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4 years ago

What is the potential difference vab between points a and b when the switch s is open?

1 answer:

3 0

Assuming is a closed circuit when the switch is closed, it would be the strength of the battery in the circuit, assuming a and b were points at each end of the circuit. Depending on the specifics of the circuit, this could change, you could add more batteries together or it could reduce based on resistance if there's a viable current path still open while the switch is open. You may have missed attaching a picture or link.

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A stone is catapulted at time t = 0, with an initial velocity of magnitude 19.9 m/s and at an angle of 39.9° above the horizonta

larisa [96]

Answer:

Part a)

$x = 15.76 m$

Part b)

$y = 7.94 m$

Part c)

$x = 26.16 m$

Part d)

$y = 7.49 m$

Part e)

$x = 83.23 m$

Part f)

$y = -75.6 m$

Explanation:

As we know that catapult is projected with speed 19.9 m/s

so here we have

$v_x = 19.9 cos39.9$

$v_x = 15.3 m/s$

similarly we have

$v_y = 19.9 sin39.9$

$v_y = 12.76 m/s$

Part a)

Horizontal displacement in 1.03 s

$x = v_x t$

$x = (15.3)(1.03)$

$x = 15.76 m$

Part b)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(1.03) - 4.9(1.03)^2$

$y = 7.94 m$

Part c)

Horizontal displacement in 1.71 s

$x = v_x t$

$x = (15.3)(1.71)$

$x = 26.16 m$

Part d)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(1.71) - 4.9(1.71)^2$

$y = 7.49 m$

Part e)

Horizontal displacement in 5.44 s

$x = v_x t$

$x = (15.3)(5.44)$

$x = 83.23 m$

Part f)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(5.44) - 4.9(5.44)^2$

$y = -75.6 m$

6 0

3 years ago

Estimate the inductance L of a coil that is 12 cm long, made of about 235 copper-wire turns and a diameter of about 1.7 cm. Show

ANTONII [103]

Answer:

Inductance as calculated is 13.12 mH

Solution:

As per the question:

Length of the coil, l = 12 cm = 0.12 m

Diameter, d = 1.7 cm = 0.017 m

No. of turns, N = 235

Now,

Area of cross-section of the wire, A = $\frac{\pi d^{2}}{4} = \frac{\pi \times 0.017^{2}}{4} = 2.269\times 10^{- 4}\ m^{2}$

We know that the inductance of the coil is given by the formula:

$L = \frac{mu_{o}AN^{2}}{l} = \frac{4\pi \times 10^{- 7}\times 2.269\times 10^{- 4}\times 235^{2}}{0.12} = 1.312\times 10^{- 4}\ H = 13.12\ mH$

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Delicious77 [7]

Dropping it on a hard surface.

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D? maybe i’m trying to. see if it’s d or b

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