

\dfrac{T_{2}}{T_{1}} = \left (\dfrac{P_{2}}{P_{1}}  \right )^{\frac{K-1}{k} }

{T_{2}}{} = T_{1} \times \left (\dfrac{P_{2}}{P_{1}}  \right )^{\frac{K-1}{k} } = 280.15 \times \left (9  \right )^{\frac{1.333-1}{1.333} } = 485.03\ K

The heat supplied = \dot {m}_f × Heating value of jet fuel

The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s

The heat supplied = \dot m · c_p(T_3 - T_2)

\dot m = 20 kg/s

The heat supplied = 20*c_p(T_3 - T_2) = 21,350 kJ/s

c_p = 1.15 kJ/kg

T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K

p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa

p₃ = p₂ = 855 kPa

T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K

T₄ = 1413.3 - 204.88 = 1208.42 K

\dfrac{T_5}{T_4}  = \dfrac{2}{1.333 + 1}

T₅ = 1208.42*(2/2.333) = 1035.94 K

C_j = \sqrt{\gamma \times R \times T_5} = √(1.333*287.3*1035.94) = 629.87 m/s

The total thrust = \dot m × C_j = 20*629.87 = 12597.4 N

Therefore;

The break force that must be applied to hold the plane stationary = 12597.4 N.

5 0
2 years ago
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