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jeyben [28]
3 years ago
10

An ideal battery would produce an extraordinarily large current if "shorted" by connecting the positive and negative terminals w

ith a short wire of very low resistance. Real batteries do not. The current of a real battery is limited by the fact that the battery itself has resistance.
What is the resistance of a 9.0 V battery that produces a 17 A current when shorted by a wire of negligible resistance?

R=____
Physics
1 answer:
liubo4ka [24]3 years ago
6 0

Answer:

Resistance, R=0.529\ \Omega

Explanation:

Given that,

Voltage of the battery, V = 9 volts

Current produced in the circuit, I = 17 A

We need to find the resistance when shorted by a wire of negligible resistance. It is a case of Ohm's law. The voltage is given by :

V=IR

R=\dfrac{V}{I}

R=\dfrac{9\ V}{17\ A}

R=0.529\ \Omega

So, the resistance in the circuit is 0.529 ohms. Hence, this is the required solution.

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Hooke's law. The distance d when a spring is stretched by a hanging objective varies directly as the weight w of the object. If
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In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that
erik [133]

Answer:

On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means

\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m

This means that the relation between the wavelength and the length of the string is

3\lambda/2 = L

By definition, this standing wave is at the third harmonic, n = 3.

Furthermore, the standing wave equation is as follows:

y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)

The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.

v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)

a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0

For this equation to be equal to zero, sin(59.94t) = 0. So,

59.94t = \pi\\t = \pi/59.94 = 0.0524~s

This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:

v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s

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3 years ago
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