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3 years ago

7

If an atom had a nucleus 1 ft in diameter, what would be the diameter of the atom, in miles? (1 mi = 5,280 ft)

1 answer:

hjlf3 years ago

5 0

In nomine patris, et filii, et spiritus sancti.

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A series AC circuit contains a resistor, an inductor of 150 mH, a capacitor of 5.00 mF, and a generator with DVmax 5 240 V opera

yanalaym [24]

Given:

Inductance, L = 150 mH

Capacitance, C = 5.00 mF

$\Delta V_{max}$ = 240 V

frequency, f = 50Hz

$I_{max}$ = 100 mA

Solution:

To calculate the parameters of the given circuit series RLC circuit:

angular frequency, $\omega$ = $2\pi f = 2\pi \times50 = 100\pi$

a). Inductive reactance, $X_{L}$ is given by:

$\X_{L} = \omega L = 100\pi \times 150\times 10^{-3} = 47.12\Omega$

$X_{L} = 47.12\Omega$

b). The capacitive reactance, $X_{C}$ is given by:

$\X_{C} = \frac{1}{\omega C} = \frac{1}{2\pi fC} = \frac{1}{2\pi \times 50\times 5.00\times 10^{-3}} = 0.636\Omega$

$X_{C} = 0.636\Omega$

c). Impedance, Z = $\frac{\Delta V_{max}}{I_{max}} = \frac{240}{100\times 10^{-3}} = 2400\Omega$

$Z = 2400\Omega$

d). Resistance, R is given by:

$Z = \sqrt {R^{2} + (X_{L} - X_{C})}$

$2400^{2} = R^{2} + (47.12 - 0.636)^{2}$

$R = \sqrt {5757839.238}$

$R = 2399.5\Omega$

e). Phase angle between current and the generator voltage is given by:

$tan\phi = \frac{X_{L} - X_{C}}{R}$

$\phi =tan^{-1}( \frac{X_{L} - X_{C}}{R})$

$\phi =tan^{-1}( \frac{47.12 - 0.636}{2399.5}) = tan^{-1}{0.0.01937}$

$\phi = 1.11^{\circ}$

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4 years ago

An apple weighs 1.00 N. When you hang it from the end of a long spring of force constant 1.50 N/m and negligible mass, it bounce

Ierofanga [76]

Answer:

2.67 m

Explanation:

k = Spring constant = 1.5 N/m

g = Acceleration due to gravity = 9.81 m/s²

l = Unstretched length

Frequency of SHM motion is given by

$f_s=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

Frequency of pendulum is given by

$f_p=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}$

Given in the question

$f_p=\dfrac{1}{2}f_s$

$\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\\\Rightarrow \sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\sqrt{\dfrac{k}{m}}\\\Rightarrow \dfrac{g}{l}=\dfrac{1}{4}\dfrac{k}{m}\\\Rightarrow l=\dfrac{4gm}{k}\\\Rightarrow l=\dfrac{4\times 9.81\times \dfrac{1}{9.81}}{1.5}\\\Rightarrow l=2.67\ m$

The unstretched length of the spring is 2.67 m

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3 years ago

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Soloha48 [4]

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3 years ago

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A 100-W lightbulb is placed in a cylinder equipped with a moveable piston. The lightbulb is turned on for 0.010 hour, and the as

Taya2010 [7]

Answer:

$w = - 508.53$ joules

$q = - 3091.47$ joules

Explanation:

Let us convert the time in hours into seconds

$0.010* 3600\\= 36$

Change in internal energy

$\delta E = p * \delta t$

where E is the internal energy in Joules

p is the power in watts

and t is the time in seconds

$\delta E = - 100 * 36\\$

$\delta E = - 3600$ Joules

Amount of work done by the system

$w = - P * \delta V$

where P is the pressure and V is the volume

Substituting the given values in above equation, we get -

$w = - 1 * ( 5.92 -0.90)\\$

$w = -5.02$ liter-atmospheres

Work done in Joules

$- 5.02 * 101.3\\$ $= 508.53$ Joules

$q = \delta E - w\\$

Substituting the given values we get -

$q = - 3600 - (-508.53)\\q = - 3091.47$

Thus

$w = - 508.53$ joules

$q = - 3091.47$ joules

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Mariana [72]

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3 years ago