A scuba diver uses a magnifying glass under water to examine a Spanish Doubloon

Question

A scuba diver uses a magnifying glass under water to examine a Spanish Doubloon found at the site of an old shipwreck. Compared to its magnifying power in air, the magnifying power of the submerged magnifying glass ...increased. ...decreased. ...remains the same. For the series circuit, R_eq = R_1+R_2+...+R_n, and for the parallel circuit, 1/R_eq = 1/R_1+1/R_2+...+1/R_n-Using the values given on the circuit diagram on the right, compute the total resistance in the circuit 34.0 Ohm 21.5 Ohm 4.0 Ohm 1.55 Ohm the current passing through the battery 0.24 A 0.37 A 2.00 A 5.16 A the voltage difference across the 3.0- Ohm resistor. 4.0 V 6.0 V 2.0 V 4.8 V 10.0 n Old-fashion glass-plate-negative cameras consist of a single lens (the objective) and the glass-plate film cartridge. The position of the objective lens is moved relative to the glass-plate film cartridge until the image plane coincides with the glass plate. If a photographer uses such a camera to take an image of distant mountains, and then decides to take a portrait of a client in his studio, he would need to... adjust the position of the objective lens by moving it closer to the glass plate to bring the client's image into focus. adjust the position of the objective lens by moving it farther from the glass plate to bring the client's image into focus. keep the objective lens at the same position. The light enters from air (nair = 1) into the substance and travels a distance of 0.96 m in 4.0 times 10'9 s in the substance. Compute the approximate Index of Refraction of the material, n = c/v and c = 3.0 times 10^8 m/sec. 1.25 1.5 0.85 1.0 If the Angle of Incidence is 30.0 degree, compute the Angle of Refraction in the substance. Snell's Law, n_1sintheta_1 = n_2sintheta_2 1 45.2 degree 30.0 degree 23.6 degree 10.4 degree Fraction grating has 500 lines per mm. The angle between the central and the 2nd order maximum is 30.0 degree. that is the wavelength lambda of the light? a sin theta = m lambda, m = 0,1, 2, 3... 700nm 500 nm 200 nm 400 nm i this light be absorbed by a metal with W_0 = 4.0eV, what is the Kinetic Energy for the liberated electrons'? n c=lambda f,and hf=W_0 + KE, h = 6.626 times 10^-34 J sec, leV = 1.602 times 10^-19 4.0J 3.98 times 10^-19 J 1.6 times 10^-19 J

Explanation / Answer

Here,

6)

i) for the total resistance of the circuit

Req = 2 + ((10 || 2.5) + 4 || 3)

Req = 2 + ((10* 2.5/(10 + 2.5) + 4 || 3)

Req = 2 + ((2 + 4 || 3) )

Req = 2 + ((6 || 3)

Req = 2 + 2

Req = 4 Ohm

the net resistance of the circuit is D) 4 Ohm

ii)

for the current in battery = V/Req

current in battery = 8/4 A

current in battery = 2 A

iii)

for the voltage across the 3 Ohm resistance

voltage across the 3 Ohm resistance = V - R * 2

voltage across the 3 Ohm resistance = 8 - 2 * 2

voltage across the 3 Ohm resistance = 4 V

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