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vovangra [49]
3 years ago
8

An aluminum (E = 10 x 106 lb/in2, ν = 0.333) bar of length (L = 49 inch) has a circular cross-section of radius (r = 0.7 inch).

The beam is fixed at one end. A torque (T = 1,666 lb-in) is applied to the free end. Calculate the twist of the bar free-end. (Units: degree)
Physics
1 answer:
romanna [79]3 years ago
8 0

Answer:

1.24 degrees

Explanation:

The twist angle can be calculated using the following formula for torsion:

T = \frac{E I \theta}{L}

Where I = \frac{\pi r^4}{2} = \frac{\pi 0.7^4}{2} = 0.377 in^4 is the polar inertia of the circular cross section.

1666 = \frac{10^7 * 0.377 \theta}{49} = 76969 \theta

\theta = 1666 / 76969 = 0.0216 rad = 0.0216 * 180 / \pi = 1.24^o

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Answer:

c. The steady-state value of the current depends on the resistance of the resistor.

Explanation:

Since all the components are connected in series, when the switch is at first open, current will not flow round the circuit. As current needs to flow through from the positive terminal of the battery through the resistor, inductor, and  switch to the negative terminal of the battery.

But the moment the switch is closed, at the initial time t = 0, the current flow through from the positive terminal of the battery through the resistor, inductor, and switch to the negative terminal of the battery. It then begins to increase at a rate that depends upon the value of the inductance of the inductor.

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a stone with a mass of 2.40 kg is moving with velocity (6.60î − 2.40ĵ) m/s. find the net work (in j) on the stone if its velocit
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W = \Delta K = \dfrac m2 ({v_2}^2 - {v_1}^2)

We have

\vec v_1 = (6.60\,\vec\imath - 2.40\,\vec\jmath)\dfrac{\rm m}{\rm s} \implies {v_1}^2 = \|\vec v_1\|^2 = 49.32 \dfrac{\rm m^2}{\rm s^2}

\vec v_2 = (8.00\,\vec\imath + 4.00\,\vec\jmath) \dfrac{\rm m}{\rm s} \implies {v_2}^2 = \|\vec v_2\|^2 = 80.0\dfrac{\mathrm m^2}{\mathrm s^2}

Then the total work is

W = \dfrac{2.40\,\rm kg}2 \left(80.0\dfrac{\rm m^2}{\rm s^2} - 49.32\dfrac{\rm m^2}{\rm s^2}\right)  \approx \boxed{36.8\,\rm J}

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2 years ago
Adam drops a ball from rest from the top floor of a building at the same time Bob throws a ball horizontally from the same locat
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Answer:

Both balls hit the ground at the same time

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In this case we have that gravity is negative, and as Adam drops the ball, v_{0y}=0

Bob throws the ball horizontally, so the movement will be a <em>parabola</em>, we can divide into horizontal direction, and vertical direction.

But we only need to analize the vertical movement, in wich again the only acceleration is gravity, and compare it with Adam's ball. Again we have that gravity is negative, and as the initial throw is horizontal, v_{0y}=0

Finally, we have that

y(t)=h-\frac{1}{2}gt^{2}

where

h=y_{0}

both for Adam's vertical drop, and for Bob's vertical component of the parabolic throw.

Now, if we put y(t)=0 (the origin of the vertical coordinate), we get for both cases that

h=\frac{1}{2}gt^{2}

where we can clear the value for the time t, of the fall, wich will be the same in both cases.

Hence, both balls hit the ground at the same time.

3 0
3 years ago
A viola string with a fundamental frequency of D4 (293 Hz) is generally tuned using a tension of 49.0 N. However, just before a
Alborosie

Answer:

Explanation:

For fundamental frequency in a vibrating string , the formula is

n = 1 / 2L  x  √ ( T /m₁ )

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For first string ,

293 =  1 / 2L  x  √ ( 49 N  /m₁ )

For second string , let mass per unit length be m₂ .

196 =  1 / 2L  x  √ ( 49 N  /m₂ ) ------ ( 1 )

To bring its frequency back to previous one let tension be T

293  =  1 / 2L  x  √ ( T  /m₂ ) ------- ( 2 )

Dividing

293 / 196 = √ ( T  /49  )

1.4948 = √ ( T  /49  )

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Answer:

b. Decreases

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8 0
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