The centripetal force acting on a moon in a circular orbit around a planet continuously changes the ____

A 1.80-m -long uniform bar that weighs 531 N is suspended in a horizontal position by two vertical wires that are attached to th

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Answer:

(a) 498.4 Hz

(b) 442 Hz

Solution:

As per the question:

Length of the wire, L = 1.80 m

Weight of the bar, W = 531 N

The position of the copper wire from the left to the right hand end, x = 0.40 m

Length of each wire, l = 0.600 m

Radius of the circular cross-section, R = 0.250 mm = $.250\times 10^{- 3}\ m$

Now,

Applying the equilibrium condition at the left end for torque:

$T_{Al}.0 + T_{C}(L - x) = W\frac{L}{2}$

$T_{C}(1.80 - 0.40) = 531\times \frac{1.80}{2}$

$T_{C} = 341.357\ Nm$

The weight of the wire balances the tension in both the wires collectively:

$W = T_{Al} + T_{C}$

$531 = T_{Al} + 341.357$

$T_{Al} = 189.643\ Nm$

Now,

The fundamental frequency is given by:

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where

$\mu = A\rho = \pi R^{2}\rho$

(a) For the fundamental frequency of Aluminium:

$f = \frac{1}{2L}\sqrt{\frac{T_{Al}}{\mu}}$

$f = \frac{1}{2L}\sqrt{\frac{T_{Al}}{\pi R^{2}\rho_{Al}}}$

where

$\rho_{l} = 2.70\times 10^{3}\ kg/m^{3}$

$f = \frac{1}{2\times 0.600}\sqrt{\frac{189.643}{\pi 0.250\times 10^{- 3}^{2}\times 2.70\times 10^{3}}} = 498.4\ Hz$

(b) For the fundamental frequency of Copper:

$f = \frac{1}{2L}\sqrt{\frac{T_{C}}{\mu}}$

$f = \frac{1}{2L}\sqrt{\frac{T_{C}}{\pi R^{2}\rho_{C}}}$

where

$\rho_{C} = 8.90\times 10^{3}\ kg/m^{3}$

$f = \frac{1}{2\times 0.600}\sqrt{\frac{341.357}{\pi 0.250\times 10^{- 3}^{2}\times 2.70\times 10^{3}}} = 442\ Hz$

7 0

4 years ago

Speed of light in air is 3 x 10 m/s, determine the speed of light in the glass fibre.

Dafna11 [192]

30 speed of light in the glass

4 0

3 years ago

Which model below shows the positions of the Sun, Moon, and Earth that have the greatest effect on ocean tides?

Lynna [10]

Answer:

Below!

Explanation:

Referring to the picture, we can conclude that the picture J will have the most effect on ocean tides.

Whichever model is similar to model J in the picture will be your answer.

Hoped this helped.

6 0

2 years ago

An engine has an energy input of 125 J, and 35 J of that energy is transformed into useful energy. What is the efficiency of the

xeze [42]

The efficiency of the engine at the given useful energy and input energy is 28%.

The given parameters:

Input energy of the engine, = 125 J
Useful energy of the engine, 35 J

The efficiency of the engine is defined as the ratio of the output energy (useful energy) to the input energy and it is calculated as follows;

$efficiency\ = \frac{0ut put \ energy}{1nput \ energy} \times \ 100\%\\\\efficiency\ = \frac{35}{125} \times 100\%\\\\efficiency\ = 28 \ \%$

Thus, the efficiency of the engine at the given useful energy and input energy is 28%.

Learn more about efficiency of engine here: brainly.com/question/6751595

7 0

2 years ago

5) [Honors]A seagull, ascending straight upward at 5.2 m/s, drops a shell when it is 12.5m above the ground. (A)

jolli1 [7]

Answer:

(B) 13.9 m

(C) 1.06 s

Explanation:

Given:

v₀ = 5.2 m/s

y₀ = 12.5 m

(A) The acceleration in free fall is -9.8 m/s².

(B) At maximum height, v = 0 m/s.

v² = v₀² + 2aΔy

(0 m/s)² = (5.2 m/s)² + 2 (-9.8 m/s²) (y − 12.5 m)

y = 13.9 m

(C) When the shell returns to a height of 12.5 m, the final velocity v is -5.2 m/s.

v = at + v₀

-5.2 m/s = (-9.8 m/s²) t + 5.2 m/s

t = 1.06 s

3 0

3 years ago

The centripetal force acting on a moon in a circular orbit around a planet continuously changes the _____ of the moon's motion.